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sequence","neu":true,"file":"index.tex","reftag":"6.48","cls":"Definition"},{"label":"Example","id":"Example-6.49","autoid":"Example-6.49","refnum":"6.49","reflabel":"Example","title":"","neu":true,"file":"index.tex","reftag":"6.49","cls":"Example"},{"label":"Theorem","id":"theorem-monotone-convergence","autoid":"Theorem-6.50","refnum":"6.50","reflabel":"Theorem","title":"Monotone Convergence Theorem","neu":true,"file":"index.tex","reftag":"6.50","cls":"Theorem"},{"label":"Proof","id":"Proof*-6.23","autoid":"Proof*-6.23","refnum":"??","reflabel":"Proof","title":"","neu":true,"file":"index.tex","reftag":"","cls":"Proof"},{"label":"Definition","id":"partial-sums","autoid":"Definition-8.1","refnum":"8.1","reflabel":"Definition","title":"Partial sums","neu":true,"file":"index.tex","reftag":"8.1","cls":"Definition"},{"label":"Definition","id":"convergent-series-1","autoid":"Definition-8.2","refnum":"8.2","reflabel":"Definition","title":"Convergent series","neu":true,"file":"index.tex","reftag":"8.2","cls":"Definition"},{"label":"Definition","id":"divergent-series","autoid":"Definition-8.3","refnum":"8.3","reflabel":"Definition","title":"Divergent series","neu":true,"file":"index.tex","reftag":"8.3","cls":"Definition"},{"label":"Example","id":"Example-8.4","autoid":"Example-8.4","refnum":"8.4","reflabel":"Example","title":"","neu":true,"file":"index.tex","reftag":"8.4","cls":"Example"},{"label":"Example","id":"Example-8.5","autoid":"Example-8.5","refnum":"8.5","reflabel":"Example","title":"","neu":true,"file":"index.tex","reftag":"8.5","cls":"Example"},{"label":"Theorem","id":"theorem-series-necessary","autoid":"Theorem-id-theorem-series-necessary","refnum":"8.6","reflabel":"Theorem","title":"","neu":true,"file":"index.tex","reftag":"8.6","cls":"Theorem"},{"label":"Proof","id":"Proof*-8.1","autoid":"Proof*-8.1","refnum":"??","reflabel":"Proof","title":"","neu":true,"file":"index.tex","reftag":"","cls":"Proof"},{"label":"Theorem","id":"theorem-series-necessary-1","autoid":"Theorem-id-theorem-series-necessary-1","refnum":"8.7","reflabel":"Theorem","title":"","neu":true,"file":"index.tex","reftag":"8.7","cls":"Theorem"},{"label":"Example","id":"Example-8.8","autoid":"Example-8.8","refnum":"8.8","reflabel":"Example","title":"","neu":true,"file":"index.tex","reftag":"8.8","cls":"Example"},{"label":"Example","id":"Example-8.9","autoid":"Example-8.9","refnum":"8.9","reflabel":"Example","title":"","neu":true,"file":"index.tex","reftag":"8.9","cls":"Example"},{"label":"Important","id":"Important*-8.2","autoid":"Important*-8.2","refnum":"??","reflabel":"Important","title":"","neu":true,"file":"index.tex","reftag":"","cls":"Important"},{"label":"Example","id":"Example-8.10","autoid":"Example-8.10","refnum":"8.10","reflabel":"Example","title":"","neu":true,"file":"index.tex","reftag":"8.10","cls":"Example"},{"label":"Remark","id":"Remark-8.11","autoid":"Remark-8.11","refnum":"8.11","reflabel":"Remark","title":"","neu":true,"file":"index.tex","reftag":"8.11","cls":"Remark"},{"label":"Example","id":"Example-8.12","autoid":"Example-8.12","refnum":"8.12","reflabel":"Example","title":"","neu":true,"file":"index.tex","reftag":"8.12","cls":"Example"}]
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-    "markdown": "::: {.content-hidden}\n$\n\\newcommand{\\R}{\\mathbb{R}}\n\\newcommand{\\Q}{\\mathbb{Q}}\n\\newcommand{\\C}{\\mathbb{C}}  \n\\newcommand{\\N}{\\mathbb{N}}\n\\newcommand{\\Z}{\\mathbb{Z}}\n\n\\renewcommand\\Re{\\operatorname{Re}}\n\\renewcommand\\Im{\\operatorname{Im}}\n\n\n\n\n\n\\newcommand{\\e}{\\varepsilon}\n\\newcommand{\\g}{\\gamma}\n\n\\newcommand{\\st}{\\, \\text{ s.t. } \\, }\n\\newcommand{\\divider}{\\, \\colon \\,}\n\n\\newcommand{\\closure}[2][2]{{}\\mkern#1mu \\overline{\\mkern-#1mu #2 \\mkern-#1mu}\\mkern#1mu {}}\n\n\n\\newcommand{\\Czero}{\\mathnormal{C}}\n\\newcommand{\\Cinf}{{\\mathnormal{C}}^{\\infty}}\n\\newcommand{\\Cinfcomp}{{\\mathnormal{C}}_0^{\\infty}}\n\\newcommand{\\Lone}{{\\mathnormal{L}}^1}\n\\newcommand{\\Loneloc}{{\\mathnormal{L}}_{loc}^1}\n\\newcommand{\\Ltwo}{{\\mathnormal{L}}^2}\n\\newcommand{\\Lp}{{\\mathnormal{L}}^p}\n\\newcommand{\\Linf}{{\\mathnormal{L}}^{\\infty}}\n\\newcommand{\\Woneone}{{\\mathnormal{W}}^{1,1}}\n\\newcommand{\\Wonetwo}{{\\mathnormal{W}}^{1,2}}\n\\newcommand{\\Wonep}{{\\mathnormal{W}}^{1,p}}\n\\newcommand{\\Woneinf}{{\\mathnormal{W}}^{1,\\infty}}\n\\newcommand{\\Wtwotwo}{{\\mathnormal{W}}^{2,2}}\n\\newcommand{\\Wktwo}{{\\mathnormal{W}}^{k,2}}\n\\newcommand{\\Wkp}{{\\mathnormal{W}}^{k,p}}\n\\newcommand{\\Lip}{\\mathnormal{Lip}}\n\n\\newcommand{\\scp}[2]{\\left\\langle #1,#2 \\right\\rangle} %prodotto scalare\n\\newcommand{\\abs}[1]{\\left| #1 \\right|}  %valore assoluto\n\\newcommand{\\norm}[1]{\\left\\| #1 \\right\\|} %norma\n\n\n\n\n\n\n\n\n\n$\n:::\n\n\n\n\n\n# Sequences in $\\mathbb{R}$\n\n\n\nA sequence is an infinite list of real numbers. For example, the following are sequences:\n\n\n- $(1,2,3,4, \\ldots)$\n\n- $(-1,1,-1,1, \\ldots)$\n\n- $\\left(1, \\frac{1}{2}, \\frac{1}{3}, \\frac{1}{4}, \\frac{1}{5}, \\ldots\\right)$\n\n\n::: Remark\n\n- The order of elements in a sequence matters. \n\n> For example\n>$$\n>(1,2,3,4,5,6, \\ldots) \\neq(2,1,4,3,6,5, \\ldots)\n>$$\n\n- A sequence is not a set. \n\n> For example \n>$$\n>\\{-1,1,-1,1,-1,1, \\ldots\\}=\\{-1,1\\}\n>$$\n>but we cannot make a similar statement for the sequence \n>$$\n>(-1,1,-1,1,-1,1, \\ldots) \\,.\n>$$\n\n- The above notation is ambiguous.\n\n> For example the sequence \n>$$\n>(1,2,3,4, \\ldots)\n>$$\ncan continue as\n>\n>$$\n>(1,2,3,4,1,2,3,4,1,2,3,4, \\ldots) \\,.\n>$$\n\n- In the sequence \n$$\n\\left(1, \\frac{1}{2}, \\frac{1}{3}, \\frac{1}{4}, \\frac{1}{5}, \\ldots\\right)\n$$\nthe elements get smaller and smaller, and closer and closer to $0$. We say that this sequence converges to $0$, or has $0$ as a limit. \n\n\n:::\n\n\nWe would like to make the notions of sequence and convergence more precise.\n\n\n## Definition of sequence\n\n\nWe start with the definition of sequence of Real numbers.\n\n\n::: Definition \n### Sequence of Real numbers\n\nA sequence $a$ in $\\R$ is a function\n$$\na \\colon \\N \\to \\R \\,.\n$$\nFor $n \\in \\N$, we denote the $n$-th element of the sequence $a$ by \n$$\na_{n}=a(n)\n$$ \nand write the sequence as \n$$\n\\left(a_{n}\\right)_{n \\in \\N} \\,.\n$$\n:::\n\n\n::: Notation\n\nWe will sometimes omit the subscript $n \\in \\N$ and simply write \n$$\n\\left(a_{n}\\right) \\,.\n$$ \nIn certain situations, we will also write\n$$\n\\left(a_{n}\\right)_{n=1}^{\\infty} \\,.\n$$\n\n\n:::\n\n\n\n\n::: Example \n\n- In general $\\left(a_{n}\\right)_{n \\in \\N}$ is the sequence \n$$\n\\left(a_{1}, a_{2}, a_{3}, \\ldots\\right)\\,.\n$$\n\n- Consider the function \n$$\na \\colon \\N \\to \\N \\, , \\quad  n \\mapsto 2 n \\,.\n$$\nThis is also a sequence of real numbers. It can be written as \n$$\n(2 n)_{n \\in \\N}\n$$\nand it represents the sequence of even numbers \n$$\n(2,4,6,8,10, \\ldots) \\,.\n$$\n\n- Let \n$$\na_{n}=(-1)^{n}\n$$\nThen $\\left(a_{n}\\right)$ is the sequence \n$$\n(-1,1,-1,1,-1,1, \\ldots) \\,.\n$$\n\n- $\\left(\\frac{1}{n}\\right)_{n \\in \\N}$ is the sequence \n$$\n\\left(1, \\frac{1}{2}, \\frac{1}{4}, \\frac{1}{5}, \\ldots\\right) \\,.\n$$\n\n\n:::\n\n\n\n## Convergent sequences\n\n\nWe have notice that the sequence $\\left(\\frac{1}{n}\\right)_{n \\in \\N}$ gets close to $0$ as $n$ gets large. We would like to say that $a_{n}$ converges to $0$ as $n$ tends to infinity.\n\nTo make this precise, we first have to say what it means for two numbers to be *close*. For this we use the notion of absolute value, and say that:\n\n- $x$ and $y$ are close if $|x-y|$ is small. \n- $|x-y|$ is called the distance between $x$ and $y$\n- For $x$ to be close to $0$, we need that $|x-0|=|x|$ is small.\n\nSaying that $|x|$ is *small* is not very precise. Let us now give the formal definition of convergent sequence.\n\n::: Definition \n### Convergent sequence  {#definition-convergent-sequence}\n\nWe say that a sequence $\\left(a_{n}\\right)_{n \\in \\N}$ in $\\mathbb{R}$ **converges** to $a \\in \\mathbb{R}$, or equivalently has limit $a$, denoted by\n$$\n\\lim_{n \\rightarrow \\infty} a_{n}=a \n$$\nif for all $\\e \\in \\mathbb{R}, \\e>0$, there exists $N \\in \\N$ such that for all $n \\in \\N, n \\geq N$ it holds that \n$$\n\\left|a_{n}-a\\right| < \\e \\,.\n$$\nUsing quantifiers, we can write this as\n$$\n\\forall \\, \\e>0 , \\, \\exists \\, N \\in \\N \\st \\forall \\, n \\geq N \\,, \\,\\left|a_{n}-a\\right| < \\e \\,.\n$$\n\nIf there exists $a \\in \\R$ such that $\\lim_{n \\rightarrow \\infty} a_{n}=a$, then we say that the sequence $\\left(a_{n}\\right)_{n \\in \\N}$ is **convergent**.\n\n:::\n\n\n::: Notation\n\nWe will often write\n$$\na_n \\to a\n$$\nin place of\n$$\n\\lim_{n \\rightarrow \\infty} a_{n}=a \\,.\n$$\n\n:::\n\n\n::: Remark\n\n- Informally, Definition \\ref{definition-convergent-sequence} says that, no matter how small we choose $\\e$ (as long as it is strictly positive), we always have that $a_{n}$ has a distance to $a$ of less than or equal to $\\e$ from a certain point onwards (i.e., from $N$ onward). The sequence $\\left(a_{n}\\right)$ may fluctuate wildly in the beginning, but from $N$ onward it should stay within a distance of $\\e$ of $a$.\n\n- In general $N$ depends on $\\e$. If $\\e$ is chosen smaller, we might have to take $N$ larger: this means we need to wait longer before the sequence stays within a distance $\\e$ from $a$.\n\n:::\n\n\nWe now prove that the sequence\n$$\na_n = \\frac1n\n$$\nconverges to $0$, according to Definition \\ref{definition-convergent-sequence}.\n\n::: {.Theorem #theorem-one-over-n}\n\nThe sequence $\\left(\\frac{1}{n}\\right)_{n \\in \\N}$ converges to $0$ , i.e.,\n\n$$\n\\lim_{n \\rightarrow \\infty} \\frac{1}{n}=0 \\,.\n$$\n\n:::\n\nWe give two proofs of the above theorem:\n\n- Long proof, with all the details.\n- Short proof, with less details, but still acceptable. \n\n::: Proof \n### Proof of Theorem \\ref{theorem-one-over-n} (Long version)\n\nWe have to show that\n\n$$\n\\lim _{n \\rightarrow \\infty} \\frac{1}{n}=0,\n$$\n\nwhich by definition is equivalent to showing that\n\n$$\n\\forall \\, \\e>0 \\,, \\, \\exists \\, N \\in \\N \\st  \\forall  \\, n \\geq N \\,, \\, \\left|\\frac{1}{n}-0\\right| < \\e \\,.\n$${#eq-one-over-n}\n\nLet $\\e \\in \\mathbb{R}$ with $\\e>0$. Choose $N \\in \\N$ such that \n$$\nN > \\frac{1}{\\e}\\,.\n$$\nSuch natural number $N$ exists thanks to the Archimedean property. The above implies\n$$\n\\frac{1}{N} < \\e \\,,\n$${#eq-one-over-n-1}\nLet $n \\in \\N$ with $n \\geq N$. By (@eq-one-over-n-1) we have\n$$\n\\frac{1}{n} \\leq \\frac{1}{N} < \\e \\,.\n$$\nFrom this we deduce\n$$\n\\left|\\frac{1}{n}-0\\right|=\\frac{1}{n} \\leq \\frac{1}{N} < \\e\n$$\n\nSince $n \\in \\N, n \\geq N$ was arbitrary, we have proven that \n$$\n\\left|\\frac{1}{n}-0\\right| < \\e \\,, \\quad \\forall \\, n \\geq N \\,.\n$${#eq-one-over-n-2}\nCondition (@eq-one-over-n-2) holds for all $\\e>0$, for the choice of $N \\in \\N$ such that\n$$\nN > \\frac{1}{\\e} \\,.\n$$\nWe have hence shown (@eq-one-over-n), and the proof is concluded.\n\n:::\n\nAs the above proof is quite long and includes lots of details, it is acceptable to shorten it. For example:\n\n- We skip some intermediate steps.\n- We do not mention the Archimedean property.\n- We leave out the conclusion when it is obvious that the statement has been proven. \n\n\n\n\n::: Proof \n### Proof of Theorem \\ref{theorem-one-over-n} (Short version)\n\nWe have to show that\n\n$$\n\\forall \\, \\e>0 \\,, \\, \\exists \\, N \\in \\N \\st  \\forall  \\, n \\geq N \\,, \\, \\left|\\frac{1}{n}-0\\right| < \\e \\,.\n$$\n\nLet $\\e>0$. Choose $N \\in \\N$ such that \n$$\nN > \\frac{1}{\\e}\\,.\n$$\nLet $n \\geq N$. Then\n$$\n\\left|\\frac{1}{n}-0\\right|=\\frac{1}{n} \\leq \\frac{1}{N} < \\e \\,.\n$$\n\n:::\n\n\nIn Theorem \\ref{theorem-one-over-n} we showed that \n$$\n\\lim_{n \\to \\infty} \\, \\frac{1}{n} = 0 \\,.\n$$\nWe can generalise this statement to prove that \n$$\n\\lim_{n \\to \\infty} \\, \\frac{1}{n^{p}} = 0\n$$\nfor any $p>0$ fixed.\n\n\n::: {.Theorem  #theorem-one-over-np}\nFor all $p>0$, the sequence $\\left(\\frac{1}{n^{p}}\\right)_{n \\in \\N}$ converges to $0$ , i.e.,\n\n$$\n\\lim _{n \\rightarrow \\infty} \\frac{1}{n^{p}}=0\n$$\n:::\n\n::: Proof\n\nLet $p>0$. We have to show that\n$$\n\\forall \\e>0 \\,, \\, \\exists \\, N \\in \\N \\st  \\forall \\, n \\geq N \\,, \\, \\left|\\frac{1}{n^{p}}-0\\right| < \\e \\,.\n$$\nLet $\\e>0$. Choose $N \\in \\N$ such that \n$$\nN > \\frac{1}{\\e^{1 / p}} \\,.\n$${#eq-theorem-one-over-np-1}\nLet $n \\geq N$. Since $p>0$, we have $n^{p} \\geq N^{p}$, which implies \n$$\n\\frac{1}{n^p} \\leq \\frac{1}{N^p} \\,.\n$$\nBy (@eq-theorem-one-over-np-1) we deduce\n$$\n\\frac{1}{N^p} < \\e \\,.\n$$\nThen\n$$\n\\left|\\frac{1}{n^{p}}-0\\right|=\\frac{1}{n^{p}} \\leq \\frac{1}{N^{p}} < \\e \\,.\n$$\n\n:::\n\n::: Question\nWhy did we choose $N \\in \\N$ such that\n$$\nN > \\frac{1}{\\e^p}\n$$\nin the above proof?\n:::\n\nThe answer is: because it works. Finding a number $N$ that makes the proof work requires some *rough work*: Specifically, such rough work consists in finding $N \\in \\N$ such that the inequality\n$$\n|a_N - a | < \\e \n$$\nis satisfied. \n\n\n::: Important \n\nAny *rough work* required to prove convergence must be shown before the formal proof (in assignments). \n\n:::\n\n\n\n\n::: Example \n\nProve that, as $n \\rightarrow \\infty$,\n\n$$\n\\frac{n}{2n+3} \\to \\frac{1}{2} \\,.\n$$\n\n*Part 1. Rough Work.*\n\nLet $\\e>0$. We want to find $N \\in \\N$ such that\n$$\n\\left|\\frac{n}{2n+3}-\\frac{1}{2}\\right| < \\e \\,, \\quad \\forall \\, n \\geq N \\,.\n$$\nTo this end, we compute:\n\\begin{align*}\n\\left|\\frac{n}{2 n+3}-\\frac{1}{2}\\right| & \n=\\left|\\frac{2n -(2n + 3)}{2(2n +3)}\\right| \\\\\n& =\\left|\\frac{- 3}{4n + 6}\\right| \\\\\n& = \\frac{3}{4n + 6} \\,.\n\\end{align*}\nTherefore \n\\begin{align*}\n\\left|\\frac{n}{2n+3}-\\frac{1}{2}\\right| < \\e\n\\quad \\iff &  \\quad\n\\frac{3}{4n + 6} < \\e \\\\\n\\quad \\iff &  \\quad\n\\frac{4n + 6}{3} > \\frac{1}{\\e} \\\\\n\\quad \\iff &  \\quad\n4n + 6 > \\frac{3}{\\e} \\\\\n\\quad \\iff &  \\quad\n4n  > \\frac{3}{\\e} - 6 \\\\\n\\quad \\iff &  \\quad\nn  > \\frac{3}{4\\e} - \\frac{6}{4} \\,.\n\\end{align*}\nLooking at the above equivalences, it is clear that $N \\in \\N$ has to be chosen so that\n$$\nN  > \\frac{3}{4\\e} - \\frac{6}{4} \\,.\n$$\n\n\n*Part 2. Formal Proof.* We have to show that\n$$\n\\forall \\e>0 \\,, \\, \\exists \\, N \\in \\N \\st  \\forall \\, n \\geq N \\,, \\, \\left|\\frac{n}{2n+3}-\\frac{1}{2}\\right| < \\e  \\,.\n$$\nLet $\\e>0$. Choose $N \\in \\N$ such that\n$$\nN > \\frac{3}{4\\e} - \\frac{6}{4}   \\,.\n$${#eq-example-limit-1}\nBy the rough work shown above, inequality (@eq-example-limit-1) is equivalent to \n$$\n \\frac{3}{4N + 6} < \\e \\,.\n$$\nLet $n \\geq N$. Then\n\\begin{align*}\n\\left|\\frac{n}{2n+3}-\\frac{1}{2}\\right| \n& = \\left|\\frac{- 3 }{2(2 n+3)}\\right| \\\\\n& = \\frac{3}{4 n+ 6 } \\\\\n& \\leq \\frac{3}{4 N+ 6 } \\\\\n& < \\e \\,,\n\\end{align*}\nwhere in the third line we used that $n \\geq N$.\n\n:::\n\n\nWe conclude by showing that constant sequences always converge.\n\n::: {.Theorem #theorem-constant-sequence}\n\nLet $c \\in \\R$ and define the constant sequence\n$$\na_n := c \\,, \\quad \\forall \\, n \\in \\N \\,.\n$$\nWe have that \n$$\n\\lim_{n \\to \\infty} \\, a_n = c \\,.\n$$\n\n:::\n\n::: Proof\n\nWe have to prove that\n$$\n\\forall \\e>0 \\,, \\, \\exists \\, N \\in \\N \\st  \\forall \\, n \\geq N \\,, \\, \\left|a_n - c \\right| < \\e  \\,.\n$${#eq-constant-sequence}\nLet $\\e>0$. We have\n$$\n\\left|a_n - c \\right| = \\left|c - c \\right| = 0 < \\e \\,, \\quad \\forall \\, n \\in \\N \\,.\n$$\nTherefore we can choose $N=1$ and (@eq-constant-sequence) is satisfied. \n:::\n\n\n\n\n## Divergent sequences\n\n\nThe opposite of convergent sequences are divergent sequences.\n\n::: Definition\n### Divergent sequence\n\nWe say that a sequence $\\left(a_{n}\\right)_{n \\in \\N}$ in $\\R$ is **divergent** if it is not convergent.\n\n:::\n\n\n\n::: Remark\n\nProving that a sequence $(a_n)$ is divergent is more complicated than showing it is convergent: To show that $(a_n)$ is divergent, we need to show that \n$(a_n)$ cannot converge to $a$ for any $a \\in \\R$.\n\nIn other words, we have to show that there does not exist an $a \\in \\mathbb{R}$ such that \n$$\n\\lim _{n \\rightarrow \\infty} \\, a_n = a  \\,.\n$$\nUsing quantifiers, this means\n$$\n\\nexists \\, a \\in \\R \\st \\forall \\, \\e>0 \\,, \\, \n\\exists \\, N \\in \\N \\st \\forall \\, n \\geq N \\,, \\,\n\\left|a_{n}-a\\right| < \\e \\,.\n$$\nThe above is equivalent to showing that\n$$\n\\forall \\, a \\in \\R \\,, \\, \\exists \\, \\e > 0 \\st  \\forall \\, N \\in \\N \\,, \\, \\exists \\, n \\geq N \\st \\left|a_{n}-a\\right| \\geq \\e \\,.\n$$\n\n:::\n\n\n\n::: {.Theorem #theorem-1-minus-n}\nLet $\\left(a_{n}\\right)$ be the sequence defined by\n$$\na_{n}=(-1)^{n} \\,.\n$$\nThen $\\left(a_{n}\\right)$ does not converge. \n\n:::\n\n::: Proof\n\nTo prove that $\\left(a_{n}\\right)$ does not converge, we have to show that \n$$\n\\forall \\, a \\in \\R \\,, \\, \\exists \\, \\e > 0 \\st  \\forall \\, N \\in \\N \\,, \\, \\exists \\, n \\geq N \\st \\left|a_{n}-a\\right| \\geq \\e \\,.\n$$\nLet $a \\in \\R$. Choose \n$$\n\\e=\\frac{1}{2} \\,.\n$$ \nLet $N \\in \\N$. We distinguish two cases:\n\n- $a \\geq 0$: Choose $n=2 N+1$. Note that $n \\geq N$. Then\n\\begin{align*}\n\\left|a_{n}-a\\right| & = \\left|a_{2 N+1}-a\\right| \\\\\n& = \\left|(-1)^{2 N+1}-a\\right| \\\\\n& =|-1-a| \\\\\n& =1+a \\\\\n& \\geq 1 \\\\\n& > \\frac{1}{2} = \\e \\,,\n\\end{align*}\nwhere we used that $a \\geq 0$, and therefore \n$$\n|-1-a|=1+a \\geq 1 \\,.\n$$\n\n- $a<0$: Choose $n=2 N$. Note that $n \\geq N$. Then\n\\begin{align*}\n\\left|a_{n}-a\\right| & = \\left|a_{2 N}-a\\right| \\\\\n& =\\left|(-1)^{2 N}-a\\right| \\\\ \n& = |1-a| \\\\\n& = 1-a \\\\\n& > 1 \\\\\n& > \\frac{1}{2} = \\e \\,,\n\\end{align*}\nwhere we used that $a < 0$, and therefore \n$$\n|1-a|=1-a > 1 \\,.\n$$\n\n\n:::\n\n\n\n## Uniqueness of limit\n\n\nIn Definition \\ref{definition-convergent-sequence} of convergence, we used the notation \n$$\n\\lim_{n \\rightarrow \\infty} a_{n}=a \\,.\n$$ \nThe above notation makes sense only if the limit is unique, that is, if we do not have that  \n$$\n\\lim_{n \\rightarrow \\infty} a_{n} = b \\,,\n$$ \nfor some \n$$\na \\neq b \\,.\n$$\nIn the next theorem we will show that the limit is unique, if it exists.\n\n::: Theorem \n### Uniqueness of limit  {#theorem-uniqueness-limit}\n\nLet $\\left(a_{n}\\right)_{n \\in \\N}$ be a sequence. Suppose that \n$$\n\\lim_{n \\rightarrow \\infty} a_{n} = a \\,, \\quad \n\\lim_{n \\rightarrow \\infty} a_{n} = b \\,.\n$$\nThen $a=b$.\n:::\n\n\n::: Proof\n\nAssume that, \n$$\n\\lim_{n \\rightarrow \\infty} a_{n} = a \\,, \\quad \n\\lim_{n \\rightarrow \\infty} a_{n} = b \\,.\n$$\nSuppose by contradiction that\n$$\na \\neq b \\,.\n$$\nChoose \n$$\n\\e := \\frac{1}{2} \\, |a-b| \\,.\n$$\nTherefore $\\e>0$, since $|a-b|>0$. By the convergence $a_{n} \\rightarrow a$,\n$$\n\\exists \\, N_1 \\in \\N \\st \\forall \\, n \\geq N_1 \\,, \\, \n\\left|a_{n}-a\\right| < \\e \\,.\n$$\nBy the convergence $a_{n} \\rightarrow b$,\n$$\n\\exists \\, N_2 \\in \\N \\st \\forall \\, n \\geq N_2 \\,, \\, \n\\left|a_{n} - b \\right| < \\e \\,.\n$$\nDefine\n$$\nN := \\max \\{ N_1, N_2 \\} \\,.\n$$\nChoose an $n \\in \\N$ such that $n \\geq N$. In particular\n$$\nn \\geq N_1 \\,, \\quad n \\geq N_2 \\,.\n$$\nThen\n\\begin{align*}\n2 \\e & = |a-b| \\\\\n& = \\left|a-a_{n}+a_{n}-b\\right| \\\\\n& \\leq\\left|a-a_{n}\\right|+\\left|a_{n}-b\\right| \\\\\n& < \\e + \\e \\\\\n& = 2 \\e \\,,\n\\end{align*}\nwhere we used the triangle inequality in the first inequality.\nHence $2 \\e < 2 \\e$, which gives a contradiction.\n\n:::\n\n\n::: Example \n\nProve that\n\n$$\n\\lim _{n \\rightarrow \\infty} \\frac{n^{2}-1}{2 n^{2}-3}=\\frac{1}{2}\n$$\n\nAccording to Theorem \\ref{theorem-uniqueness-limit}, it suffices to show that the sequence \n$$\n\\left(\\frac{n^{2}-1}{2 n^{2}-3}\\right)_{n \\in \\N}\n$$ \nconverges to $\\frac{1}{2}$, since then $\\frac{1}{2}$ can be the only limit.\n\n\n*Part 1. Rough Work.*\n\nLet $\\e>0$. We want to find $N \\in \\N$ such that\n$$\n\\left|\\frac{n^{2}-1}{2 n^{2}-3} - \\frac{1}{2}\\right| < \\e \\,, \\quad \\forall \\, n \\geq N \\,.\n$$\nTo this end, we compute:\n\\begin{align*}\n\\left|\\frac{n^{2}-1}{2 n^{2}-3}-\\frac{1}{2}\\right| \n& =\\left|\\frac{2\\left(n^{2}-1\\right)-\\left(2 n^{2}-3\\right)}{2\\left(2 n^{2}-3\\right)}\\right| \\\\\n& =\\left|\\frac{1}{4 n^{2}-6}\\right| \\\\\n& =\\frac{1}{4 n^{2}-6} \\\\\n& =\\frac{1}{3 n^{2}+n^{2}-6} \\\\\n& \\leq \\frac{1}{3 n^{2}} \n\\end{align*}\nwhich holds if $n \\geq 3$, since in this case $n^2 - 6 \\geq 0$.\nTherefore \n\\begin{align*}\n\\left|\\frac{n^{2}-1}{2 n^{2}-3}-\\frac{1}{2}\\right|  < \\e\n\\quad \\impliedby &  \\quad\n\\frac{1}{3n^2} < \\e \\\\\n\\quad \\iff &  \\quad\n3n^2  > \\frac{1}{\\e} \\\\\n\\quad \\iff &  \\quad\nn^2  > \\frac{1}{3\\e} \\\\\n\\quad \\iff &  \\quad\nn  > \\frac{1}{\\sqrt{3\\e}} \\,.\n\\end{align*}\nLooking at the above implications, it is clear that $N \\in \\N$ has to be chosen so that\n$$\nN  > \\frac{1}{\\sqrt{3\\e}} \\,.\n$$\nMoreover we need to recall that $N$ has to satisfy \n$$\nN \\geq 3\n$$ \nfor the estimates to hold.\n\n\n*Part 2. Formal Proof.* We have to show that\n$$\n\\forall \\e>0 \\,, \\, \\exists \\, N \\in \\N \\st  \\forall \\, n \\geq N \\,, \\, \\left|\\frac{n^{2}-1}{2 n^{2}-3}-\\frac{1}{2}\\right| < \\e  \\,.\n$$\nLet $\\e>0$. Choose $N \\in \\N$ such that\n$$\nN > \\max \\left\\{ \\frac{1}{\\sqrt{3\\e}}, 3 \\right\\}   \\,.\n$$\nLet $n \\geq N$. Then\n\\begin{align*}\n\\left|\\frac{n^{2}-1}{2 n^{2}-3}-\\frac{1}{2}\\right| \n& =\\frac{1}{4 n^{2}-6} \\\\\n& =\\frac{1}{3 n^{2}+n^{2}-6} \\\\\n& \\leq \\frac{1}{3 n^{2}} \\\\\n& \\leq \\frac{1}{3 N^{2}} \\\\\n& < \\e \\,,\n\\end{align*}\nwhere we used that \n$$\nn \\geq N \\geq 3 \n$$ \nwhich implies \n$$\nn^2 - 6 \\geq 0\\,,\n$$\nin the third line. The last inequality holds, since it is equivalent \nto \n$$\nN > \\frac{1}{\\sqrt{3 \\e}} \\,.\n$$\n\n:::\n\n\n\n\n## Bounded sequences\n\n\nAn important property of sequences is boundedness. \n\n\n::: Definition \n### Bounded sequence  {#definition-bounded-sequence}\n\nA sequence $\\left(a_{n}\\right)_{n \\in \\N}$ is called **bounded** if there exists a constant $M \\in \\R$, with $M>0$, such that\n$$\n\\left|a_{n}\\right| \\leq M \\,, \\quad \\forall \\, n \\in \\N \\,.\n$$\n\n:::\n\nDefinition \\ref{definition-bounded-sequence} says that a sequence is bounded, if we can find some constant $M>0$ (possibly very large), such that for all elements of the sequence it holds that \n$$\n\\left|a_{n}\\right| \\leq M \\,,\n$$\nor equivalently, that \n$$\n-M \\leq a_{n} \\leq M \\,.\n$$\n\n\nWe now show that any sequence that converges is also bounded\n\n::: {.Theorem  #theorem-convergent-bounded}\n\nIf a sequence $\\left(a_{n}\\right)_{n \\in \\mathbb{N}}$ converges, then the sequence is bounded.\n\n:::\n\n\n::: Proof \n\nSuppose the sequence $\\left(a_{n}\\right)_{n \\in \\mathbb{N}}$ converges and let\n$$\na: = \\lim_{n \\rightarrow \\infty} a_{n}\n$$\nBy definition of convergence we have that\n$$\n\\forall \\, \\e >0 \\,, \\, \\exists N \\in \\N  \\st  \\forall \\, n \\geq N \\,, \\, \\left|a_{n}-a \\right| < \\e \\,.\n$$\nIn particular, we can choose\n$$\n\\e = 1\n$$ \nand let $N \\in \\N$ be that value such that\n$$\n\\left|a_{n}-a \\right| < 1 \\,, \\quad \\forall \\, n \\geq N \\,.\n$$\nIf $n \\geq N$ we have, by the triangle inequality,\n\\begin{align*}\n\\left|a_{n}\\right| & = \\left|a_{n}-a+a\\right| \\\\\n                   & \\leq \\left|a_{n}-a\\right|+|a| \\\\\n                   & < 1+|a| \\,.\n\\end{align*}\nSet\n$$\nM:=\\max \\left\\{\\left|a_{1}\\right|,\\left|a_{2}\\right|, \\ldots,\\left|a_{N-1}\\right|, 1+|a|\\right\\} \\,.\n$$\nNote that such maximum exists, being the set finite. Then \n$$\n\\left|a_{n}\\right| \\leq M \\,, \\quad \\forall \\, n \\in \\N \\,,\n$$\nshowing that $(a_n)$ is bounded.\n\n:::\n\n\nThe choice of $M$ in the above proof says that the sequence can behave wildly for a finite number of terms. After that, it will stay close to the value of the limit, if the latter exists.\n\n::: Example\n\nIn Theorem \\ref{theorem-one-over-n} we have shown that \n$$\n\\lim_{n \\to \\infty} \\,  \\frac{1}{n} = 0 \n$$\nHence, it follows from Theorem \\ref{theorem-convergent-bounded} that the sequence $(1/n)$ is bounded.\n\nIndeed, we have that\n$$\n\\left|\\frac{1}{n}\\right|=\\frac{1}{n} \\leq 1 \\,, \\quad \\forall \\, n \\in \\N \\,,\n$$\nsince $n \\geq 1$ for all $n \\in \\N$.\n\n:::\n\n\n::: Warning\n\nThe converse of Theorem \\ref{theorem-convergent-bounded} does not hold: There exist sequences $(a_n)$ which are bounded, but not convergent.\n\n:::\n\n\n::: Example\n\nDefine the sequence\n$$\na_n = (-1)^n \\,.\n$$\nWe have proven in Theorem \\ref{theorem-1-minus-n} that $(a_n)$ is not convergent. However\n$(a_n)$ is bounded, with $M=1$, since\n$$\n|a_n| = |(-1)^n| = 1 = M  \\,, \\quad \\forall \\, n \\in \\N \\,.\n$$\n\n:::\n\n\n\nTaking the contrapositive of the statement in Theorem \\ref{theorem-convergent-bounded} we get the following corollary:\n\n\n::: {.Corollary   #corollary-convergent-bounded}\n\nIf a sequence $\\left(a_{n}\\right)_{n \\in \\mathbb{N}}$ is not bounded, then the sequence does not converge.\n\n:::\n\n\n::: Remark\n\nFor a sequence $(a_n)$ to be unbounded, it means that\n$$\n\\forall \\, M > 0 \\,, \\,\\,  \\exists \\, n \\in \\N \\st \\left|a_{n}\\right| > M \\,.\n$$\nThe above is saying that no real number $M>0$ can be a bound for $|a_n|$, since there is always an index $n \\in \\N$ such that \n$$\n|a_n| > M \\,.\n$$\n\n:::\n\n\nWe can use Corollary \\ref{corollary-convergent-bounded} to show that certain sequences do not converge.\n\n\n::: Theorem \n\nFor all $p>0$, the sequence $\\left(n^{p}\\right)_{n \\in \\mathbb{N}}$ does not converge.\n\n:::\n\n\n::: Proof\nLet $p>0$. We prove that the sequence $\\left(n^{p}\\right)_{n \\in \\mathbb{N}}$ is unbounded, that is, \n$$\n\\forall \\, M > 0 \\,, \\,\\,  \\exists \\, n \\in \\N \\st \\left|a_{n}\\right| > M \\,.\n$$\nTo this end, let $M > 0$. Choose $n \\in \\N$ such that \n$$\nn > M^{1/p} \\,.\n$$\nThen\n$$\na_n = n^{p}>\\left(M^{1/p}\\right)^{p}=M \\,.\n$$\nThis proves that the sequence $(n^p)$ is unbounded. Hence $(n^p)$ cannot converge, by Corollary \\ref{corollary-convergent-bounded}.\n\n:::\n\n\n\n::: Theorem\n\nThe sequence $(\\log n)_{n \\in \\mathbb{N}}$ does not converge.\n\n:::\n\n\n::: Proof \n\nLet us show that $(\\log n)_{n \\in \\mathbb{N}}$ is unbounded, that is, \n$$\n\\forall \\, M > 0  \\,, \\,\\,  \\exists \\, n \\in \\N \\st \\left|a_{n}\\right| > M \\,.\n$$\nTo this end let $M > 0$. Choose $n \\in \\N$ such that \n$$\nn \\geq e^{M+1} \\,.\n$$\nThen\n$$\n|a_n| = |\\log n| \\geq \\left|   \\log e^{M+1}   \\right| = M + 1 > M \\,.\n$$\nThis proves that the sequence $(\\log n)$ is unbounded. Hence $(\\log n)$ cannot converge, by Corollary \\ref{corollary-convergent-bounded}.\n\n:::\n\n\n\n\n## Algebra of limits\n\nProving convergence using Definition \\ref{definition-convergent-sequence} can be a tedious task. In this section we discuss how to prove convergence, starting from known convergence results.\n\n\n::: Theorem \n### Algebra of limits {#theorem-algebra-limits} \n\nLet $\\left(a_{n}\\right)_{n \\in \\mathbb{N}}$ and $\\left(b_{n}\\right)_{n \\in \\mathbb{N}}$ be sequences in $\\mathbb{R}$. Suppose that\n$$\n\\lim_{n \\rightarrow \\infty} a_{n}=a \\,, \\quad  \\lim_{n \\rightarrow \\infty} b_{n}=b \\,,\n$$\nfor some $a,b \\in \\R$. Then,\n\n1. Limit of sum is the sum of limits:\n$$\n\\lim_{n \\rightarrow \\infty}\\left(a_{n} \\pm b_{n}\\right)=a \\pm b\n$$\n\n2. Limit of product is the product of limits: \n$$\n\\lim _{n \\rightarrow \\infty}\\left(a_{n}  b_{n}\\right) = a  b \n$$\n\n3. If $b_{n} \\neq 0$ for all $n \\in \\mathbb{N}$ and $b \\neq 0$, then \n$$\n\\lim_{n \\rightarrow \\infty} \\left(\\frac{a_{n}}{b_{n}}\\right)=\\frac{a}{b} \n$$\n\n:::\n\n\n\n\n::: Proof \n\n\n Let $\\left(a_{n}\\right)_{n \\in \\mathbb{N}}$ and $\\left(b_{n}\\right)_{n \\in \\mathbb{N}}$ be sequences in $\\mathbb{R}$ and let $c \\in \\mathbb{R}$. Suppose that, for some $a, b \\in \\mathbb{R}$\n$$\n\\lim _{n \\rightarrow \\infty} a_{n}=a \\,, \\quad  \\lim _{n \\rightarrow \\infty} b_{n}=b \\,.\n$$\n\n\n*Proof of Point 1.* \n\nWe need to show that \n$$\n\\lim_{n \\to \\infty} (a_n \\pm  b_n) = a \\pm b \\,.\n$$\nWe only give a proof of the formula with $+$, since the case with $-$ follows with a very similar proof. Hence, we need to show that\n$$\n\\forall \\, \\e > 0 \\,, \\, \\exists  \\, N \\in \\N \\st \\forall \\, n \\geq N \\,, \\, \\left|(a_{n} + b_n) - (a+b)  \\right| < \\e \\,.\n$$\nLet $\\e>0$ and set\n$$\n\\widetilde{\\e} := \\frac{\\e}{2} \\,.\n$$\nSince $a_{n} \\rightarrow a$ and $\\widetilde{\\e}>0$, there exists $N_1 \\in \\N$ such that\n$$\n|a_n - a|< \\widetilde{\\e} \\,, \\quad \\forall \\, n \\geq N_1 \\,.\n$$\nSince $b_{n} \\rightarrow b$ and $\\widetilde{\\e}>0$, there exists $N_2 \\in \\N$ such that\n$$\n|b_n - b|< \\widetilde{\\e} \\,, \\quad \\forall \\, n \\geq N_2 \\,.\n$$\nDefine\n$$\nN : = \\max  \\{ N_1, N_2 \\} \\,.\n$$\nFor all $n \\geq N$ we have, by the triangle inequality,\n\\begin{align*}\n\\left|\\left(a_{n}+b_{n}\\right)-(a+b)\\right| \n& = \\left|\\left(a_{n}-a\\right) + \\left(b_{n}-b\\right)\\right| \\\\\n& \\leq \\left|a_{n}-a\\right|+\\left|b_{n}-b\\right| \\\\\n& < \\widetilde{\\e}+\\widetilde{\\e} \\\\\n& =\\e \\,.\n\\end{align*}\n\n\n\n*Proof of Point 2.*\n\n\nWe need to show that \n$$\n\\lim_{n \\to \\infty} (a_n  b_n) = a b  \\,,\n$$\nwhich is equivalent to \n$$\n\\forall \\, \\e > 0 \\,, \\, \\exists \\, N \\in \\N \\st  \\forall \\, n \\geq N \\,, \\, \\left|a_{n} b_{n} - a b \\right| < \\e \\,.\n$$\nLet $\\e>0$. The sequence $\\left(a_{n}\\right)$ converges, and hence is bounded, by Theorem \\ref{theorem-convergent-bounded}. This means there exists some $M>0$ such that \n$$\n\\left|a_{n}\\right| \\leq M \\,, \\quad \\forall \\, n \\in \\N \\,.\n$$\nDefine \n$$\n\\widetilde{\\e} = \\frac{\\e}{ M + |b| } \\,. \n$$\nSince $a_{n} \\rightarrow a$ and $\\widetilde{\\e} > 0$, there exists \n$N_1 \\in \\N$ such that\n$$\n| a_n  - a | < \\widetilde{\\e} \\,, \\quad \\forall \\, n \\geq N_1 \\,.\n$$\nSince $b_{n} \\rightarrow b$ and $\\widetilde{\\e} > 0$, there exists \n$N_2 \\in \\N$ such that\n$$\n| b_n  - b | < \\widetilde{\\e} \\,, \\quad \\forall \\, n \\geq N_2 \\,.\n$$\nLet \n$$\nN:= \\max\\{ N_1, N_2 \\} \\,.\n$$\nFor all $n \\geq N$ we have\n\\begin{align*}\n\\left|a_{n}b_{n}-a b\\right| \n& = \\left|a_{n} b_{n} - a_n  b + a_n  b - a  b \\right| \\\\\n& \\leq \\left|a_{n}  b_{n} - a_n  b \\right| + \\left|a_n  b - a b \\right| \\\\\n& =\\left|a_{n}\\right| \\left|b_{n}-b\\right|+ |b|  \\left|a_{n}-a\\right| \\\\\n& < M \\, \\widetilde{\\e} +|b| \\, \\widetilde{\\e} \\\\\n& = (M + |b|) \\, \\widetilde{\\e} \\\\\n& = \\e  \\,.\n\\end{align*}\n\n\n*Proof of Point 3.* \n\nSuppose in addition that $b_n \\neq 0$ and $b \\neq 0$. We need to show that\n$$\n\\lim_{n \\to \\infty} \\, \\frac{a_n}{b_n} = \\frac{a}{b} \\,,\n$$\nwhich is equivalent to\n$$\n\\forall \\, \\e > 0 \\,, \\, \\exists \\, N \\in \\N \\st  \\forall \\, n \\geq N \\,, \\, \\left|\\frac{a_{n}}{ b_{n}} - \\frac{a}{b} \\right| < \\e \\,.\n$$\nWe suppose in addition that $b>0$. The proof is very similar for the case $b <0$. Let $\\e > 0$. Set\n$$\n\\delta := \\frac{b}{2} \\,.\n$$\nSince $b_n \\to b$ and $\\delta>0$, there exists $N_1 \\in \\N$ such that\n$$\n|b_n - b| < \\delta \\, \\quad \\forall \\, n \\geq N_1 \\,.\n$$\nIn particular we have\n$$\nb_n > b - \\delta = b - \\frac{b}{2} = \\frac{b}{2}   \\, \\quad \\forall \\, n \\geq N_1 \\,.\n$$\nDefine\n$$\n\\widetilde{\\e} := \\frac{ b^2 }{ 2 (b + |a|)} \\, \\e \\,.\n$$\nSince $\\widetilde{\\e}>0$ and $a_n \\to a$, there exists $N_2 \\in \\N$ such that\n$$\n|a_n - a |< \\widetilde{\\e} \\,, \\quad \\forall \\, n \\geq N_2 \\,.\n$$\nSince $\\widetilde{\\e}>0$ and $b_n \\to b$, there exists $N_3 \\in \\N$ such that\n$$\n|b_n - b |< \\widetilde{\\e} \\,, \\quad \\forall \\, n \\geq N_3 \\,.\n$$\nDefine\n$$\nN:= \\max\\{ N_1, N_2, N_3 \\} \\,.\n$$\nFor all $n \\geq N$ we have\n\\begin{align*}\n\\left|\\frac{a_{n}}{ b_{n}} - \\frac{a}{b} \\right| & = \n\\left|\\frac{a_{n} b - a b_n}{ b_{n} b }  \\right| \\\\\n& = \\frac{1}{ \\left|b_{n} b \\right|} \\,  \n\\left| a_{n}b - a b + a b - a b_n    \\right| \\\\\n& = \\frac{1}{ \\left|b_{n} b \\right|} \\,  \n\\left| (a_{n} - a) b + a(b-b_n)   \\right| \\\\\n& \\leq \\frac{1}{ \\left|b_{n} b \\right|} \\,  \n\\left( |a_{n} - a| |b| + |a| |b-b_n| \\right) \\\\\n& < \\frac{1}{ \\dfrac{b}{2} \\, b } \\, \\left(  \\widetilde{\\e} \\, b + \\widetilde{\\e} |a|   \\right) \\\\\n& = \\frac{2 (b + |a|) }{b^2} \\, \\widetilde{\\e} \\\\\n& = \\e \\,.\n\\end{align*}\n\n:::\n\n\n\nIn the future we will refer to Theorem \\ref{theorem-algebra-limits} as the *Algebra of Limits*. We now show how to use Theorem \\ref{theorem-algebra-limits} for computing certain limits. \n\n\n::: {.Example  #example-limit-ploynomial}\nProve that \n$$\n\\lim_{n \\to \\infty} \\, \\frac{3 n}{7 n+4} = \\frac{3}{7} \\,.\n$$\n\n\n> *Proof*. We can rewrite\n$$\n\\frac{3 n}{7 n+4}=\\frac{3}{7+\\dfrac{4}{n}}\n$$\nBy Theorem \\ref{theorem-constant-sequence} we know that\n$$\n3 \\rightarrow 3\\,,  \\quad 4 \\to 4 \\,, \\quad 7 \\to 7 \\,.\n$$\nFrom Theorem \\ref{theorem-one-over-n}  we know that \n$$\n\\frac{1}{n} \\rightarrow 0 \\,.\n$$\nHence, it follows from Theorem \\ref{theorem-algebra-limits} Point 2 that \n$$\n\\frac{4}{n} = 4 \\cdot \\frac1n \\rightarrow 4 \\cdot 0 = 0 \\,.\n$$\nBy Theorem \\ref{theorem-algebra-limits} Point 1 we have\n$$\n7 + \\frac{4}{n} \\rightarrow  7 + 0 = 7 \\,.\n$$\nFinally we can use Theorem \\ref{theorem-algebra-limits} Point 3 to infer\n$$\n\\frac{3 n}{7 n+4}=\\frac{3}{7+\\dfrac{4}{n}} \\rightarrow \\frac{3}{7} \\,.\n$$\n\n:::\n\n\n::: Important \n\nThe technique shown in Example \\ref{example-limit-ploynomial} is useful to compute limits of fractions of polynomials. To identify the possible limit, if it exists, it is often best to divide by the largest power of $n$ in the denominator.\n\n:::\n\n\n\n::: {.Example #example-algebra-of-limits-2}\nProve that\n$$\n\\lim_{n \\rightarrow \\infty} \\frac{n^{2}-1}{2 n^{2}-3} = \\frac{1}{2} \\,.\n$$\n\n> *Proof*. Factor $n^2$ to obtain\n$$\n\\frac{n^{2}-1}{2 n^{2}-3} = \\frac{1-\\dfrac{1}{n^{2}}}{2-\\dfrac{3}{n^{2}}} \\,.\n$$\nBy Theorem \\ref{theorem-one-over-np} we have\n$$\n\\frac{1}{n^2} \\to 0 \\,.\n$$\nWe can then use the Algebra of Limits Theorem \\ref{theorem-algebra-limits} Point 2 to infer\n$$\n\\frac{3}{n^2} \\to 3 \\cdot 0 = 0\n$$\nand Theorem \\ref{theorem-algebra-limits} Point 1 to get\n$$\n1 - \\frac{1}{n^2} \\to 1 - 0 = 1 \\,, \\quad \n2 - \\frac{3}{n^2} \\to 2 - 0 = 2 \\,.\n$$\nFinally we use Theorem \\ref{theorem-algebra-limits} Point 3 and conclude\n$$\n \\frac{1-\\dfrac{1}{n^{2}}}{2-\\dfrac{3}{n^{2}}} \\to \\frac{1}{2} \\,.\n$$\nTherefore\n$$\n\\lim_{n \\to \\infty } \\, \\frac{n^{2}-1}{2 n^{2}-3} = \\lim_{n \\to \\infty} \\, \\frac{1-\\dfrac{1}{n^{2}}}{2-\\dfrac{3}{n^{2}}} = \\frac{1}{2} \\,.\n$$\n\n:::\n\n\n\nWe can also use the Algebra of Limits to prove that certain limits do not exist.\n\n\n\n::: Example \n\nProve that the sequence\n$$\na_n = \\frac{4 n^{3}+8 n+1}{7 n^{2}+2 n+1}\n$$\ndoes not converge.\n\n> *Proof*. To show that the sequence $\\left(a_n\\right)$ does not converge, we divide by the largest power in the denominator, which in this case is $n^2$\n\\begin{align*}\na_n & = \\frac{4 n^{3}+8 n+1}{7 n^{2}+2 n+1} \\\\\n    & =\\frac{4 n+ \\dfrac{8}{n}  +  \\dfrac{1}{n^{2}} }{7+  \\dfrac{2}{n}  +  \\dfrac{1}{n^{2}} } \\\\\n    & = \\frac{b_n}{c_n}\n\\end{align*}\nwhere we set \n$$\nb_n := 4 n+ \\dfrac{8}{n}  +  \\dfrac{1}{n^{2}}\\,, \\quad\nc_n := 7 + \\dfrac{2}{n}  +  \\dfrac{1}{n^{2}} \\,.\n$$\nUsing the Algebra of Limits Theorem \\ref{theorem-algebra-limits} we see that\n$$\nc_n = 7+  \\dfrac{2}{n}  +  \\dfrac{1}{n^{2}}  \\to 7 \\,.\n$$\nSuppose by contradiction that\n$$\na_n \\to a\n$$\nfor some $a \\in \\R$. Then, by the Algebra of Limits Theorem \\ref{theorem-algebra-limits} we would infer\n$$\nb_n = c_n \\cdot a_n \\to 7 a \\,,\n$$\nconcluding that $b_n$ is convergent to $7a$. We have that\n$$\nb_n = 4n + d_n \\,, \\quad d_n := \\dfrac{8}{n}  +  \\dfrac{1}{n^{2}} \\,.\n$$\nAgain by the Algebra of Limits Theorem \\ref{theorem-algebra-limits} we get that \n$$\nd_n =  \\dfrac{8}{n}  +  \\dfrac{1}{n^{2}} \\to 0 \\,,\n$$\nand hence\n$$\n4n = b_n - d_n \\to 7a - 0 = 7a \\,.\n$$\nThis is a contradiction, since the sequence $(4n)$ is unbounded, and hence cannot be convergent. Hence $(a_n)$ is not convergent.\n:::\n\n\n\n::: Warning\n\nConsider the sequence \n$$\na_n = \\frac{4 n^{3}+8 n+1}{7 n^{2}+2 n+1}\n$$\nfrom the previous example. We have proven that $(a_n)$ is not convergent, by making use of the Algebra of Limits. \n\nLet us review a **faulty** argument to conclude that $(a_n)$ is not convergent. Write\n$$\na_n = \\frac{b_n}{c_n} \\,, \\quad b_n:= 4 n^{3}+8 n+1\\,, \n\\quad c_n : =7 n^{2}+2 n+1 \\,.\n$$\nThe numerator \n$$\nb_n = 4 n^{3}+8 n+1\n$$\nand denominator \n$$\nc_n  =7 n^{2}+2 n+1 \n$$\nare both unbounded, and hence $(b_n)$ and $(c_n)$ do not converge. One might be tempted to conclude that $(a_n)$ does not converge. However this is **false** in general: as seen in \nExample \\ref{example-algebra-of-limits-2}, we have\n$$\n\\lim_{n \\rightarrow \\infty} \\frac{n^{2}-1}{2 n^{2}-3} = \\frac{1}{2} \\,,\n$$\nwhile numerator and denominator are unbounded.\n\n:::\n\n\n\n\n\nSometimes it is useful to rearrange the terms of a sequence, before applying the Algebra of Limits.\n\n\n\n::: Example \n\nDefine \n$$\n a_n := \\frac{2 n^{3}+7 n+1}{5 n+9} \\cdot \\frac{8 n+9}{6 n^{3}+8 n^{2}+3} \\,.\n$$\nProve that \n$$\n\\lim_{n \\to \\infty}  a_n = \\frac{8}{15} \\,.\n$$\n\n\n> *Proof.* \nThe first fraction in $(a_n)$ does not converge, as it is unbounded. Therefore we cannot use Point 2 in Theorem \\ref{theorem-algebra-limits} directly. However, we note that\n\\begin{align*}\na_{n} & =\\frac{2 n^{3}+7 n+1}{5 n+9} \\cdot \\frac{8 n+9}{6 n^{3}+8 n^{2}+3} \\\\\n& = \\frac{8 n+9}{5 n+9} \\cdot \\frac{2 n^{3}+7 n+1}{6 n^{3}+8 n^{2}+3} \\,.\n\\end{align*}\nFactoring out $n$ and $n^3$, respectively, and using the Algebra of Limits, we see that\n$$\n\\frac{8 n+9}{5 n+9}=\\frac{8+9 / n}{5+9 / n} \\to \\frac{8+0}{5+0}=\\frac{8}{5}\n$$\nand\n$$\n\\frac{2+7 / n^{2}+1 / n^{3}}{6+8 / n+3 / n^{3}} \\to \\frac{2+0+0}{6+0+0}=\\frac{1}{3}\n$$\nTherefore Theorem \\ref{theorem-algebra-limits} Point 2 ensures that\n$$\na_{n} \\to \\frac{8}{5} \\cdot \\frac{1}{3}=\\frac{8}{15} \\,.\n$$\n\n:::\n\n\n\n\n\n## Fractional powers\n\n\nThe Algebra of Limits Theorem \\ref{theorem-algebra-limits} can also be used when fractional powers of $n$ are involved.\n\n::: Example\n\nProve that \n$$\na_n = \\frac{n^{7 / 3}+2 \\sqrt{n}+7}{4 n^{3 / 2}+5 n}\n$$\ndoes not converge.\n\n> *Proof*. The largest power of $n$ in the denominator is $n^{3/2}$. Hence we factor out $n^{3/2}$\n\\begin{align*}\na_n & = \\frac{n^{7 / 3}+2 \\sqrt{n}+7}{4 n^{3 / 2}+5 n} \\\\\n    & = \\frac{n^{7 / 3-3 / 2}+2 n^{1/2 - 3/2} + 7  n^{-3 / 2}}{4 + 5 n^{-3/2} } \\\\\n    & = \\frac{n^{5/6}+2 n^{- 1} + 7  n^{-3 / 2}}{4 + 5 n^{-3/2} } \\\\\n    & = \\frac{b_n}{c_n}\n\\end{align*}\nwhere we set\n$$\nb_n := n^{5/6}+2 n^{- 1} + 7  n^{-3 / 2} \\,, \\quad \nc_n := 4 + 5 n^{-3/2} \\,.\n$$\nWe see that $b_n$ is unbounded while $c_n \\to 4$. By the Algebra of Limits (and usual contradiction argument) we conclude that $(a_n)$ is divergent.\n\n:::\n\n\n\nWe now present a general result about the square root of a sequence.\n\n::: {.Theorem #theorem-square-root-limit}\n\nLet $\\left(a_{n}\\right)_{n \\in \\mathbb{N}}$ be a sequence in $\\mathbb{R}$ such that \n$$\n\\lim_{n \\to \\infty} \\, a_n = a \\,,\n$$\nfor some $a \\in \\R$. If $a_{n} \\geq 0$ for all $n \\in \\mathbb{N}$ and $a \\geq 0$, then\n$$\n\\lim _{n \\rightarrow \\infty} \\sqrt{a_{n}}=\\sqrt{a} \\,.\n$$\n\n:::\n\n\n::: Proof \n\nLet $\\e>0$. We the two cases $a>0$ and $a=0$:\n\n- $a>0$: Define\n$$\n\\delta := \\frac{a}{2} \\,.\n$$\nSince $\\delta > 0$ and $a_n \\to a$, there exists $N_1 \\in \\N$ such that\n$$\n\\left|a_{n}-a\\right| < \\delta \\,, \\quad \\forall \\, n \\geq N_1 \\,. \n$$\nIn particular\n$$\na_n > a - \\delta = a - \\frac{a}{2} = \\frac{a}{2} \\,, \\quad \\forall \\, n \\geq N_1 \\,,\n$$\nfrom which we infer\n$$\n\\sqrt{a_n} > \\sqrt{a/2} \\,, \\quad \\forall \\, n \\geq N_1 \\,,\n$$\nNow set\n$$\n\\widetilde{\\e} := \\left(\\sqrt{a/2} + \\sqrt{a} \\right) \\, \\e \\,.\n$$\nSince $\\widetilde{\\e}>0$ and $a_n \\to a$, there exists $N_2 \\in \\mathbb{N}$ such that \n$$\n\\left|a_{n}-a\\right| < \\widetilde{\\e} \\,, \\quad \\forall \\, n \\geq N_2 \\,. \n$$\nLet \n$$\nN:= \\max\\{  N_1, N_2 \\} \\,.\n$$\nFor $n \\geq N$ we have\n\\begin{align*}\n\\left|\\sqrt{a_{n}}-\\sqrt{a}\\right| \n& = \\left|\\frac{\\left(\\sqrt{a_{n}}-\\sqrt{a}\\right)\\left(\\sqrt{a_{n}}+\\sqrt{a}\\right)}{\\sqrt{a_{n}}+\\sqrt{a}}\\right| \\\\\n& = \\frac{\\left|a_{n}-a\\right|}{\\sqrt{a_{n}}+\\sqrt{a}} \\\\\n& < \\frac{ \\widetilde{\\e} }{\\sqrt{a/2} + \\sqrt{a}}  \\\\\n& = \\e \\,.\n\\end{align*}\n\n\n- $a=0$: In this case\n$$\na_n \\to a = 0 \\,.\n$$\nSince $\\e^2>0$, there exists $N \\in \\N$ such that\n$$\n|a_n - 0 | = |a_n| < \\e^2 \\,, \\quad \\forall \\, n \\geq N \\,.\n$$\nTherefore\n$$\n|\\sqrt{a_n} - \\sqrt{0} | = | \\sqrt{a_n}| < \\sqrt{\\e^2} = \\e \\,, \\quad \\forall \\, n \\geq N \\,.\n$$\n\n\n:::\n\n\nLet us show an application of Theorem \\ref{theorem-square-root-limit}.\n\n\n::: Example \n\nDefine the sequence\n$$\na_n  = \\sqrt{9 n^{2}+3 n+1}-3 n \\,.\n$$\nProve that\n$$\n\\lim_{n \\to \\infty} \\, a_n = \\frac12 \\,.\n$$\n\n\n> *Proof*. We first rewrite\n\\begin{align*}\na_n & = \\sqrt{9 n^{2}+3 n+1}-3 n \\\\\n& = \\frac{\\left(\\sqrt{9 n^{2}+3 n+1}-3 n\\right)\\left(\\sqrt{9 n^{2}+3 n+1}+3 n\\right)}{\\sqrt{9 n^{2}+3 n+1}+3 n} \\\\\n& =\\frac{9 n^{2}+3 n+1-(3 n)^{2}}{\\sqrt{9 n^{2}+3 n+1}+3 n} \\\\\n& =\\frac{3 n+1}{\\sqrt{9 n^{2}+3 n+1}+3 n} \\, .\n\\end{align*}\nThe biggest power of $n$ in the denominator is $n$. Therefore we factor out $n$:\n\\begin{align*}\na_n & = \\sqrt{9 n^{2}+3 n+1}-3 n \\\\\n& =\\frac{3 n+1}{\\sqrt{9 n^{2}+3 n+1}+3 n} \\\\\n& =\\frac{3+ \\dfrac{1}{n} }{\\sqrt{9+ \\dfrac{3}{n} + \\dfrac{1}{n^{2}}} + 3 } \\,.\n\\end{align*}\nBy the Algebra of Limits we have\n$$\n9+ \\frac{3}{n} + \\frac{1}{n^{2}} \\to 9 + 0 + 0 = 9 \\,.\n$$\nTherefore we can use Theorem \\ref{theorem-square-root-limit} to infer\n$$\n\\sqrt{ 9 + \\frac{3}{n} + \\frac{1}{n^{2}} } \\to \\sqrt{9}  \\,.\n$$\nBy the Algebra of Limits we conclude:\n$$\na_n = \\frac{3+ \\dfrac{1}{n} }{\\sqrt{9+ \\dfrac{3}{n}  + \\dfrac{1}{n^{2}} }+ 3 }   \\to  \\frac{ 3 + 0 }{ \\sqrt{9} + 3 } = \\frac12 \\,.\n$$\n\n\n:::\n\n\n::: Example \n\nProve that the sequence\n$$\na_n = \\sqrt{9 n^{2}+3 n+1}-2 n\n$$\ndoes not converge.\n\n> *Proof.* We rewrite $a_n$ as\n\\begin{align*}\na_n & = \\sqrt{9 n^{2}+3 n+1}-2 n  \\\\\n& =\\frac{  (\\sqrt{9 n^{2}+3 n+1} - 2 n)  (\\sqrt{9 n^{2}+3 n+1}+2 n) }{\\sqrt{9 n^{2}+3 n+1}+2 n} \\\\\n& =\\frac{9 n^{2}+3 n+1-(2 n)^{2}}{\\sqrt{9 n^{2}+3 n+1}+2 n} \\\\\n& =\\frac{5 n^{2}+3 n+1}{\\sqrt{9 n^{2}+3 n+1}+2 n} \\\\\n& =\\frac{5 n + 3 +  \\dfrac{1}{n} }{\\sqrt{9+ \\dfrac{3}{n}  + \\dfrac{1}{ n^2 } } + 2 } \\\\\n& = \\frac{b_n}{c_n} \\,,\n\\end{align*}\nwhere we factored $n$, being it the largest power of $n$ in the denominator, and defined\n$$\nb_n : = 5 n + 3 +  \\dfrac{1}{n}\\,, \\quad \nc_n := \\sqrt{9+ \\dfrac{3}{n}  + \\dfrac{1}{ n^2 } } + 2 \\,.\n$$\nNote that \n$$\n9+ \\dfrac{3}{n}  + \\dfrac{1}{ n^2 }  \\to 9\n$$\nby the Algebra of Limits. Therefore \n$$\n\\sqrt{9+ \\dfrac{3}{n}  + \\dfrac{1}{ n^2 }}  \\to \\sqrt{9} = 3 \n$$\nby Theorem \\ref{theorem-square-root-limit}. Hence \n$$\nc_n = \\sqrt{9+ \\dfrac{3}{n}  + \\dfrac{1}{ n^2 }} + 2 \\to 3 + 2 = 5  \\,.\n$$\nThe numerator \n$$\nb_n = 5 n + 3 +  \\dfrac{1}{n}\n$$\nis instead unbounded. Therefore $(a_n)$ is not convergent, by the Algebra of Limits and the usual contradiction argument.\n:::\n\n\n\n\n\n\n\n## Limit Tests\n\n\nIn this section we discuss a number of *Tests* to determine whether a sequence converges or not. These are known as **Limit Tests**.\n\n\n### Squeeze Theorem\n\n\nWhen a sequence $(a_n)$ oscillates, it is difficult to compute the limit. Examples of terms which produce oscillations are\n$$\n(-1)^n \\,, \\quad  \\sin(n) \\,, \\quad \\cos(n) \\,. \n$$\nIn such instance it might be useful to compare $(a_n)$ with other sequences whose limit is known. If we can prove that $(a_n)$ is *squeezed* between two other sequences with the same limiting value, then we can show that also $(a_n)$ converges to this value.\n\n\n\n::: Theorem \n### Squeeze theorem {#theorem-squeeze}\n\nLet $\\left(a_{n}\\right), \\left(b_{n}\\right)$ and $\\left(c_{n}\\right)$ be sequences in $\\R$. Suppose that\n$$\nb_{n} \\leq a_{n} \\leq c_{n} \\,, \\quad \\forall  \\, n \\in \\N \\,,\n$$\nand that\n$$\n\\lim_{n \\rightarrow \\infty} b_{n} = \\lim_{n \\rightarrow \\infty} c_{n} = L \\, .\n$$\nThen\n$$\n\\lim_{n \\rightarrow \\infty} a_{n}= L \\, .\n$$\n\n:::\n\n\n::: Proof\n\nLet $\\e>0$. Since $b_{n} \\to L$ and $c_n \\to L$ , there exist $N_1, N_2 \\in \\N$ such that \n\\begin{align*}\n-\\e < b_n - L < \\e \\,, \\,\\, \\forall \\, n \\geq N_1 \\,,  \\\\\n- \\e < c_n - L < \\e \\,, \\,\\, \\forall \\, n \\geq N_2 \\,. \n\\end{align*}\nSet\n$$\nN := \\max\\{N_1,N_2\\} \\,.\n$$\nLet $n \\geq N$. Using the assumption that $b_n \\leq a_{n} \\leq c_{n}$, we get\n$$\nb_n - L  \\leq a_{n} - L \\leq c_{n} - L \\,.\n$$\nIn particular\n$$\n- \\e < b_n - L  \\leq a_n - L \\leq b_n - L < \\e \\,.\n$$\nThe above implies \n$$\n- \\e < a_n - L < \\e  \\quad \\implies \\quad \\left|a_{n}-L\\right| < \\e \\,.\n$$\n\n:::\n\n\n\n::: Example \n\nProve that\n$$\n\\lim_{n \\to \\infty} \\, \\frac{(-1)^{n}}{n} = 0 \\,.\n$$\n\n> *Proof.*\nFor all $n \\in \\N$ we can estimate\n$$\n-1 \\leq(-1)^{n} \\leq 1 \\,.\n$$\nTherefore\n$$\n\\frac{-1}{n} \\leq \\frac{(-1)^{n}}{n} \\leq \\frac{1}{n} \\,, \\quad \\forall \\, n \\in \\N \\,.\n$$\nMoreover\n$$\n\\lim_{n \\to \\infty} \\frac{-1}{n}= -1 \\cdot 0=0 \\,, \\quad \n\\lim_{n \\to \\infty} \\frac{1}{n}=0 \\,.\n$$\nBy the Squeeze Theorem \\ref{theorem-squeeze} we conclude\n$$\n\\lim_{n \\to \\infty} \\frac{(-1)^{n}}{n}=0 \\,.\n$$\n\n:::\n\n\n\n::: {.Example #example-squeeze}\n\nProve that \n$$\n\\lim_{n \\to \\infty} \\frac{\\cos (3 n) + 9 n^{2}}{11 n^{2}+15 \\sin (17 n)} = \\frac{9}{11} \\,.\n$$\n\n\n> *Proof.* \nWe know that \n$$\n-1 \\leq \\cos(x) \\leq 1 \\,, \\quad - 1 \\leq \\sin(x) \\leq 1 \\,, \\quad \\forall \\, x \\in \\R \\,.\n$$\nTherefore, for all $n \\in \\N$\n$$\n- 1 \\leq \\cos(3n) \\leq 1 \\,, \\quad -1 \\leq \\sin(17n) \\leq 1 \\,.\n$$\nWe can use the above to estimate the numerator in the given sequence:\n$$\n-1 + 9 n^{2} \\leq \\cos (3 n)+9 n^{2} \\leq 1+ 9 n^{2} \\,.\n$${#eq-squeeze-example-1}\nConcerning the denominator, we have\n$$\n11 n^{2}-15 \\leq 11 n^{2}+15 \\sin (17 n) \\leq  11 n^{2} + 15\n$$\nand therefore\n$$\n\\frac{1}{11 n^{2} + 15} \\leq \\frac{1}{11 n^{2}+15 \\sin (17 n)} \\leq \\frac{1}{11 n^{2}-15} \\,.\n$${#eq-squeeze-example-2}\nPutting together (@eq-squeeze-example-1)-(@eq-squeeze-example-2) we obtain \n$$\n\\frac{-1 + 9 n^{2}}{11 n^{2} + 15} \\leq \\frac{\\cos (3 n)+9 n^{2}}{11 n^{2}+15 \\sin (17 n)} \\leq \\frac{1+ 9 n^{2}}{11 n^{2}-15} \\,.\n$$\nBy the Algebra of Limits we infer\n$$\n\\frac{-1+9 n^{2}}{11 n^{2}+15}=\\frac{-\\dfrac{1}{n^{2}} + 9}{11 + \\dfrac{15}{n^{2}}} \\to \\frac{0+9}{11+0}=\\frac{9}{11}\n$$\nand\n$$\n\\frac{1+9 n^{2}}{11 n^{2} - 15}=\\frac{ \\dfrac{1}{n^{2}} + 9}{ 11 - \\dfrac{15}{n^{2}}} \\to \\frac{0+9}{11+0}=\\frac{9}{11} \\,.\n$$\nApplying the Squeeze Theorem \\ref{theorem-squeeze} we conclude\n$$\n\\lim_{n \\to \\infty} \\frac{\\cos (3 n)+ 9 n^{2}}{11 n^{2}+15 \\sin (17 n)}=\\frac{9}{11} \\,.\n$$\n\n:::\n \n::: Warning\n\nSuppose that the sequences $(a_n), (b_n), (c_n)$ satisfy\n$$\nb_n \\leq a_n \\leq c_n \\,, \\quad \\forall n \\in \\N \\,,\n$$\nand\n$$\nb_n \\to L_1 \\,, \\quad c_n \\to L_2 \\,, \\quad L_1 \\neq L_2 \\,.\n$$\nIn general, we cannot conclude that $a_n$ converges.\n\n:::\n\n\n::: Example \n\nConsider the sequence\n$$\na_n = \\left( 1 + \\frac{1}{n} \\right) (-1)^n  \\,.\n$$\nFor all $n \\in \\N$ we can bound\n$$\n- 1 - \\frac{1}{n} \\leq \\left( 1 + \\frac{1}{n} \\right) (-1)^n   \\leq 1 + \\frac{1}{n}  \\,.\n$$\nHowever \n$$\n- 1 - \\frac{1}{n} \\longrightarrow - 1 - 0 = -1 \n$$\nand\n$$\n1 + \\frac{1}{n}  \\longrightarrow  1 + 0 = 1 \\,.\n$$\nSince \n$$\n- 1 \\neq 1 \\,,\n$$\nwe cannot apply the Squeeze Theorem \\ref{theorem-squeeze} to conclude convergence of $(a_n)$. Indeed, $(a_n)$ is a divergent sequence.\n\n> *Proof.* Suppose by contradiction that $a_n \\to a$. We have\n$$\na_n = (-1)^n + \\frac{(-1)^n}{n} = b_n + c_n\n$$\nwhere\n$$\nb_n := (-1)^n \\,, \\quad c_n := \\frac{(-1)^n}{n} \\,.\n$$\nWe have seen in Example \\ref{example-squeeze} that $c_n \\to 0$. Therefore, \nby the Algebra of Limits, we have\n$$\nb_n = a_n - c_n \\longrightarrow a - 0 = a \\,.\n$$\nHowever, Theorem \\ref{theorem-1-minus-n} says that the sequence\n$b_n = (-1)^n$ diverges. Contradiction. Hence $(a_n)$ diverges.\n\n:::\n\n\n\n\n\n### Geometric sequences\n\n\n::: Definition \n\nA sequence $\\left(a_{n}\\right)$ is called a **geometric sequence** if\n$$\na_{n}=x^{n} \\,,\n$$\nfor some $x \\in \\R$.\n\n:::\n\n\nThe value of $|x|$ determines whether or not a geometric sequence converges, as shown in the following theorem.\n\n\n::: Theorem \n### Geometric Sequence Test   {#theorem-geometric-sequence}\n\nLet $x \\in \\R$ and let $\\left(a_{n}\\right)$ be the sequence defined by\n$$\na_{n}:=x^{n} \\,.\n$$\nWe have:\n\n1. If $|x|<1$, then \n$$\n\\lim_{n \\to \\infty} a_{n} = 0 \\,.\n$$\n\n2. If $|x|>1$, then sequence $\\left(a_{n}\\right)$ is unbounded, and hence divergent.\n\n:::\n\n\n\n::: Warning\n\nThe Geometric Sequence Test in Theorem \\ref{theorem-geometric-sequence} does not address the case\n$$\n|x|=1 \\,.\n$$\nThis is because, in this case, the sequence \n$$\na_n = x^n \n$$\nmight converge or diverge, depending on the value of $x$. Indeed, \n$$\n|x| = 1 \\quad \\implies \\quad x = \\pm 1 \\,.\n$$\nWe therefore have two cases:\n\n- $x = 1$: Then \n$$\na_n = 1^n = 1\n$$\nso that $a_n \\to 1$ and $(a_n)$ is convergent.\n\n- $x=-1$: Then\n$$\na_n = x^n = (-1)^n\n$$\nwhich is divergent by Theorem \\ref{theorem-1-minus-n}.\n\n:::\n\n\n\nTo prove Theorem \\ref{theorem-geometric-sequence} we need the following inequality, known as Bernoulli's inequality.\n\n\n::: Lemma \n### Bernoulli's inequality  {#lemma-bernoulli}\n\nLet $x \\in \\R$ with $x>-1$. Then\n$$\n(1+x)^{n} \\geq 1+n x \\,, \\quad \\forall \\, n \\in \\N \\,.\n$${#eq-bernoulli-inequality}\n\n:::\n\n\n::: Proof\n\nLet $x \\in \\mathbb{R}, x>-1$. We prove the statement by induction:\n\n- Base case: (@eq-bernoulli-inequality) holds with equality when $n=1$.\n\n- Induction hypothesis: Let $k \\in \\N$ and suppose that (@eq-bernoulli-inequality) holds for $n=k$, i.e.,\n$$\n(1+x)^{k} \\geq 1+k x \\,.\n$$\nThen\n\\begin{align*}\n(1+x)^{k+1} & = (1+x)^{k}(1+x) \\\\\n            & \\geq(1+k x)(1+x) \\\\\n            & =1+k x+x+k x^{2} \\\\\n            & \\geq 1+(k+1) x \\,,\n\\end{align*}\nwhere we used that $kx^2 \\geq 0$. Then (@eq-bernoulli-inequality) holds for $n=k+1$.\n\nBy induction we conclude (@eq-bernoulli-inequality).\n\n:::\n\n\nWe are ready to prove Theorem \\ref{theorem-geometric-sequence}.\n\n\n::: Proof \n### Proof of Theorem \\ref{theorem-geometric-sequence}\n\n*Part 1. The case $|x|<1$*. \n\nIf $x=0$, then\n$$\na_n = x^n = 0 \n$$\nso that $a_n \\to 0$. Hence assume $x \\neq 0$. We need to prove that\n$$\n\\forall \\, \\e> 0 \\,, \\, \\exists \\, N \\in \\N \\st \n\\forall \\, n \\geq N \\,, \\,\\, |x^n - 0| < \\e \\,.\n$$\nLet $\\e>0$. We have \n$$\n|x| < 1 \\quad \\implies \\quad \\frac{1}{|x|} > 1 \\,.\n$$\nTherefore \n$$\n|x|= \\frac{1}{1+u} \\,, \\quad u:=\\frac{1}{|x|} - 1 > 0 \\,.\n$$\nLet $N \\in \\N$ be such that \n$$\nN > \\frac{1}{\\e u} \\,,\n$$\nso that \n$$\n\\frac{1}{N u} < \\e \\,.\n$$\nLet $n \\geq N$. Then\n\\begin{align*}\n\\left|x^{n}-0\\right| & =|x|^{n} \\\\\n                     & = \\left(\\frac{1}{1+u}\\right)^{n} \\\\\n                     & =\\frac{1}{(1+u)^{n}} \\\\\n                     & \\leq \\frac{1}{1+n u} \\\\\n                     & \\leq \\frac{1}{n u} \\\\\n                     & \\leq \\frac{1}{N u} \\\\\n                     & < \\e \\,,\n\\end{align*}\nwhere we used Bernoulli's inequality (@eq-bernoulli-inequality) in the first inequality.\n\n\n*Part 2. The case $|x|>1$*. \n\nTo prove that (a_n) does not converge, we prove that it is unbounded. \nThis means showing that\n$$\n\\forall \\, M > 0 \\,, \\, \\exists n \\in \\N \\st  \\left| a_{n}\\right| >M \\,.\n$$\nLet $M > 0$. We have two cases: \n\n- $0 < M \\leq 1$: Choose $n=1$. Then\n$$\n\\left|a_{1}\\right|=|x|>1 \\geq M \\,.\n$$\n\n- $M>1$: Choose $n \\in \\N$ such that \n$$\nn> \\frac{\\log M}{\\log |x|} \\,.\n$$ \nNote that $\\log |x|>0$ since $|x|>1$. Therefore\n\\begin{align*}\nn>\\frac{\\log M}{\\log |x|} & \\iff n \\log |x|>\\log M \\\\\n                          & \\iff \\log |x|^n>\\log M \\\\\n                          & \\iff |x|^{n}>M \\,.\n\\end{align*}\nThen \n$$\n\\left|a_{n}\\right|=\\left|x^{n}\\right|=|x|^{n} > M \\,.\n$$\n\n\nHence $(a_n)$ is unbounded. By Corollary \\ref{corollary-convergent-bounded} we conclude that $(a_n)$ is divergent. \n\n\n:::\n\n\n\n\n\n\n::: Example \n\nWe can apply Theorem \\ref{theorem-geometric-sequence} to prove convergence\nor divergence for the following sequences.\n\n1. We have\n$$\n\\left(\\frac{1}{2}\\right)^{n} \\longrightarrow 0\n$$\nsince \n$$\n\\left|\\frac{1}{2}\\right|=\\frac{1}{2}<1 \\,.\n$$\n\n2. We have\n$$\n\\left(\\frac{-1}{2}\\right)^{n} \\longrightarrow 0\n$$\nsince \n$$\n\\left|\\frac{-1}{2}\\right|=\\frac{1}{2}<1 \\,.\n$$\n\n3. The sequence \n$$\na_n = \\left(\\frac{-3}{2}\\right)^{n}\n$$ \ndoes not converge, since \n$$\n\\left|\\frac{-3}{2}\\right|=\\frac{3}{2}>1 \\,.\n$$\n\n4. As $n \\rightarrow \\infty$,\n$$\n\\frac{3^{n}}{(-5)^{n}}=\\left(-\\frac{3}{5}\\right)^{n} \\longrightarrow 0\n$$\nsince \n$$\n\\left|-\\frac{3}{5}\\right|=\\frac{3}{5}<1 \\,.\n$$\n\n5. The sequence \n$$\na_{n}=\\frac{(-7)^{n}}{2^{2 n}}\n$$ \ndoes not converge, since\n$$\n\\frac{(-7)^{n}}{2^{2 n}}=\\frac{(-7)^{n}}{\\left(2^{2}\\right)^{n}}=\\left(-\\frac{7}{4}\\right)^{n}\n$$\nand \n$$\n\\left|-\\frac{7}{4}\\right|=\\frac{7}{4}>1  \\,.\n$$\n\n:::\n\n\n\n\n\n\n\n### Ratio Test\n\n\n\n::: Theorem \n### Ratio Test {#theorem-ratio-test}\n\nLet $\\left(a_{n}\\right)$ be a sequence in $\\R$ such that\n$$\na_{n} \\neq 0 \\,, \\quad \\forall \\, n \\in \\N \\,.\n$$ \n\n\n1. Suppose that the following limit exists:\n$$\nL:=\\lim_{n \\to \\infty}\\left|\\frac{a_{n+1}}{a_{n}}\\right| \\,.\n$$\nThen,\n\n    - If $L<1$ we have\n    $$\n    \\lim_{n \\to\\infty} a_{n}=0 \\,.\n    $$\n\n    - If $L>1$, the sequence $\\left(a_{n}\\right)$ is unbounded, and hence does not converge.\n\n2. Suppose that there exists $N \\in \\N$ and $L>1$ such that\n$$\n\\left|\\frac{a_{n+1}}{a_{n}}\\right| \\geq L \\,, \\quad \\forall \\, n \\geq N \\,. \n$$\nThen the sequence $\\left(a_{n}\\right)$ is unbounded, and hence does not converge.\n\n:::\n\n\n\n::: Proof\n\nDefine the sequence $b_{n}=\\left|a_{n}\\right|$. Then,\n\n$$\n\\left|\\frac{a_{n+1}}{a_{n}}\\right|=\\frac{\\left|a_{n+1}\\right|}{\\left|a_{n}\\right|}=\\frac{b_{n+1}}{b_{n}}\n$$\n\n\n*Part 1.* Suppose that there exists the limit\n$$\nL:=\\lim_{n \\to \\infty}\\left|\\frac{a_{n+1}}{a_{n}}\\right| \\,.\n$$\nTherefore \n$$\n\\lim_{n \\to \\infty} \\frac{b_{n+1}}{b_{n}} = L \\,.\n$${#eq-ratio-test-proof}\n\n- $L<1$: Choose $r>0$ such that \n$$\nL<r<1 \\,.\n$$\nSet \n$$\n\\e := r-L\n$$\nBy the convergence at (@eq-ratio-test-proof) there exists $N \\in \\N$ such that\n$$\n\\left|\\frac{b_{n+1}}{b_{n}}-L\\right| < \\e = r - L \\,, \\quad \\forall \\, n \\geq N \\,.\n$$\nIn particular \n$$\n\\frac{b_{n+1}}{b_{n}}-L < r-L \\,, \\quad \\forall \\, n \\geq N \\,,\n$$\nwhich implies \n$$\nb_{n+1} <  r \\,{b_{n}} \\,, \\quad \\forall \\, n \\geq N \\,.\n$${#eq-ratio-test-proof-1}\nLet $n \\geq N$, we can use (@eq-ratio-test-proof-1) recursively and obtain\n$$\n0 \\leq b_{n} < \\, r b_{n-1} < \\ldots < r^{n-N} \\, b_{N} = r^{n} \\, \\frac{b_{N}}{r^{N}} \\,.\n$$\nIn particular, we have proven that\n$$\n0 \\leq b_{n} < r^{n} \\, \\frac{b_{N}}{r^{N}} \\,, \\quad \\forall \\, n \\in \\N \\,.\n$${#eq-ratio-test-proof-2}\nSince $|r|<1$, by the Geometric Sequence Test Theorem \\ref{theorem-geometric-sequence} we infer\n$$\nr^{n} \\rightarrow 0 \\,.\n$$\nThe Algebra of Limits the yields\n$$\nr^{n} \\, \\frac{b_{N}}{r^{N}} \\rightarrow 0 \\,.\n$$\nBy the Squeeze Theorem \\ref{theorem-squeeze} applied to (@eq-ratio-test-proof-2), it follows that \n$$   \nb_{n} = |a_n| \\to 0 \\,.\n$$\nSince \n$$\n-\\left|a_{n}\\right| \\leq a_{n} \\leq\\left|a_{n}\\right| \\,,\n$$\nand \n$$\n-|a_n| \\to 0 \\,, \\quad |a_n| \\to 0 \\,, \n$$\nwe can again apply the Squeeze Theorem \\ref{theorem-squeeze} to infer\n$$\na_{n} \\rightarrow 0 \\,.\n$$\n\n\n- $L>1$: Choose $r>0$ such that \n$$\n1<r<L  \\,.\n$$\nDefine\n$$\n\\e := L - r > 0 \\,.\n$$\nBy the convergence (@eq-ratio-test-proof), there exists $N \\in \\N$ such that\n$$\n\\left|\\frac{b_{n+1}}{b_{n}}-L\\right| < \\e = L - r \\,, \\quad \\forall \\, n \\geq N \\,.\n$$\nIn particular,\n$$\n-(L-r) < \\frac{b_{n+1}}{b_{n}}-L \\,, \\quad \\forall \\, n \\geq N \\,,\n$$\nwhich implies\n$$\nb_{n+1} > r \\, b_{n} \\,, \\quad  \\forall \\, n \\geq N \\,.\n$${#eq-ratio-test-proof-4}\nLet $n \\geq N$. Applying (@eq-ratio-test-proof-4) recursively we get\n$$\nb_{n} > r^{n-N} \\, b_{N} = r^{n} \\, \\frac{b_{N}}{r^{N}} \\,, \\quad \\forall \\, n \\geq N \\,.\n$${#eq-ratio-test-proof-5}\nSince $|r|>1$, by the Geometric Sequence Test we have that the sequence\n$$\n(r^n)\n$$\nis unbounded. Therefore also the right hand side of (@eq-ratio-test-proof-5) is unbounded, proving that $(b_n)$ is unbounded. Since\n$$\nb_n = |a_n|\\,,\n$$\nwe conclude that $(a_n)$ is unbounded. By Corollary \\ref{corollary-convergent-bounded} we conclude that $(a_n)$ does not converge.\n\n\n*Part 2.* Suppose that there exists $N \\in \\N$ and $M>1$ such that\n$$\n\\left|\\frac{a_{n+1}}{a_{n}}\\right| \\geq M \\,, \\quad \\forall \\, n \\geq N \\,. \n$$\nSince $b_n = |a_n|$, we infer\n$$\nb_{n+1} \\geq L \\, b_n \\,, \\quad \\forall \\, n \\geq N \\,. \n$$\nArguing as above, we obtain\n$$\nb_n \\geq L^n \\, \\frac{b_N}{L^N} \\,, \\quad \\forall \\, n \\geq N \\,.\n$$\nSince $L>1$, we have that the sequence \n$$  \nL^n \\, \\frac{b_N}{L^N}\n$$\nis unbounded, by the Geometric Sequence Test. Hence also $(b_n)$ is unbounded, from which we conclude that $(a_n)$ is unbounded. By Corollary \\ref{corollary-convergent-bounded} we conclude that $(a_n)$ does not converge.\n\n\n:::\n\n\nLet us apply the Ratio Test to some concrete examples.\n\n\n\n::: Example\n\n\nLet \n$$\na_{n}=\\frac{3^{n}}{n !} \\,,\n$$\nwhere we recall that $n!$ (pronounced $n$ factorial) is defined by\n$$\nn! := n \\cdot (n-1) \\cdot (n-2) \\cdot  \\ldots \\cdot  3 \\cdot 2 \\cdot 1 \\,. \n$$\nProve that\n$$\n\\lim_{n \\to \\infty} a_n  = 0 \\,.\n$$\n\n> *Proof*. We have\n\\begin{align*}\n\\left|\\frac{a_{n+1}}{a_{n}}\\right| & = \\dfrac{\\left( \\dfrac{3^{n+1}}{(n+1) !} \\right) }{  \\left( \\dfrac{3^{n}}{n !} \\right) } \\\\\n& = \\frac{3^{n+1}}{3^{n}} \\, \\frac{n !}{(n+1) !} \\\\\n& = \\frac{3 \\cdot 3^n}{3^n} \\, \\frac{n!}{(n+1) n!} \\\\\n& =\\frac{3}{n+1} \\longrightarrow L = 0 \\,.\n\\end{align*}\nHence, $L=0<1$ so $a_{n} \\to 0$ by the Ratio Test in Theorem \\ref{theorem-ratio-test}.\n\n:::\n\n\n::: Example\n\nConsider the sequence\n$$\na_{n}=\\frac{n ! \\cdot 3^{n}}{\\sqrt{(2 n) !}} \\,.\n$$\nProve that $(a_n)$ is divergent.\n\n> *Proof.* We have\n\\begin{align*}\n\\left|\\frac{a_{n+1}}{a_{n}}\\right| & =\\frac{(n+1) ! \\cdot 3^{n+1}}{\\sqrt{(2(n+1)) !}}  \\frac{\\sqrt{(2 n) !}}{n ! \\cdot 3^{n}} \\\\\n& =\\frac{(n+1) !}{n !} \\cdot \\frac{3^{n+1}}{3^{n}} \\cdot \\frac{\\sqrt{(2 n) !}}{\\sqrt{(2(n+1)) !}}\n\\end{align*}\nFor the first two fractions we have\n$$\n\\frac{(n+1) !}{n !} \\cdot \\frac{3^{n+1}}{3^{n}} = 3(n+1) \\,,\n$$\nwhile for the third fraction\n\\begin{align*}\n\\frac{\\sqrt{(2 n) !}}{\\sqrt{(2(n+1)) !}} & =\\sqrt{\\frac{(2 n) !}{(2 n+2) !}} \\\\\n& = \\sqrt{\\frac{ (2n)! }{ (2n+2) \\cdot (2n+1) \\cdot (2n)! }} \\\\\n& = \\frac{1}{\\sqrt{(2 n+1)(2 n+2)}} \\,.\n\\end{align*}\nTherefore, using the Algebra of Limits,\n\\begin{align*}\n\\left|\\frac{a_{n+1}}{a_{n}}\\right| & = \\frac{3(n+1)}{\\sqrt{(2 n+1)(2 n+2)}}\\\\\n& = \\frac{3n \\left(1+ \\dfrac{1}{n} \\right)}{\\sqrt{n^2 \\left( 2 + \\dfrac{1}{n}\\right) \\left(2 + \\dfrac{2}{n} \\right)}} \\\\\n& = \\frac{3 \\left(1+ \\dfrac{1}{n} \\right)}{\\sqrt{ \\left( 2 + \\dfrac{1}{n}\\right) \\left(2 + \\dfrac{2}{n} \\right)}}  \\longrightarrow \\frac{3}{\\sqrt{4}} = \\frac{3}{2} > 1 \\,.\n\\end{align*}\nBy the Ratio Test we conclude that $(a_n)$ is divergent.\n\n:::\n\n\n\n::: Example \n\nLet \n$$\na_{n}=\\frac{n !}{100^{n}} \\,.\n$$\nProve that $(a_n)$ is divergent.\n\n> *Proof.*\n$$\n\\left|\\frac{a_{n+1}}{a_{n}}\\right| \n= \\frac{100^{n}}{100^{n+1}} \\frac{(n+1) !}{n !}\n=\\frac{n+1}{100} \\,.\n$$\nChoose $N=101$. Then for all $n \\geq N$,\n$$\n\\left|\\frac{a_{n+1}}{a_{n}}\\right| \\geq \\frac{101}{100}>1 \\,.\n$$\nHence $a_{n}$ is divergent by the Ratio Test.\n\n:::\n\n\n\n\n::: Warning\n\nThe Ratio Test in Theorem \\ref{theorem-ratio-test} does not address the case\n$$\nL=1 \\,.\n$$\nThis is because, in this case, the sequence $(a_n)$ \nmight converge or diverge. \n\nFor example:\n\n- Define the sequence\n$$\na_n = \\frac1n \\,.\n$$\nWe have\n$$\n\\left|\\frac{a_{n+1}}{a_{n}}\\right| = \\frac{n}{n+1} \\rightarrow L = 1 \\,. \n$$\nHence we cannot apply the Ratio Test. However we know that \n$$\n\\lim_{n \\to \\infty} \\, \\frac1n = 0 \\,.\n$$\n\n\n- Consider the sequence\n$$\na_n = n \\,.\n$$\nWe have\n$$\n\\frac{|a_{n+1}|}{|a_n|} = \\frac{n+1}{n} \\rightarrow L = 1 \\,. \n$$\nHence we cannot apply the Ratio Test. However we know that $(a_n)$ is unbounded, and thus divergent.\n\n:::\n\n\n\nIf the sequence $(a_n)$ is geometric, the Ratio Test of Theorem \\ref{theorem-ratio-test} will give the same\nanswer as the Geometric Sequence Test of Theorem \\ref{theorem-geometric-sequence}. This is the content of the following remark.\n\n\n::: Remark\n\nLet $x \\in \\R$ and define the geometric sequence \n$$\na_{n}=x^{n} \\,.\n$$\nThen\n$$\n\\left|\\frac{a_{n+1}}{a_{n}}\\right|  = \\frac{\\left|x^{n+1}\\right|}{\\left|x^{n}\\right|} \n= \\frac{|x|^{n+1}}{|x|^{n}} \n =|x| \\rightarrow |x| \\,.\n$$\nHence: \n\n- If $|x|<1$, the sequence $(a_n)$ converges by the Ratio Test\n- If $|x|>1$, the sequence $(a_n)$ diverges by the Ratio Test. \n- If $|x|=1$, the sequence $(a_n)$ might be convergent or divergent.\n\nThese results are in agreement with the Geometric Sequence Test. \n\n:::\n\n\n\n\n\n\n## Monotone sequences\n\n\nWe showed in Theorem \\ref{theorem-convergent-bounded} that convergent sequences are bounded. We noted that the converse statement is not true. \nFor example the sequence\n$$\na_n = (-1)^n\n$$\nis bounded but not convergent, as shown in Theorem \\ref{theorem-1-minus-n}.\nOn the other hand, if a bounded sequence is **monotone**, then it is convergent.\n\n\n::: Definition\n### Monotone sequence\n\nLet $(a_n)$ be a real sequence. We say that:\n\n1. $(a_n)$ is **increasing** if \n$$\na_n \\leq a_{n+1} \\,, \\quad \\forall \\, n \\geq N \\,.\n$$\n\n2. $(a_n)$ is **decreasing** if \n$$\na_n \\geq a_{n+1} \\,, \\quad \\forall \\, n \\geq N \\,.\n$$\n\n3. $(a_n)$ is **monotone** if it is either increasing or decreasing.\n\n:::\n\n\n\n::: Example\n\n- The sequence \n$$\na_n = \\frac1n\n$$\nis decreasing.\n\n  > We have\n  $$\n  a_n = \\frac1n > \\frac{1}{n+1} = a_{n+1} \\,.\n  $$\n\n- The sequence \n$$\nb_n = \\frac{n-1}{n} \n$$\nis increasing. \n\n  > We have\n  $$\n  b_{n+1} = \\frac{n}{n+1} > \\frac{n-1}{n} = b_n \\,,\n  $$\n  where the inequality holds because\n  \\begin{align*}\n  \\frac{n}{n+1} > \\frac{n-1}{n} \\quad & \\iff \\quad \n  n^2 > (n-1)(n+1) \\quad  \\\\\n  & \\iff \\quad n^2 > n^2 - 1 \\\\\n  & \\iff \\quad 0 >  - 1\n  \\end{align*}\n\n:::\n\n\nThe main result about monotone sequences is the Monotone Convergence\nTheorem.\n\n\n::: Theorem\n### Monotone Convergence Theorem  {#theorem-monotone-convergence}\n\nIf a sequence is bounded and monotone, then it converges.\n\n:::\n\n\n::: Proof\n\nAssume $(a_n)$ is bounded and monotone. Since $(a_n)$ is bounded, the set \n$$\nA:=\\{ a_n \\divider n \\in \\N \\}  \\subseteq \\R \n$$\nis bounded below and above. By the Axiom of Completeness of $\\R$ there exist $i,s \\in \\R$ such that\n$$\ni = \\inf A \\,, \\quad s = \\sup A \\,.\n$$\n\nWe have two cases:\n\n1. $(a_n)$ is increasing: We are going to prove that\n$$\n\\lim_{n \\to \\infty} a_n = s \\,.\n$$\nEquivalently, we need to prove that\n$$\n\\forall \\, \\e>0 \\,, \\, \\exists \\, N \\in \\N \\st \\forall \\, n \\geq N \\,, \\,\\,\n|a_n - s| < \\e \\,. \n$$\nLet $\\e > 0$. Since $s$ is the smallest upper bound for $A$, this means\n$$\ns - \\e\n$$\nis not an upper bound. Therefore there exists $N \\in \\N$ such that \n$$\ns - \\e < a_N \\,.\n$${#eq-monotone-convergence-proof-1}\nLet $n \\geq N$. Since $a_n$ is increasing, we have\n$$\na_N \\leq a_n \\,, \\quad \\forall \\, n \\geq N \\,.\n$${#eq-monotone-convergence-proof-2}\nMoreover $s$ is the supremum of $A$, so that\n$$\na_n \\leq s < s + \\e \\,, \\quad \\forall \\, n \\in \\N \\,.\n$${#eq-monotone-convergence-proof-3}\nPutting together estimates (@eq-monotone-convergence-proof-1)-(@eq-monotone-convergence-proof-2)-(@eq-monotone-convergence-proof-3)\nwe get\n$$\ns - \\e < a_N \\leq a_n \\leq s < s + \\e \\,, \\quad \\forall \\, n \\geq N \\,.\n$$\nTherefore, for all $n \\geq N$, we have\n$$\ns - \\e < a_n < s + \\e \\quad \\implies \\quad |a_n - s | < \\e \\,.\n$$\n\n\n\n\n2. $(a_n)$ is decreasing: With a similar proof, one can show that\n$$\n\\lim_{n \\to \\infty} a_n = i \\,.\n$$\nThis is left as an exercise.\n\n\n:::\n\n\n\n\n::: {.content-hidden}\n\n\n\n### Example: Euler's Number\n\n\nAs an application of the Monotone Convergence Theorem we can give\na formal definition for the Euler's Number\n$$\ne = 2.71828182845904523536 \\dots\n$$\n\n\n::: Theorem \n\nConsider the sequence\n$$\na_n = \\left(  1 + \\frac{1}{n}  \\right)^n  \\,.\n$$\n\nWe have that:\n\n1. $(a_n)$ is monotone increasing,\n2. $(a_n)$ is bounded.\n\nIn particular $(a_n)$ is convergent.\n\n:::\n\n\n::: Proof\n\nCopiala da Marcellini Sbordone, pagina 106\n\n:::\n\n\nThanks to Theorem \\ref{theorem-euler-number} we can define the Euler's Number $e$.\n\n\n::: Definition\n### Euler's Number\n\nThe Euler's number is defined as\n$$\ne := \\lim_{n \\to \\infty } \\, \\left(  1 + \\frac{1}{n}  \\right)^n \\,.\n$$\n\n:::\n\n\nAlready for $n=1000$ we have a good approximation of $e$:\n$$\na_{1000} = 2.7169 \\,.\n$$\n\n\n\n\n\n\n\n\n\nUse abbot, marcellini, and analisi sapienza note. For next year remove integral test and add cauchy condensation test.\n\n\n## Divergence to infinity\n\nDo this from abbott or marcellini. For the moment we have only done\ndivergent as non-convergent. Would need definition of Divergence to $\\pm \\infty$. Vedi anche forme indeterminate a pagina 96 Marcellini.\n\n\n## Some notable limits\n\nMarcellini Page 101. Also  see what is the english name for limiti notevoli.\n\n\n\n\n\n## Recurrence relations\n\nMarcellini Page 110\n\n\n\n## Example: Heron's Method\n\n\nThe first explicit algorithm for approximating \n$$\n\\sqrt{x}\n$$ \nfor $x > 0$ is known as **Heron's method**, after the first-century Greek mathematician [Heron of Alexandria](https://en.wikipedia.org/wiki/Hero_of_Alexandria) who described the method in his AD 60 work Metrica, see reference to\n[Wikipedia page](https://en.wikipedia.org/wiki/Methods_of_computing_square_roots#Babylonian_method).\n\nLet us see what is the idea of the algorithm:\n\n- Suppose that $a_1$ is an approximation of $\\sqrt{x}$ from above, that is,\n$$\n \\sqrt{x} < a_1 \\,.\n$$ {#eq-heron}\n- Multiplying (@eq-heron) by $\\sqrt{x}/a_1$ we obtain\n$$\n\\frac{x}{a_1} < \\sqrt{x} \\,,\n$${#eq-heron-1}\nobtaining an approximation of $\\sqrt{x}$ from below.\n- Therefore, putting together the above inequalities,\n$$\n\\frac{x}{a_1} < \\sqrt{x} < a_1 \n$$ {#eq-heron-2}\n- If we take the average of the points $x/a_1$ and $a_1$, it is reasonable to think that we find a better approximation of $\\sqrt{x}$. Thus our next approximation is\n$$\na_2 := \\frac{1}{2} \\left( a_1 + \\frac{x}{a_1} \\right) \\,,\n$$\nsee figure below.\n\n\n![Heron's Algorithm for approximating $\\sqrt{x}$](/images/heron.png){width=70%}\n\n\nIterating, we define by recurrence the sequence \n$$\na_{n+1} := \\frac12  \\left( a_n + \\frac{x}{a_n} \\right)\n$$\nfor all $n \\in \\N$, where the initial guess $a_1$ has to satisfy (@eq-heron). The aim of the section is to show that\n$$\n\\lim_{n \\to \\infty } \\ a_n = \\sqrt{x} \\,.\n$${#eq-heron-convergence}\nWe start by showing that (@eq-heron-2) holds for all $n \\in \\N$.\n\n::: {.Proposition  #proposition-heron}\nWe have\n$$\n\\frac{x}{a_n} < \\sqrt{x} < a_n \n$${#eq-heron-3}\nfor all $n \\in \\N$.\n:::\n\n\n::: Proof\nWe prove it by induction:\n\n1. By (@eq-heron) and (@eq-heron-1) we know that (@eq-heron-3) holds for $n=1$. \n\n2. Suppose now that (@eq-heron-3) holds for $n$. Then\n\\begin{align}\na_{n+1} - \\sqrt{x} & = \\frac12  \\left( a_n + \\frac{x}{a_n} \\right) - \\sqrt{x} \\\\\n& = \\frac{1}{2 a_n} ( a_n^2 + x - 2 a_n \\sqrt{x} ) \\\\\n& = \\frac{1}{2 a_n} ( a_n - \\sqrt{x} )^2 > 0 \\,,\n\\end{align}\nsince we are assuming that $a_n > \\sqrt{x}$. Therefore\n$$\n\\sqrt{x} < a_{n+1} \\,.\n$${#eq-heron-proof-1}\nMultiplying the above by $\\sqrt{x}/a_{n+1}$ we get\n$$\n\\frac{x}{a_{n+1}} < \\sqrt{x} \\,.\n$${#eq-heron-proof-2}\nInequalities (@eq-heron-proof-1) and (@eq-heron-proof-2) show that (@eq-heron-3) holds for $n+1$.\n\nTherefore we conclude (@eq-heron-3) by the Principle of Induction.\n:::\n\n\nWe are now ready to prove error estimates, that is, estimating how far away $a_n$ is from $\\sqrt{x}$. \n\n\n::: Proposition\n### Error estimate {#proposition-heron-error}\nFor all $n \\in \\N$ we have\n$$\na_{n+1} - \\sqrt{x} < \\frac12 (a_{n} - \\sqrt{x}) \\,.\n$${#eq-heron-half}\n:::\n\n\n::: Proof\nBy Proposition \\ref{proposition-heron} we know that \n$$\n\\frac{x}{a_n} < \\sqrt{x}\n$$\nfor all $n \\in \\N$. Therefore\n\\begin{align}\na_{n+1} & = \\frac12 \\left(  a_n +  \\frac{x}{a_n}   \\right) \\\\\n        & < \\frac12 \\left(  a_n +  \\sqrt{x}   \\right) \\,.\n\\end{align}\nSubtracting $\\sqrt{x}$ from both members in the above inequality we get the thesis.\n:::\n\n\nInequality (@eq-heron-half) is saying that the error halves at each step. Therefore we can prove that after $n$ steps the error is exponentially lower, as detailed in the following proposition.\n\n\n::: {.Proposition #proposition-heron-error-exp}\nFor all $n \\in \\N$ we have\n$$\na_{n+1} - \\sqrt{x} < \\frac{1}{2^n} (a_1 - \\sqrt{x})\n$$ {#eq-heron-error}\n:::\n\n\n\n::: Proof\nWe prove (@eq-heron-error) by induction:\n\n1. For $n=1$ we have that (@eq-heron-error) is satisfied, since it coincides with (@eq-heron-half) for $n=1$.\n2. Suppose that (@eq-heron-error) holds for $n$. By (@eq-heron-half) with $n$ replaced by $n+1$ we have\n\\begin{align}\na_{n+2} - \\sqrt{x} & < \\frac12 (a_{n+1} - \\sqrt{x}) \\\\\n& < \\frac12 \\,\\cdot \\, \\frac{1}{2^n} (a_{1} - \\sqrt{x}) \\\\\n& = \\frac{1}{2^{n+1}} (a_{1} - \\sqrt{x})\n\\end{align}\nwhere in the second inequality we used the induction hypothesis (@eq-heron-error). Hence (@eq-heron-error) holds for $n+1$.\n\nBy invoking the Induction Principle we conclude the proof.\n:::\n\nLet us comment estimate (@eq-heron-error). Denote the error at step $n$ by\n$$\ne_n := a_n - \\sqrt{x}\\,.\n$$\nThe initial error $e_1$ depends on how far the initial guess is from $\\sqrt{x}$. The estimate in (@eq-heron-error) is telling us that $e_n$ is a fraction of $e_1$, and actually\n$$\n\\lim_{n \\to \\infty} \\ e_n = 0\n$$\nexponentially fast. From this fact we are finally able to prove (@eq-heron-convergence).\n\n::: Theorem\n### Convergence of Heron's Algorithm\nWe have that \n$$\n\\lim_{n \\to \\infty} \\ a_n = \\sqrt{x} \\,.\n$$\n:::\n\n::: Proof\nBy Proposition \\ref{proposition-heron-error-exp} we have that\n$$\na_{n+1} - \\sqrt{x} < \\frac{1}{2^n} (a_1 - \\sqrt{x})\n$$\nMoreover Proposition \\ref{proposition-heron} tells us that\n$$\n\\sqrt{x} < a_{n+1} \\,.\n$$\nPutting together the two inequalities above we infer\n$$\n\\sqrt{x} < a_{n+1} < \\sqrt{x} + \\frac{1}{2^n} (a_1 - \\sqrt{x}) \\,.\n$${#eq-heron-final}\nNow note that\n$$\n\\lim_{n \\to \\infty} \\ \\frac{1}{2^n} =\n\\lim_{n \\to \\infty} \\ \\left( \\frac{1}{2} \\right)^n = 0 \\,.\n$$\nTherefore the RHS of (@eq-heron-final) converges to $\\sqrt{x}$ as $n \\to \\infty$. Applying the Squeeze Theorem to (@eq-heron-final) we conclude that $a_n \\to \\sqrt{x}$ as $n \\to \\infty$.\n:::\n\n\n\n### Coding the Algorithm\n\nHeron's Algorithm can be easily coded in Python. For example, see the function below:\n\n```python\n# x is the number for which to compute sqrt(x)\n# guess is the point a_1\n# a_1 must be strictly larger than sqrt(x)\n# n is the number of iterations\n# the function returns a_{n+1}\n\ndef herons_algorithm(x, guess, n):\n    for i in range(n):\n        guess = (guess + x / guess) / 2.0\n    return guess\n```\n\nFor example let us use the Algorithm to compute $\\sqrt{2}$ after $3$ iterations. For initial guess we take $a_1 = 2$. \n\n```python\n# Calculate sqrt(2) with 3 iterations and guess 2\nsqrt_2 = herons_algorithm(2, 2, 3)\n\nprint(f\"The sqrt(2) is approximately {sqrt_2}\")\n```\n\n::: {.cell execution_count=1}\n\n::: {.cell-output .cell-output-stdout}\n```\nThe sqrt(2) is approximately 1.4142156862745097\n```\n:::\n:::\n\n\nThat is a pretty good approximation in just $3$ iterations!\n\n\n\n\n\n## Fibonacci Sequence\n\n\n\n\n:::\n\n",
+    "markdown": "::: {.content-hidden}\n$\n\\newcommand{\\R}{\\mathbb{R}}\n\\newcommand{\\Q}{\\mathbb{Q}}\n\\newcommand{\\C}{\\mathbb{C}}  \n\\newcommand{\\N}{\\mathbb{N}}\n\\newcommand{\\Z}{\\mathbb{Z}}\n\n\\renewcommand\\Re{\\operatorname{Re}}\n\\renewcommand\\Im{\\operatorname{Im}}\n\n\n\n\n\n\\newcommand{\\e}{\\varepsilon}\n\\newcommand{\\g}{\\gamma}\n\n\\newcommand{\\st}{\\, \\text{ s.t. } \\, }\n\\newcommand{\\divider}{\\, \\colon \\,}\n\n\\newcommand{\\closure}[2][2]{{}\\mkern#1mu \\overline{\\mkern-#1mu #2 \\mkern-#1mu}\\mkern#1mu {}}\n\n\n\\newcommand{\\Czero}{\\mathnormal{C}}\n\\newcommand{\\Cinf}{{\\mathnormal{C}}^{\\infty}}\n\\newcommand{\\Cinfcomp}{{\\mathnormal{C}}_0^{\\infty}}\n\\newcommand{\\Lone}{{\\mathnormal{L}}^1}\n\\newcommand{\\Loneloc}{{\\mathnormal{L}}_{loc}^1}\n\\newcommand{\\Ltwo}{{\\mathnormal{L}}^2}\n\\newcommand{\\Lp}{{\\mathnormal{L}}^p}\n\\newcommand{\\Linf}{{\\mathnormal{L}}^{\\infty}}\n\\newcommand{\\Woneone}{{\\mathnormal{W}}^{1,1}}\n\\newcommand{\\Wonetwo}{{\\mathnormal{W}}^{1,2}}\n\\newcommand{\\Wonep}{{\\mathnormal{W}}^{1,p}}\n\\newcommand{\\Woneinf}{{\\mathnormal{W}}^{1,\\infty}}\n\\newcommand{\\Wtwotwo}{{\\mathnormal{W}}^{2,2}}\n\\newcommand{\\Wktwo}{{\\mathnormal{W}}^{k,2}}\n\\newcommand{\\Wkp}{{\\mathnormal{W}}^{k,p}}\n\\newcommand{\\Lip}{\\mathnormal{Lip}}\n\n\\newcommand{\\scp}[2]{\\left\\langle #1,#2 \\right\\rangle} %prodotto scalare\n\\newcommand{\\abs}[1]{\\left| #1 \\right|}  %valore assoluto\n\\newcommand{\\norm}[1]{\\left\\| #1 \\right\\|} %norma\n\n\n\n\n\n\n\n\n\n$\n:::\n\n\n\n\n\n# Sequences in $\\mathbb{R}$\n\n\n\nA sequence is an infinite list of real numbers. For example, the following are sequences:\n\n\n- $(1,2,3,4, \\ldots)$\n\n- $(-1,1,-1,1, \\ldots)$\n\n- $\\left(1, \\frac{1}{2}, \\frac{1}{3}, \\frac{1}{4}, \\frac{1}{5}, \\ldots\\right)$\n\n\n::: Remark\n\n- The order of elements in a sequence matters. \n\n> For example\n>$$\n>(1,2,3,4,5,6, \\ldots) \\neq(2,1,4,3,6,5, \\ldots)\n>$$\n\n- A sequence is not a set. \n\n> For example \n>$$\n>\\{-1,1,-1,1,-1,1, \\ldots\\}=\\{-1,1\\}\n>$$\n>but we cannot make a similar statement for the sequence \n>$$\n>(-1,1,-1,1,-1,1, \\ldots) \\,.\n>$$\n\n- The above notation is ambiguous.\n\n> For example the sequence \n>$$\n>(1,2,3,4, \\ldots)\n>$$\ncan continue as\n>\n>$$\n>(1,2,3,4,1,2,3,4,1,2,3,4, \\ldots) \\,.\n>$$\n\n- In the sequence \n$$\n\\left(1, \\frac{1}{2}, \\frac{1}{3}, \\frac{1}{4}, \\frac{1}{5}, \\ldots\\right)\n$$\nthe elements get smaller and smaller, and closer and closer to $0$. We say that this sequence converges to $0$, or has $0$ as a limit. \n\n\n:::\n\n\nWe would like to make the notions of sequence and convergence more precise.\n\n\n## Definition of sequence\n\n\nWe start with the definition of sequence of Real numbers.\n\n\n::: Definition \n### Sequence of Real numbers\n\nA sequence $a$ in $\\R$ is a function\n$$\na \\colon \\N \\to \\R \\,.\n$$\nFor $n \\in \\N$, we denote the $n$-th element of the sequence $a$ by \n$$\na_{n}=a(n)\n$$ \nand write the sequence as \n$$\n\\left(a_{n}\\right)_{n \\in \\N} \\,.\n$$\n:::\n\n\n::: Notation\n\nWe will sometimes omit the subscript $n \\in \\N$ and simply write \n$$\n\\left(a_{n}\\right) \\,.\n$$ \nIn certain situations, we will also write\n$$\n\\left(a_{n}\\right)_{n=1}^{\\infty} \\,.\n$$\n\n\n:::\n\n\n\n\n::: Example \n\n- In general $\\left(a_{n}\\right)_{n \\in \\N}$ is the sequence \n$$\n\\left(a_{1}, a_{2}, a_{3}, \\ldots\\right)\\,.\n$$\n\n- Consider the function \n$$\na \\colon \\N \\to \\N \\, , \\quad  n \\mapsto 2 n \\,.\n$$\nThis is also a sequence of real numbers. It can be written as \n$$\n(2 n)_{n \\in \\N}\n$$\nand it represents the sequence of even numbers \n$$\n(2,4,6,8,10, \\ldots) \\,.\n$$\n\n- Let \n$$\na_{n}=(-1)^{n}\n$$\nThen $\\left(a_{n}\\right)$ is the sequence \n$$\n(-1,1,-1,1,-1,1, \\ldots) \\,.\n$$\n\n- $\\left(\\frac{1}{n}\\right)_{n \\in \\N}$ is the sequence \n$$\n\\left(1, \\frac{1}{2}, \\frac{1}{4}, \\frac{1}{5}, \\ldots\\right) \\,.\n$$\n\n\n:::\n\n\n\n## Convergent sequences\n\n\nWe have notice that the sequence $\\left(\\frac{1}{n}\\right)_{n \\in \\N}$ gets close to $0$ as $n$ gets large. We would like to say that $a_{n}$ converges to $0$ as $n$ tends to infinity.\n\nTo make this precise, we first have to say what it means for two numbers to be *close*. For this we use the notion of absolute value, and say that:\n\n- $x$ and $y$ are close if $|x-y|$ is small. \n- $|x-y|$ is called the distance between $x$ and $y$\n- For $x$ to be close to $0$, we need that $|x-0|=|x|$ is small.\n\nSaying that $|x|$ is *small* is not very precise. Let us now give the formal definition of convergent sequence.\n\n::: Definition \n### Convergent sequence  {#definition-convergent-sequence}\n\nWe say that a sequence $\\left(a_{n}\\right)_{n \\in \\N}$ in $\\mathbb{R}$ **converges** to $a \\in \\mathbb{R}$, or equivalently has limit $a$, denoted by\n$$\n\\lim_{n \\rightarrow \\infty} a_{n}=a \n$$\nif for all $\\e \\in \\mathbb{R}, \\e>0$, there exists $N \\in \\N$ such that for all $n \\in \\N, n \\geq N$ it holds that \n$$\n\\left|a_{n}-a\\right| < \\e \\,.\n$$\nUsing quantifiers, we can write this as\n$$\n\\forall \\, \\e>0 , \\, \\exists \\, N \\in \\N \\st \\forall \\, n \\geq N \\,, \\,\\left|a_{n}-a\\right| < \\e \\,.\n$$\n\nIf there exists $a \\in \\R$ such that $\\lim_{n \\rightarrow \\infty} a_{n}=a$, then we say that the sequence $\\left(a_{n}\\right)_{n \\in \\N}$ is **convergent**.\n\n:::\n\n\n::: Notation\n\nWe will often write\n$$\na_n \\to a\n$$\nin place of\n$$\n\\lim_{n \\rightarrow \\infty} a_{n}=a \\,.\n$$\n\n:::\n\n\n::: Remark\n\n- Informally, Definition \\ref{definition-convergent-sequence} says that, no matter how small we choose $\\e$ (as long as it is strictly positive), we always have that $a_{n}$ has a distance to $a$ of less than or equal to $\\e$ from a certain point onwards (i.e., from $N$ onward). The sequence $\\left(a_{n}\\right)$ may fluctuate wildly in the beginning, but from $N$ onward it should stay within a distance of $\\e$ of $a$.\n\n- In general $N$ depends on $\\e$. If $\\e$ is chosen smaller, we might have to take $N$ larger: this means we need to wait longer before the sequence stays within a distance $\\e$ from $a$.\n\n:::\n\n\nWe now prove that the sequence\n$$\na_n = \\frac1n\n$$\nconverges to $0$, according to Definition \\ref{definition-convergent-sequence}.\n\n::: {.Theorem #theorem-one-over-n}\n\nThe sequence $\\left(\\frac{1}{n}\\right)_{n \\in \\N}$ converges to $0$ , i.e.,\n\n$$\n\\lim_{n \\rightarrow \\infty} \\frac{1}{n}=0 \\,.\n$$\n\n:::\n\nWe give two proofs of the above theorem:\n\n- Long proof, with all the details.\n- Short proof, with less details, but still acceptable. \n\n::: Proof \n### Proof of Theorem \\ref{theorem-one-over-n} (Long version)\n\nWe have to show that\n\n$$\n\\lim _{n \\rightarrow \\infty} \\frac{1}{n}=0,\n$$\n\nwhich by definition is equivalent to showing that\n\n$$\n\\forall \\, \\e>0 \\,, \\, \\exists \\, N \\in \\N \\st  \\forall  \\, n \\geq N \\,, \\, \\left|\\frac{1}{n}-0\\right| < \\e \\,.\n$${#eq-one-over-n}\n\nLet $\\e \\in \\mathbb{R}$ with $\\e>0$. Choose $N \\in \\N$ such that \n$$\nN > \\frac{1}{\\e}\\,.\n$$\nSuch natural number $N$ exists thanks to the Archimedean property. The above implies\n$$\n\\frac{1}{N} < \\e \\,,\n$${#eq-one-over-n-1}\nLet $n \\in \\N$ with $n \\geq N$. By (@eq-one-over-n-1) we have\n$$\n\\frac{1}{n} \\leq \\frac{1}{N} < \\e \\,.\n$$\nFrom this we deduce\n$$\n\\left|\\frac{1}{n}-0\\right|=\\frac{1}{n} \\leq \\frac{1}{N} < \\e\n$$\n\nSince $n \\in \\N, n \\geq N$ was arbitrary, we have proven that \n$$\n\\left|\\frac{1}{n}-0\\right| < \\e \\,, \\quad \\forall \\, n \\geq N \\,.\n$${#eq-one-over-n-2}\nCondition (@eq-one-over-n-2) holds for all $\\e>0$, for the choice of $N \\in \\N$ such that\n$$\nN > \\frac{1}{\\e} \\,.\n$$\nWe have hence shown (@eq-one-over-n), and the proof is concluded.\n\n:::\n\nAs the above proof is quite long and includes lots of details, it is acceptable to shorten it. For example:\n\n- We skip some intermediate steps.\n- We do not mention the Archimedean property.\n- We leave out the conclusion when it is obvious that the statement has been proven. \n\n\n\n\n::: Proof \n### Proof of Theorem \\ref{theorem-one-over-n} (Short version)\n\nWe have to show that\n\n$$\n\\forall \\, \\e>0 \\,, \\, \\exists \\, N \\in \\N \\st  \\forall  \\, n \\geq N \\,, \\, \\left|\\frac{1}{n}-0\\right| < \\e \\,.\n$$\n\nLet $\\e>0$. Choose $N \\in \\N$ such that \n$$\nN > \\frac{1}{\\e}\\,.\n$$\nLet $n \\geq N$. Then\n$$\n\\left|\\frac{1}{n}-0\\right|=\\frac{1}{n} \\leq \\frac{1}{N} < \\e \\,.\n$$\n\n:::\n\n\nIn Theorem \\ref{theorem-one-over-n} we showed that \n$$\n\\lim_{n \\to \\infty} \\, \\frac{1}{n} = 0 \\,.\n$$\nWe can generalise this statement to prove that \n$$\n\\lim_{n \\to \\infty} \\, \\frac{1}{n^{p}} = 0\n$$\nfor any $p>0$ fixed.\n\n\n::: {.Theorem  #theorem-one-over-np}\nFor all $p>0$, the sequence $\\left(\\frac{1}{n^{p}}\\right)_{n \\in \\N}$ converges to $0$ , i.e.,\n\n$$\n\\lim _{n \\rightarrow \\infty} \\frac{1}{n^{p}}=0\n$$\n:::\n\n::: Proof\n\nLet $p>0$. We have to show that\n$$\n\\forall \\e>0 \\,, \\, \\exists \\, N \\in \\N \\st  \\forall \\, n \\geq N \\,, \\, \\left|\\frac{1}{n^{p}}-0\\right| < \\e \\,.\n$$\nLet $\\e>0$. Choose $N \\in \\N$ such that \n$$\nN > \\frac{1}{\\e^{1 / p}} \\,.\n$${#eq-theorem-one-over-np-1}\nLet $n \\geq N$. Since $p>0$, we have $n^{p} \\geq N^{p}$, which implies \n$$\n\\frac{1}{n^p} \\leq \\frac{1}{N^p} \\,.\n$$\nBy (@eq-theorem-one-over-np-1) we deduce\n$$\n\\frac{1}{N^p} < \\e \\,.\n$$\nThen\n$$\n\\left|\\frac{1}{n^{p}}-0\\right|=\\frac{1}{n^{p}} \\leq \\frac{1}{N^{p}} < \\e \\,.\n$$\n\n:::\n\n::: Question\nWhy did we choose $N \\in \\N$ such that\n$$\nN > \\frac{1}{\\e^p}\n$$\nin the above proof?\n:::\n\nThe answer is: because it works. Finding a number $N$ that makes the proof work requires some *rough work*: Specifically, such rough work consists in finding $N \\in \\N$ such that the inequality\n$$\n|a_N - a | < \\e \n$$\nis satisfied. \n\n\n::: Important \n\nAny *rough work* required to prove convergence must be shown before the formal proof (in assignments). \n\n:::\n\n\n\n\n::: Example \n\nProve that, as $n \\rightarrow \\infty$,\n\n$$\n\\frac{n}{2n+3} \\to \\frac{1}{2} \\,.\n$$\n\n*Part 1. Rough Work.*\n\nLet $\\e>0$. We want to find $N \\in \\N$ such that\n$$\n\\left|\\frac{n}{2n+3}-\\frac{1}{2}\\right| < \\e \\,, \\quad \\forall \\, n \\geq N \\,.\n$$\nTo this end, we compute:\n\\begin{align*}\n\\left|\\frac{n}{2 n+3}-\\frac{1}{2}\\right| & \n=\\left|\\frac{2n -(2n + 3)}{2(2n +3)}\\right| \\\\\n& =\\left|\\frac{- 3}{4n + 6}\\right| \\\\\n& = \\frac{3}{4n + 6} \\,.\n\\end{align*}\nTherefore \n\\begin{align*}\n\\left|\\frac{n}{2n+3}-\\frac{1}{2}\\right| < \\e\n\\quad \\iff &  \\quad\n\\frac{3}{4n + 6} < \\e \\\\\n\\quad \\iff &  \\quad\n\\frac{4n + 6}{3} > \\frac{1}{\\e} \\\\\n\\quad \\iff &  \\quad\n4n + 6 > \\frac{3}{\\e} \\\\\n\\quad \\iff &  \\quad\n4n  > \\frac{3}{\\e} - 6 \\\\\n\\quad \\iff &  \\quad\nn  > \\frac{3}{4\\e} - \\frac{6}{4} \\,.\n\\end{align*}\nLooking at the above equivalences, it is clear that $N \\in \\N$ has to be chosen so that\n$$\nN  > \\frac{3}{4\\e} - \\frac{6}{4} \\,.\n$$\n\n\n*Part 2. Formal Proof.* We have to show that\n$$\n\\forall \\e>0 \\,, \\, \\exists \\, N \\in \\N \\st  \\forall \\, n \\geq N \\,, \\, \\left|\\frac{n}{2n+3}-\\frac{1}{2}\\right| < \\e  \\,.\n$$\nLet $\\e>0$. Choose $N \\in \\N$ such that\n$$\nN > \\frac{3}{4\\e} - \\frac{6}{4}   \\,.\n$${#eq-example-limit-1}\nBy the rough work shown above, inequality (@eq-example-limit-1) is equivalent to \n$$\n \\frac{3}{4N + 6} < \\e \\,.\n$$\nLet $n \\geq N$. Then\n\\begin{align*}\n\\left|\\frac{n}{2n+3}-\\frac{1}{2}\\right| \n& = \\left|\\frac{- 3 }{2(2 n+3)}\\right| \\\\\n& = \\frac{3}{4 n+ 6 } \\\\\n& \\leq \\frac{3}{4 N+ 6 } \\\\\n& < \\e \\,,\n\\end{align*}\nwhere in the third line we used that $n \\geq N$.\n\n:::\n\n\nWe conclude by showing that constant sequences always converge.\n\n::: {.Theorem #theorem-constant-sequence}\n\nLet $c \\in \\R$ and define the constant sequence\n$$\na_n := c \\,, \\quad \\forall \\, n \\in \\N \\,.\n$$\nWe have that \n$$\n\\lim_{n \\to \\infty} \\, a_n = c \\,.\n$$\n\n:::\n\n::: Proof\n\nWe have to prove that\n$$\n\\forall \\e>0 \\,, \\, \\exists \\, N \\in \\N \\st  \\forall \\, n \\geq N \\,, \\, \\left|a_n - c \\right| < \\e  \\,.\n$${#eq-constant-sequence}\nLet $\\e>0$. We have\n$$\n\\left|a_n - c \\right| = \\left|c - c \\right| = 0 < \\e \\,, \\quad \\forall \\, n \\in \\N \\,.\n$$\nTherefore we can choose $N=1$ and (@eq-constant-sequence) is satisfied. \n:::\n\n\n\n\n## Divergent sequences\n\n\nThe opposite of convergent sequences are divergent sequences.\n\n::: Definition\n### Divergent sequence\n\nWe say that a sequence $\\left(a_{n}\\right)_{n \\in \\N}$ in $\\R$ is **divergent** if it is not convergent.\n\n:::\n\n\n\n::: Remark\n\nProving that a sequence $(a_n)$ is divergent is more complicated than showing it is convergent: To show that $(a_n)$ is divergent, we need to show that \n$(a_n)$ cannot converge to $a$ for any $a \\in \\R$.\n\nIn other words, we have to show that there does not exist an $a \\in \\mathbb{R}$ such that \n$$\n\\lim _{n \\rightarrow \\infty} \\, a_n = a  \\,.\n$$\nUsing quantifiers, this means\n$$\n\\nexists \\, a \\in \\R \\st \\forall \\, \\e>0 \\,, \\, \n\\exists \\, N \\in \\N \\st \\forall \\, n \\geq N \\,, \\,\n\\left|a_{n}-a\\right| < \\e \\,.\n$$\nThe above is equivalent to showing that\n$$\n\\forall \\, a \\in \\R \\,, \\, \\exists \\, \\e > 0 \\st  \\forall \\, N \\in \\N \\,, \\, \\exists \\, n \\geq N \\st \\left|a_{n}-a\\right| \\geq \\e \\,.\n$$\n\n:::\n\n\n\n::: {.Theorem #theorem-1-minus-n}\nLet $\\left(a_{n}\\right)$ be the sequence defined by\n$$\na_{n}=(-1)^{n} \\,.\n$$\nThen $\\left(a_{n}\\right)$ does not converge. \n\n:::\n\n::: Proof\n\nTo prove that $\\left(a_{n}\\right)$ does not converge, we have to show that \n$$\n\\forall \\, a \\in \\R \\,, \\, \\exists \\, \\e > 0 \\st  \\forall \\, N \\in \\N \\,, \\, \\exists \\, n \\geq N \\st \\left|a_{n}-a\\right| \\geq \\e \\,.\n$$\nLet $a \\in \\R$. Choose \n$$\n\\e=\\frac{1}{2} \\,.\n$$ \nLet $N \\in \\N$. We distinguish two cases:\n\n- $a \\geq 0$: Choose $n=2 N+1$. Note that $n \\geq N$. Then\n\\begin{align*}\n\\left|a_{n}-a\\right| & = \\left|a_{2 N+1}-a\\right| \\\\\n& = \\left|(-1)^{2 N+1}-a\\right| \\\\\n& =|-1-a| \\\\\n& =1+a \\\\\n& \\geq 1 \\\\\n& > \\frac{1}{2} = \\e \\,,\n\\end{align*}\nwhere we used that $a \\geq 0$, and therefore \n$$\n|-1-a|=1+a \\geq 1 \\,.\n$$\n\n- $a<0$: Choose $n=2 N$. Note that $n \\geq N$. Then\n\\begin{align*}\n\\left|a_{n}-a\\right| & = \\left|a_{2 N}-a\\right| \\\\\n& =\\left|(-1)^{2 N}-a\\right| \\\\ \n& = |1-a| \\\\\n& = 1-a \\\\\n& > 1 \\\\\n& > \\frac{1}{2} = \\e \\,,\n\\end{align*}\nwhere we used that $a < 0$, and therefore \n$$\n|1-a|=1-a > 1 \\,.\n$$\n\n\n:::\n\n\n\n## Uniqueness of limit\n\n\nIn Definition \\ref{definition-convergent-sequence} of convergence, we used the notation \n$$\n\\lim_{n \\rightarrow \\infty} a_{n}=a \\,.\n$$ \nThe above notation makes sense only if the limit is unique, that is, if we do not have that  \n$$\n\\lim_{n \\rightarrow \\infty} a_{n} = b \\,,\n$$ \nfor some \n$$\na \\neq b \\,.\n$$\nIn the next theorem we will show that the limit is unique, if it exists.\n\n::: Theorem \n### Uniqueness of limit  {#theorem-uniqueness-limit}\n\nLet $\\left(a_{n}\\right)_{n \\in \\N}$ be a sequence. Suppose that \n$$\n\\lim_{n \\rightarrow \\infty} a_{n} = a \\,, \\quad \n\\lim_{n \\rightarrow \\infty} a_{n} = b \\,.\n$$\nThen $a=b$.\n:::\n\n\n::: Proof\n\nAssume that, \n$$\n\\lim_{n \\rightarrow \\infty} a_{n} = a \\,, \\quad \n\\lim_{n \\rightarrow \\infty} a_{n} = b \\,.\n$$\nSuppose by contradiction that\n$$\na \\neq b \\,.\n$$\nChoose \n$$\n\\e := \\frac{1}{2} \\, |a-b| \\,.\n$$\nTherefore $\\e>0$, since $|a-b|>0$. By the convergence $a_{n} \\rightarrow a$,\n$$\n\\exists \\, N_1 \\in \\N \\st \\forall \\, n \\geq N_1 \\,, \\, \n\\left|a_{n}-a\\right| < \\e \\,.\n$$\nBy the convergence $a_{n} \\rightarrow b$,\n$$\n\\exists \\, N_2 \\in \\N \\st \\forall \\, n \\geq N_2 \\,, \\, \n\\left|a_{n} - b \\right| < \\e \\,.\n$$\nDefine\n$$\nN := \\max \\{ N_1, N_2 \\} \\,.\n$$\nChoose an $n \\in \\N$ such that $n \\geq N$. In particular\n$$\nn \\geq N_1 \\,, \\quad n \\geq N_2 \\,.\n$$\nThen\n\\begin{align*}\n2 \\e & = |a-b| \\\\\n& = \\left|a-a_{n}+a_{n}-b\\right| \\\\\n& \\leq\\left|a-a_{n}\\right|+\\left|a_{n}-b\\right| \\\\\n& < \\e + \\e \\\\\n& = 2 \\e \\,,\n\\end{align*}\nwhere we used the triangle inequality in the first inequality.\nHence $2 \\e < 2 \\e$, which gives a contradiction.\n\n:::\n\n\n::: Example \n\nProve that\n\n$$\n\\lim _{n \\rightarrow \\infty} \\frac{n^{2}-1}{2 n^{2}-3}=\\frac{1}{2}\n$$\n\nAccording to Theorem \\ref{theorem-uniqueness-limit}, it suffices to show that the sequence \n$$\n\\left(\\frac{n^{2}-1}{2 n^{2}-3}\\right)_{n \\in \\N}\n$$ \nconverges to $\\frac{1}{2}$, since then $\\frac{1}{2}$ can be the only limit.\n\n\n*Part 1. Rough Work.*\n\nLet $\\e>0$. We want to find $N \\in \\N$ such that\n$$\n\\left|\\frac{n^{2}-1}{2 n^{2}-3} - \\frac{1}{2}\\right| < \\e \\,, \\quad \\forall \\, n \\geq N \\,.\n$$\nTo this end, we compute:\n\\begin{align*}\n\\left|\\frac{n^{2}-1}{2 n^{2}-3}-\\frac{1}{2}\\right| \n& =\\left|\\frac{2\\left(n^{2}-1\\right)-\\left(2 n^{2}-3\\right)}{2\\left(2 n^{2}-3\\right)}\\right| \\\\\n& =\\left|\\frac{1}{4 n^{2}-6}\\right| \\\\\n& =\\frac{1}{4 n^{2}-6} \\\\\n& =\\frac{1}{3 n^{2}+n^{2}-6} \\\\\n& \\leq \\frac{1}{3 n^{2}} \n\\end{align*}\nwhich holds if $n \\geq 3$, since in this case $n^2 - 6 \\geq 0$.\nTherefore \n\\begin{align*}\n\\left|\\frac{n^{2}-1}{2 n^{2}-3}-\\frac{1}{2}\\right|  < \\e\n\\quad \\impliedby &  \\quad\n\\frac{1}{3n^2} < \\e \\\\\n\\quad \\iff &  \\quad\n3n^2  > \\frac{1}{\\e} \\\\\n\\quad \\iff &  \\quad\nn^2  > \\frac{1}{3\\e} \\\\\n\\quad \\iff &  \\quad\nn  > \\frac{1}{\\sqrt{3\\e}} \\,.\n\\end{align*}\nLooking at the above implications, it is clear that $N \\in \\N$ has to be chosen so that\n$$\nN  > \\frac{1}{\\sqrt{3\\e}} \\,.\n$$\nMoreover we need to recall that $N$ has to satisfy \n$$\nN \\geq 3\n$$ \nfor the estimates to hold.\n\n\n*Part 2. Formal Proof.* We have to show that\n$$\n\\forall \\e>0 \\,, \\, \\exists \\, N \\in \\N \\st  \\forall \\, n \\geq N \\,, \\, \\left|\\frac{n^{2}-1}{2 n^{2}-3}-\\frac{1}{2}\\right| < \\e  \\,.\n$$\nLet $\\e>0$. Choose $N \\in \\N$ such that\n$$\nN > \\max \\left\\{ \\frac{1}{\\sqrt{3\\e}}, 3 \\right\\}   \\,.\n$$\nLet $n \\geq N$. Then\n\\begin{align*}\n\\left|\\frac{n^{2}-1}{2 n^{2}-3}-\\frac{1}{2}\\right| \n& =\\frac{1}{4 n^{2}-6} \\\\\n& =\\frac{1}{3 n^{2}+n^{2}-6} \\\\\n& \\leq \\frac{1}{3 n^{2}} \\\\\n& \\leq \\frac{1}{3 N^{2}} \\\\\n& < \\e \\,,\n\\end{align*}\nwhere we used that \n$$\nn \\geq N \\geq 3 \n$$ \nwhich implies \n$$\nn^2 - 6 \\geq 0\\,,\n$$\nin the third line. The last inequality holds, since it is equivalent \nto \n$$\nN > \\frac{1}{\\sqrt{3 \\e}} \\,.\n$$\n\n:::\n\n\n\n\n## Bounded sequences\n\n\nAn important property of sequences is boundedness. \n\n\n::: Definition \n### Bounded sequence  {#definition-bounded-sequence}\n\nA sequence $\\left(a_{n}\\right)_{n \\in \\N}$ is called **bounded** if there exists a constant $M \\in \\R$, with $M>0$, such that\n$$\n\\left|a_{n}\\right| \\leq M \\,, \\quad \\forall \\, n \\in \\N \\,.\n$$\n\n:::\n\nDefinition \\ref{definition-bounded-sequence} says that a sequence is bounded, if we can find some constant $M>0$ (possibly very large), such that for all elements of the sequence it holds that \n$$\n\\left|a_{n}\\right| \\leq M \\,,\n$$\nor equivalently, that \n$$\n-M \\leq a_{n} \\leq M \\,.\n$$\n\n\nWe now show that any sequence that converges is also bounded\n\n::: {.Theorem  #theorem-convergent-bounded}\n\nIf a sequence $\\left(a_{n}\\right)_{n \\in \\mathbb{N}}$ converges, then the sequence is bounded.\n\n:::\n\n\n::: Proof \n\nSuppose the sequence $\\left(a_{n}\\right)_{n \\in \\mathbb{N}}$ converges and let\n$$\na: = \\lim_{n \\rightarrow \\infty} a_{n}\n$$\nBy definition of convergence we have that\n$$\n\\forall \\, \\e >0 \\,, \\, \\exists N \\in \\N  \\st  \\forall \\, n \\geq N \\,, \\, \\left|a_{n}-a \\right| < \\e \\,.\n$$\nIn particular, we can choose\n$$\n\\e = 1\n$$ \nand let $N \\in \\N$ be that value such that\n$$\n\\left|a_{n}-a \\right| < 1 \\,, \\quad \\forall \\, n \\geq N \\,.\n$$\nIf $n \\geq N$ we have, by the triangle inequality,\n\\begin{align*}\n\\left|a_{n}\\right| & = \\left|a_{n}-a+a\\right| \\\\\n                   & \\leq \\left|a_{n}-a\\right|+|a| \\\\\n                   & < 1+|a| \\,.\n\\end{align*}\nSet\n$$\nM:=\\max \\left\\{\\left|a_{1}\\right|,\\left|a_{2}\\right|, \\ldots,\\left|a_{N-1}\\right|, 1+|a|\\right\\} \\,.\n$$\nNote that such maximum exists, being the set finite. Then \n$$\n\\left|a_{n}\\right| \\leq M \\,, \\quad \\forall \\, n \\in \\N \\,,\n$$\nshowing that $(a_n)$ is bounded.\n\n:::\n\n\nThe choice of $M$ in the above proof says that the sequence can behave wildly for a finite number of terms. After that, it will stay close to the value of the limit, if the latter exists.\n\n::: Example\n\nIn Theorem \\ref{theorem-one-over-n} we have shown that \n$$\n\\lim_{n \\to \\infty} \\,  \\frac{1}{n} = 0 \n$$\nHence, it follows from Theorem \\ref{theorem-convergent-bounded} that the sequence $(1/n)$ is bounded.\n\nIndeed, we have that\n$$\n\\left|\\frac{1}{n}\\right|=\\frac{1}{n} \\leq 1 \\,, \\quad \\forall \\, n \\in \\N \\,,\n$$\nsince $n \\geq 1$ for all $n \\in \\N$.\n\n:::\n\n\n::: Warning\n\nThe converse of Theorem \\ref{theorem-convergent-bounded} does not hold: There exist sequences $(a_n)$ which are bounded, but not convergent.\n\n:::\n\n\n::: Example\n\nDefine the sequence\n$$\na_n = (-1)^n \\,.\n$$\nWe have proven in Theorem \\ref{theorem-1-minus-n} that $(a_n)$ is not convergent. However\n$(a_n)$ is bounded, with $M=1$, since\n$$\n|a_n| = |(-1)^n| = 1 = M  \\,, \\quad \\forall \\, n \\in \\N \\,.\n$$\n\n:::\n\n\n\nTaking the contrapositive of the statement in Theorem \\ref{theorem-convergent-bounded} we get the following corollary:\n\n\n::: {.Corollary   #corollary-convergent-bounded}\n\nIf a sequence $\\left(a_{n}\\right)_{n \\in \\mathbb{N}}$ is not bounded, then the sequence does not converge.\n\n:::\n\n\n::: Remark\n\nFor a sequence $(a_n)$ to be unbounded, it means that\n$$\n\\forall \\, M > 0 \\,, \\,\\,  \\exists \\, n \\in \\N \\st \\left|a_{n}\\right| > M \\,.\n$$\nThe above is saying that no real number $M>0$ can be a bound for $|a_n|$, since there is always an index $n \\in \\N$ such that \n$$\n|a_n| > M \\,.\n$$\n\n:::\n\n\nWe can use Corollary \\ref{corollary-convergent-bounded} to show that certain sequences do not converge.\n\n\n::: Theorem \n\nFor all $p>0$, the sequence $\\left(n^{p}\\right)_{n \\in \\mathbb{N}}$ does not converge.\n\n:::\n\n\n::: Proof\nLet $p>0$. We prove that the sequence $\\left(n^{p}\\right)_{n \\in \\mathbb{N}}$ is unbounded, that is, \n$$\n\\forall \\, M > 0 \\,, \\,\\,  \\exists \\, n \\in \\N \\st \\left|a_{n}\\right| > M \\,.\n$$\nTo this end, let $M > 0$. Choose $n \\in \\N$ such that \n$$\nn > M^{1/p} \\,.\n$$\nThen\n$$\na_n = n^{p}>\\left(M^{1/p}\\right)^{p}=M \\,.\n$$\nThis proves that the sequence $(n^p)$ is unbounded. Hence $(n^p)$ cannot converge, by Corollary \\ref{corollary-convergent-bounded}.\n\n:::\n\n\n\n::: Theorem\n\nThe sequence $(\\log n)_{n \\in \\mathbb{N}}$ does not converge.\n\n:::\n\n\n::: Proof \n\nLet us show that $(\\log n)_{n \\in \\mathbb{N}}$ is unbounded, that is, \n$$\n\\forall \\, M > 0  \\,, \\,\\,  \\exists \\, n \\in \\N \\st \\left|a_{n}\\right| > M \\,.\n$$\nTo this end let $M > 0$. Choose $n \\in \\N$ such that \n$$\nn \\geq e^{M+1} \\,.\n$$\nThen\n$$\n|a_n| = |\\log n| \\geq \\left|   \\log e^{M+1}   \\right| = M + 1 > M \\,.\n$$\nThis proves that the sequence $(\\log n)$ is unbounded. Hence $(\\log n)$ cannot converge, by Corollary \\ref{corollary-convergent-bounded}.\n\n:::\n\n\n\n\n## Algebra of limits\n\nProving convergence using Definition \\ref{definition-convergent-sequence} can be a tedious task. In this section we discuss how to prove convergence, starting from known convergence results.\n\n\n::: Theorem \n### Algebra of limits {#theorem-algebra-limits} \n\nLet $\\left(a_{n}\\right)_{n \\in \\mathbb{N}}$ and $\\left(b_{n}\\right)_{n \\in \\mathbb{N}}$ be sequences in $\\mathbb{R}$. Suppose that\n$$\n\\lim_{n \\rightarrow \\infty} a_{n}=a \\,, \\quad  \\lim_{n \\rightarrow \\infty} b_{n}=b \\,,\n$$\nfor some $a,b \\in \\R$. Then,\n\n1. Limit of sum is the sum of limits:\n$$\n\\lim_{n \\rightarrow \\infty}\\left(a_{n} \\pm b_{n}\\right)=a \\pm b\n$$\n\n2. Limit of product is the product of limits: \n$$\n\\lim _{n \\rightarrow \\infty}\\left(a_{n}  b_{n}\\right) = a  b \n$$\n\n3. If $b_{n} \\neq 0$ for all $n \\in \\mathbb{N}$ and $b \\neq 0$, then \n$$\n\\lim_{n \\rightarrow \\infty} \\left(\\frac{a_{n}}{b_{n}}\\right)=\\frac{a}{b} \n$$\n\n:::\n\n\n\n\n::: Proof \n\n\n Let $\\left(a_{n}\\right)_{n \\in \\mathbb{N}}$ and $\\left(b_{n}\\right)_{n \\in \\mathbb{N}}$ be sequences in $\\mathbb{R}$ and let $c \\in \\mathbb{R}$. Suppose that, for some $a, b \\in \\mathbb{R}$\n$$\n\\lim _{n \\rightarrow \\infty} a_{n}=a \\,, \\quad  \\lim _{n \\rightarrow \\infty} b_{n}=b \\,.\n$$\n\n\n*Proof of Point 1.* \n\nWe need to show that \n$$\n\\lim_{n \\to \\infty} (a_n \\pm  b_n) = a \\pm b \\,.\n$$\nWe only give a proof of the formula with $+$, since the case with $-$ follows with a very similar proof. Hence, we need to show that\n$$\n\\forall \\, \\e > 0 \\,, \\, \\exists  \\, N \\in \\N \\st \\forall \\, n \\geq N \\,, \\, \\left|(a_{n} + b_n) - (a+b)  \\right| < \\e \\,.\n$$\nLet $\\e>0$ and set\n$$\n\\widetilde{\\e} := \\frac{\\e}{2} \\,.\n$$\nSince $a_{n} \\rightarrow a, b_n \\to b$, and $\\widetilde{\\e}>0$, there exist $N_1, N_2 \\in \\N$ such that\n\\begin{align*}\n|a_n - a| & < \\widetilde{\\e} \\,, \\quad \\forall \\, n \\geq N_1 \\,, \\\\\n|b_n - b| & < \\widetilde{\\e} \\,, \\quad \\forall \\, n \\geq N_2 \\,.\n\\end{align*}\nDefine\n$$\nN : = \\max  \\{ N_1, N_2 \\} \\,.\n$$\nFor all $n \\geq N$ we have, by the triangle inequality,\n\\begin{align*}\n\\left|\\left(a_{n}+b_{n}\\right)-(a+b)\\right| \n& = \\left|\\left(a_{n}-a\\right) + \\left(b_{n}-b\\right)\\right| \\\\\n& \\leq \\left|a_{n}-a\\right|+\\left|b_{n}-b\\right| \\\\\n& < \\widetilde{\\e}+\\widetilde{\\e} \\\\\n& =\\e \\,.\n\\end{align*}\n\n\n\n*Proof of Point 2.*\n\n\nWe need to show that \n$$\n\\lim_{n \\to \\infty} (a_n  b_n) = a b  \\,,\n$$\nwhich is equivalent to \n$$\n\\forall \\, \\e > 0 \\,, \\, \\exists \\, N \\in \\N \\st  \\forall \\, n \\geq N \\,, \\, \\left|a_{n} b_{n} - a b \\right| < \\e \\,.\n$$\nLet $\\e>0$. The sequence $\\left(a_{n}\\right)$ converges, and hence is bounded, by Theorem \\ref{theorem-convergent-bounded}. This means there exists some $M>0$ such that \n$$\n\\left|a_{n}\\right| \\leq M \\,, \\quad \\forall \\, n \\in \\N \\,.\n$$\nDefine \n$$\n\\widetilde{\\e} = \\frac{\\e}{ M + |b| } \\,. \n$$\nSince $a_{n} \\rightarrow a, b_n \\to b$, and $\\widetilde{\\e} > 0$, there exist \n$N_1, N_2 \\in \\N$ such that\n\\begin{align*}\n| a_n  - a | & < \\widetilde{\\e} \\,, \\quad \\forall \\, n \\geq N_1 \\,. \\\\\n| b_n  - b | & < \\widetilde{\\e} \\,, \\quad \\forall \\, n \\geq N_2 \\,.\n\\end{align*}\nLet \n$$\nN:= \\max\\{ N_1, N_2 \\} \\,.\n$$\nFor all $n \\geq N$ we have\n\\begin{align*}\n\\left|a_{n}b_{n}-a b\\right| \n& = \\left|a_{n} b_{n} - a_n  b + a_n  b - a  b \\right| \\\\\n& \\leq \\left|a_{n}  b_{n} - a_n  b \\right| + \\left|a_n  b - a b \\right| \\\\\n& =\\left|a_{n}\\right| \\left|b_{n}-b\\right|+ |b|  \\left|a_{n}-a\\right| \\\\\n& < M \\, \\widetilde{\\e} +|b| \\, \\widetilde{\\e} \\\\\n& = (M + |b|) \\, \\widetilde{\\e} \\\\\n& = \\e  \\,.\n\\end{align*}\n\n\n*Proof of Point 3.* \n\nSuppose in addition that $b_n \\neq 0$ and $b \\neq 0$. We need to show that\n$$\n\\lim_{n \\to \\infty} \\, \\frac{a_n}{b_n} = \\frac{a}{b} \\,,\n$$\nwhich is equivalent to\n$$\n\\forall \\, \\e > 0 \\,, \\, \\exists \\, N \\in \\N \\st  \\forall \\, n \\geq N \\,, \\, \\left|\\frac{a_{n}}{ b_{n}} - \\frac{a}{b} \\right| < \\e \\,.\n$$\nWe suppose in addition that $b>0$. The proof is very similar for the case $b <0$, and is hence omitted. Let $\\e > 0$. Set\n$$\n\\delta := \\frac{b}{2} \\,.\n$$\nSince $b_n \\to b$ and $\\delta>0$, there exists $N_1 \\in \\N$ such that\n$$\n|b_n - b| < \\delta \\, \\quad \\forall \\, n \\geq N_1 \\,.\n$$\nIn particular we have\n$$\nb_n > b - \\delta = b - \\frac{b}{2} = \\frac{b}{2}   \\, \\quad \\forall \\, n \\geq N_1 \\,.\n$$\nDefine\n$$\n\\widetilde{\\e} := \\frac{ b^2 }{ 2 (b + |a|)} \\, \\e \\,.\n$$\nSince $\\widetilde{\\e}>0$ and $a_n \\to a, b_n \\to b$, there exist $N_2, N_3 \\in \\N$ such that\n\\begin{align*}\n|a_n - a | & < \\widetilde{\\e} \\,, \\quad \\forall \\, n \\geq N_2 \\,, \\\\\n|b_n - b | & < \\widetilde{\\e} \\,, \\quad \\forall \\, n \\geq N_3 \\,.\n\\end{align*}\nDefine\n$$\nN:= \\max\\{ N_1, N_2, N_3 \\} \\,.\n$$\nFor all $n \\geq N$ we have\n\\begin{align*}\n\\left|\\frac{a_{n}}{ b_{n}} - \\frac{a}{b} \\right| & = \n\\left|\\frac{a_{n} b - a b_n}{ b_{n} b }  \\right| \\\\\n& = \\frac{1}{ \\left|b_{n} b \\right|} \\,  \n\\left| a_{n}b - a b + a b - a b_n    \\right| \\\\\n& = \\frac{1}{ \\left|b_{n} b \\right|} \\,  \n\\left| (a_{n} - a) b + a(b-b_n)   \\right| \\\\\n& \\leq \\frac{1}{ \\left|b_{n} b \\right|} \\,  \n\\left( |a_{n} - a| |b| + |a| |b-b_n| \\right) \\\\\n& < \\frac{1}{ \\dfrac{b}{2} \\, b } \\, \\left(  \\widetilde{\\e} \\, b + \\widetilde{\\e} |a|   \\right) \\\\\n& = \\frac{2 (b + |a|) }{b^2} \\, \\widetilde{\\e} \\\\\n& = \\e \\,.\n\\end{align*}\n\n:::\n\n\n\nIn the future we will refer to Theorem \\ref{theorem-algebra-limits} as the *Algebra of Limits*. We now show how to use Theorem \\ref{theorem-algebra-limits} for computing certain limits. \n\n\n::: {.Example  #example-limit-ploynomial}\nProve that \n$$\n\\lim_{n \\to \\infty} \\, \\frac{3 n}{7 n+4} = \\frac{3}{7} \\,.\n$$\n\n\n> *Proof*. We can rewrite\n$$\n\\frac{3 n}{7 n+4}=\\frac{3}{7+\\dfrac{4}{n}}\n$$\nBy Theorem \\ref{theorem-constant-sequence} we know that\n$$\n3 \\rightarrow 3\\,,  \\quad 4 \\to 4 \\,, \\quad 7 \\to 7 \\,.\n$$\nFrom Theorem \\ref{theorem-one-over-n}  we know that \n$$\n\\frac{1}{n} \\rightarrow 0 \\,.\n$$\nHence, it follows from Theorem \\ref{theorem-algebra-limits} Point 2 that \n$$\n\\frac{4}{n} = 4 \\cdot \\frac1n \\rightarrow 4 \\cdot 0 = 0 \\,.\n$$\nBy Theorem \\ref{theorem-algebra-limits} Point 1 we have\n$$\n7 + \\frac{4}{n} \\rightarrow  7 + 0 = 7 \\,.\n$$\nFinally we can use Theorem \\ref{theorem-algebra-limits} Point 3 to infer\n$$\n\\frac{3 n}{7 n+4}=\\frac{3}{7+\\dfrac{4}{n}} \\rightarrow \\frac{3}{7} \\,.\n$$\n\n:::\n\n\n::: Important \n\nThe technique shown in Example \\ref{example-limit-ploynomial} is useful to compute limits of fractions of polynomials. To identify the possible limit, if it exists, it is often best to divide by the largest power of $n$ in the denominator.\n\n:::\n\n\n\n::: {.Example #example-algebra-of-limits-2}\nProve that\n$$\n\\lim_{n \\rightarrow \\infty} \\frac{n^{2}-1}{2 n^{2}-3} = \\frac{1}{2} \\,.\n$$\n\n> *Proof*. Factor $n^2$ to obtain\n$$\n\\frac{n^{2}-1}{2 n^{2}-3} = \\frac{1-\\dfrac{1}{n^{2}}}{2-\\dfrac{3}{n^{2}}} \\,.\n$$\nBy Theorem \\ref{theorem-one-over-np} we have\n$$\n\\frac{1}{n^2} \\to 0 \\,.\n$$\nWe can then use the Algebra of Limits Theorem \\ref{theorem-algebra-limits} Point 2 to infer\n$$\n\\frac{3}{n^2} \\to 3 \\cdot 0 = 0\n$$\nand Theorem \\ref{theorem-algebra-limits} Point 1 to get\n$$\n1 - \\frac{1}{n^2} \\to 1 - 0 = 1 \\,, \\quad \n2 - \\frac{3}{n^2} \\to 2 - 0 = 2 \\,.\n$$\nFinally we use Theorem \\ref{theorem-algebra-limits} Point 3 and conclude\n$$\n \\frac{1-\\dfrac{1}{n^{2}}}{2-\\dfrac{3}{n^{2}}} \\to \\frac{1}{2} \\,.\n$$\nTherefore\n$$\n\\lim_{n \\to \\infty } \\, \\frac{n^{2}-1}{2 n^{2}-3} = \\lim_{n \\to \\infty} \\, \\frac{1-\\dfrac{1}{n^{2}}}{2-\\dfrac{3}{n^{2}}} = \\frac{1}{2} \\,.\n$$\n\n:::\n\n\n\nWe can also use the Algebra of Limits to prove that certain limits do not exist.\n\n\n\n::: Example \n\nProve that the sequence\n$$\na_n = \\frac{4 n^{3}+8 n+1}{7 n^{2}+2 n+1}\n$$\ndoes not converge.\n\n> *Proof*. To show that the sequence $\\left(a_n\\right)$ does not converge, we divide by the largest power in the denominator, which in this case is $n^2$\n\\begin{align*}\na_n & = \\frac{4 n^{3}+8 n+1}{7 n^{2}+2 n+1} \\\\\n    & =\\frac{4 n+ \\dfrac{8}{n}  +  \\dfrac{1}{n^{2}} }{7+  \\dfrac{2}{n}  +  \\dfrac{1}{n^{2}} } \\\\\n    & = \\frac{b_n}{c_n}\n\\end{align*}\nwhere we set \n$$\nb_n := 4 n+ \\dfrac{8}{n}  +  \\dfrac{1}{n^{2}}\\,, \\quad\nc_n := 7 + \\dfrac{2}{n}  +  \\dfrac{1}{n^{2}} \\,.\n$$\nUsing the Algebra of Limits Theorem \\ref{theorem-algebra-limits} we see that\n$$\nc_n = 7+  \\dfrac{2}{n}  +  \\dfrac{1}{n^{2}}  \\to 7 \\,.\n$$\nSuppose by contradiction that\n$$\na_n \\to a\n$$\nfor some $a \\in \\R$. Then, by the Algebra of Limits Theorem \\ref{theorem-algebra-limits} we would infer\n$$\nb_n = c_n \\cdot a_n \\to 7 a \\,,\n$$\nconcluding that $b_n$ is convergent to $7a$. We have that\n$$\nb_n = 4n + d_n \\,, \\quad d_n := \\dfrac{8}{n}  +  \\dfrac{1}{n^{2}} \\,.\n$$\nAgain by the Algebra of Limits Theorem \\ref{theorem-algebra-limits} we get that \n$$\nd_n =  \\dfrac{8}{n}  +  \\dfrac{1}{n^{2}} \\to 0 \\,,\n$$\nand hence\n$$\n4n = b_n - d_n \\to 7a - 0 = 7a \\,.\n$$\nThis is a contradiction, since the sequence $(4n)$ is unbounded, and hence cannot be convergent. Hence $(a_n)$ is not convergent.\n:::\n\n\n\n::: Warning\n\nConsider the sequence \n$$\na_n = \\frac{4 n^{3}+8 n+1}{7 n^{2}+2 n+1}\n$$\nfrom the previous example. We have proven that $(a_n)$ is not convergent, by making use of the Algebra of Limits. \n\nLet us review a **faulty** argument to conclude that $(a_n)$ is not convergent. Write\n$$\na_n = \\frac{b_n}{c_n} \\,, \\quad b_n:= 4 n^{3}+8 n+1\\,, \n\\quad c_n : =7 n^{2}+2 n+1 \\,.\n$$\nThe numerator \n$$\nb_n = 4 n^{3}+8 n+1\n$$\nand denominator \n$$\nc_n  =7 n^{2}+2 n+1 \n$$\nare both unbounded, and hence $(b_n)$ and $(c_n)$ do not converge. One might be tempted to conclude that $(a_n)$ does not converge. However this is **false** in general: as seen in \nExample \\ref{example-algebra-of-limits-2}, we have\n$$\n\\lim_{n \\rightarrow \\infty} \\frac{n^{2}-1}{2 n^{2}-3} = \\frac{1}{2} \\,,\n$$\nwhile numerator and denominator are unbounded.\n\n:::\n\n\n\n\n\nSometimes it is useful to rearrange the terms of a sequence, before applying the Algebra of Limits.\n\n\n\n::: Example \n\nDefine \n$$\n a_n := \\frac{2 n^{3}+7 n+1}{5 n+9} \\cdot \\frac{8 n+9}{6 n^{3}+8 n^{2}+3} \\,.\n$$\nProve that \n$$\n\\lim_{n \\to \\infty}  a_n = \\frac{8}{15} \\,.\n$$\n\n\n> *Proof.* \nThe first fraction in $(a_n)$ does not converge, as it is unbounded. Therefore we cannot use Point 2 in Theorem \\ref{theorem-algebra-limits} directly. However, we note that\n\\begin{align*}\na_{n} & =\\frac{2 n^{3}+7 n+1}{5 n+9} \\cdot \\frac{8 n+9}{6 n^{3}+8 n^{2}+3} \\\\\n& = \\frac{8 n+9}{5 n+9} \\cdot \\frac{2 n^{3}+7 n+1}{6 n^{3}+8 n^{2}+3} \\,.\n\\end{align*}\nFactoring out $n$ and $n^3$, respectively, and using the Algebra of Limits, we see that\n$$\n\\frac{8 n+9}{5 n+9}=\\frac{8+9 / n}{5+9 / n} \\to \\frac{8+0}{5+0}=\\frac{8}{5}\n$$\nand\n$$\n\\frac{2+7 / n^{2}+1 / n^{3}}{6+8 / n+3 / n^{3}} \\to \\frac{2+0+0}{6+0+0}=\\frac{1}{3}\n$$\nTherefore Theorem \\ref{theorem-algebra-limits} Point 2 ensures that\n$$\na_{n} \\to \\frac{8}{5} \\cdot \\frac{1}{3}=\\frac{8}{15} \\,.\n$$\n\n:::\n\n\n\n\n\n## Fractional powers\n\n\nThe Algebra of Limits Theorem \\ref{theorem-algebra-limits} can also be used when fractional powers of $n$ are involved.\n\n::: Example\n\nProve that \n$$\na_n = \\frac{n^{7 / 3}+2 \\sqrt{n}+7}{4 n^{3 / 2}+5 n}\n$$\ndoes not converge.\n\n> *Proof*. The largest power of $n$ in the denominator is $n^{3/2}$. Hence we factor out $n^{3/2}$\n\\begin{align*}\na_n & = \\frac{n^{7 / 3}+2 \\sqrt{n}+7}{4 n^{3 / 2}+5 n} \\\\\n    & = \\frac{n^{7 / 3-3 / 2}+2 n^{1/2 - 3/2} + 7  n^{-3 / 2}}{4 + 5 n^{-3/2} } \\\\\n    & = \\frac{n^{5/6}+2 n^{- 1} + 7  n^{-3 / 2}}{4 + 5 n^{-3/2} } \\\\\n    & = \\frac{b_n}{c_n}\n\\end{align*}\nwhere we set\n$$\nb_n := n^{5/6}+2 n^{- 1} + 7  n^{-3 / 2} \\,, \\quad \nc_n := 4 + 5 n^{-3/2} \\,.\n$$\nWe see that $b_n$ is unbounded while $c_n \\to 4$. By the Algebra of Limits (and usual contradiction argument) we conclude that $(a_n)$ is divergent.\n\n:::\n\n\n\nWe now present a general result about the square root of a sequence.\n\n::: {.Theorem #theorem-square-root-limit}\n\nLet $\\left(a_{n}\\right)_{n \\in \\mathbb{N}}$ be a sequence in $\\mathbb{R}$ such that \n$$\n\\lim_{n \\to \\infty} \\, a_n = a \\,,\n$$\nfor some $a \\in \\R$. If $a_{n} \\geq 0$ for all $n \\in \\mathbb{N}$ and $a \\geq 0$, then\n$$\n\\lim _{n \\rightarrow \\infty} \\sqrt{a_{n}}=\\sqrt{a} \\,.\n$$\n\n:::\n\n\n::: Proof \n\nLet $\\e>0$. We the two cases $a>0$ and $a=0$:\n\n- $a>0$: Define\n$$\n\\delta := \\frac{a}{2} \\,.\n$$\nSince $\\delta > 0$ and $a_n \\to a$, there exists $N_1 \\in \\N$ such that\n$$\n\\left|a_{n}-a\\right| < \\delta \\,, \\quad \\forall \\, n \\geq N_1 \\,. \n$$\nIn particular\n$$\na_n > a - \\delta = a - \\frac{a}{2} = \\frac{a}{2} \\,, \\quad \\forall \\, n \\geq N_1 \\,,\n$$\nfrom which we infer\n$$\n\\sqrt{a_n} > \\sqrt{a/2} \\,, \\quad \\forall \\, n \\geq N_1 \\,,\n$$\nNow set\n$$\n\\widetilde{\\e} := \\left(\\sqrt{a/2} + \\sqrt{a} \\right) \\, \\e \\,.\n$$\nSince $\\widetilde{\\e}>0$ and $a_n \\to a$, there exists $N_2 \\in \\mathbb{N}$ such that \n$$\n\\left|a_{n}-a\\right| < \\widetilde{\\e} \\,, \\quad \\forall \\, n \\geq N_2 \\,. \n$$\nLet \n$$\nN:= \\max\\{  N_1, N_2 \\} \\,.\n$$\nFor $n \\geq N$ we have\n\\begin{align*}\n\\left|\\sqrt{a_{n}}-\\sqrt{a}\\right| \n& = \\left|\\frac{\\left(\\sqrt{a_{n}}-\\sqrt{a}\\right)\\left(\\sqrt{a_{n}}+\\sqrt{a}\\right)}{\\sqrt{a_{n}}+\\sqrt{a}}\\right| \\\\\n& = \\frac{\\left|a_{n}-a\\right|}{\\sqrt{a_{n}}+\\sqrt{a}} \\\\\n& < \\frac{ \\widetilde{\\e} }{\\sqrt{a/2} + \\sqrt{a}}  \\\\\n& = \\e \\,.\n\\end{align*}\n\n\n- $a=0$: In this case\n$$\na_n \\to a = 0 \\,.\n$$\nSince $\\e^2>0$, there exists $N \\in \\N$ such that\n$$\n|a_n - 0 | = |a_n| < \\e^2 \\,, \\quad \\forall \\, n \\geq N \\,.\n$$\nTherefore\n$$\n|\\sqrt{a_n} - \\sqrt{0} | = | \\sqrt{a_n}| < \\sqrt{\\e^2} = \\e \\,, \\quad \\forall \\, n \\geq N \\,.\n$$\n\n\n:::\n\n\nLet us show an application of Theorem \\ref{theorem-square-root-limit}.\n\n\n::: Example \n\nDefine the sequence\n$$\na_n  = \\sqrt{9 n^{2}+3 n+1}-3 n \\,.\n$$\nProve that\n$$\n\\lim_{n \\to \\infty} \\, a_n = \\frac12 \\,.\n$$\n\n\n> *Proof*. We first rewrite\n\\begin{align*}\na_n & = \\sqrt{9 n^{2}+3 n+1}-3 n \\\\\n& = \\frac{\\left(\\sqrt{9 n^{2}+3 n+1}-3 n\\right)\\left(\\sqrt{9 n^{2}+3 n+1}+3 n\\right)}{\\sqrt{9 n^{2}+3 n+1}+3 n} \\\\\n& =\\frac{9 n^{2}+3 n+1-(3 n)^{2}}{\\sqrt{9 n^{2}+3 n+1}+3 n} \\\\\n& =\\frac{3 n+1}{\\sqrt{9 n^{2}+3 n+1}+3 n} \\, .\n\\end{align*}\nThe biggest power of $n$ in the denominator is $n$. Therefore we factor out $n$:\n\\begin{align*}\na_n & = \\sqrt{9 n^{2}+3 n+1}-3 n \\\\\n& =\\frac{3 n+1}{\\sqrt{9 n^{2}+3 n+1}+3 n} \\\\\n& =\\frac{3+ \\dfrac{1}{n} }{\\sqrt{9+ \\dfrac{3}{n} + \\dfrac{1}{n^{2}}} + 3 } \\,.\n\\end{align*}\nBy the Algebra of Limits we have\n$$\n9+ \\frac{3}{n} + \\frac{1}{n^{2}} \\to 9 + 0 + 0 = 9 \\,.\n$$\nTherefore we can use Theorem \\ref{theorem-square-root-limit} to infer\n$$\n\\sqrt{ 9 + \\frac{3}{n} + \\frac{1}{n^{2}} } \\to \\sqrt{9}  \\,.\n$$\nBy the Algebra of Limits we conclude:\n$$\na_n = \\frac{3+ \\dfrac{1}{n} }{\\sqrt{9+ \\dfrac{3}{n}  + \\dfrac{1}{n^{2}} }+ 3 }   \\to  \\frac{ 3 + 0 }{ \\sqrt{9} + 3 } = \\frac12 \\,.\n$$\n\n\n:::\n\n\n::: Example \n\nProve that the sequence\n$$\na_n = \\sqrt{9 n^{2}+3 n+1}-2 n\n$$\ndoes not converge.\n\n> *Proof.* We rewrite $a_n$ as\n\\begin{align*}\na_n & = \\sqrt{9 n^{2}+3 n+1}-2 n  \\\\\n& =\\frac{  (\\sqrt{9 n^{2}+3 n+1} - 2 n)  (\\sqrt{9 n^{2}+3 n+1}+2 n) }{\\sqrt{9 n^{2}+3 n+1}+2 n} \\\\\n& =\\frac{9 n^{2}+3 n+1-(2 n)^{2}}{\\sqrt{9 n^{2}+3 n+1}+2 n} \\\\\n& =\\frac{5 n^{2}+3 n+1}{\\sqrt{9 n^{2}+3 n+1}+2 n} \\\\\n& =\\frac{5 n + 3 +  \\dfrac{1}{n} }{\\sqrt{9+ \\dfrac{3}{n}  + \\dfrac{1}{ n^2 } } + 2 } \\\\\n& = \\frac{b_n}{c_n} \\,,\n\\end{align*}\nwhere we factored $n$, being it the largest power of $n$ in the denominator, and defined\n$$\nb_n : = 5 n + 3 +  \\dfrac{1}{n}\\,, \\quad \nc_n := \\sqrt{9+ \\dfrac{3}{n}  + \\dfrac{1}{ n^2 } } + 2 \\,.\n$$\nNote that \n$$\n9+ \\dfrac{3}{n}  + \\dfrac{1}{ n^2 }  \\to 9\n$$\nby the Algebra of Limits. Therefore \n$$\n\\sqrt{9+ \\dfrac{3}{n}  + \\dfrac{1}{ n^2 }}  \\to \\sqrt{9} = 3 \n$$\nby Theorem \\ref{theorem-square-root-limit}. Hence \n$$\nc_n = \\sqrt{9+ \\dfrac{3}{n}  + \\dfrac{1}{ n^2 }} + 2 \\to 3 + 2 = 5  \\,.\n$$\nThe numerator \n$$\nb_n = 5 n + 3 +  \\dfrac{1}{n}\n$$\nis instead unbounded. Therefore $(a_n)$ is not convergent, by the Algebra of Limits and the usual contradiction argument.\n:::\n\n\n\n\n\n\n\n## Limit Tests\n\n\nIn this section we discuss a number of *Tests* to determine whether a sequence converges or not. These are known as **Limit Tests**.\n\n\n### Squeeze Theorem\n\n\nWhen a sequence $(a_n)$ oscillates, it is difficult to compute the limit. Examples of terms which produce oscillations are\n$$\n(-1)^n \\,, \\quad  \\sin(n) \\,, \\quad \\cos(n) \\,. \n$$\nIn such instance it might be useful to compare $(a_n)$ with other sequences whose limit is known. If we can prove that $(a_n)$ is *squeezed* between two other sequences with the same limiting value, then we can show that also $(a_n)$ converges to this value.\n\n\n\n::: Theorem \n### Squeeze theorem {#theorem-squeeze}\n\nLet $\\left(a_{n}\\right), \\left(b_{n}\\right)$ and $\\left(c_{n}\\right)$ be sequences in $\\R$. Suppose that\n$$\nb_{n} \\leq a_{n} \\leq c_{n} \\,, \\quad \\forall  \\, n \\in \\N \\,,\n$$\nand that\n$$\n\\lim_{n \\rightarrow \\infty} b_{n} = \\lim_{n \\rightarrow \\infty} c_{n} = L \\, .\n$$\nThen\n$$\n\\lim_{n \\rightarrow \\infty} a_{n}= L \\, .\n$$\n\n:::\n\n\n::: Proof\n\nLet $\\e>0$. Since $b_{n} \\to L$ and $c_n \\to L$ , there exist $N_1, N_2 \\in \\N$ such that \n\\begin{align*}\n-\\e < b_n - L < \\e \\,, \\,\\, \\forall \\, n \\geq N_1 \\,,  \\\\\n- \\e < c_n - L < \\e \\,, \\,\\, \\forall \\, n \\geq N_2 \\,. \n\\end{align*}\nSet\n$$\nN := \\max\\{N_1,N_2\\} \\,.\n$$\nLet $n \\geq N$. Using the assumption that $b_n \\leq a_{n} \\leq c_{n}$, we get\n$$\nb_n - L  \\leq a_{n} - L \\leq c_{n} - L \\,.\n$$\nIn particular\n$$\n- \\e < b_n - L  \\leq a_n - L \\leq b_n - L < \\e \\,.\n$$\nThe above implies \n$$\n- \\e < a_n - L < \\e  \\quad \\implies \\quad \\left|a_{n}-L\\right| < \\e \\,.\n$$\n\n:::\n\n\n\n::: Example \n\nProve that\n$$\n\\lim_{n \\to \\infty} \\, \\frac{(-1)^{n}}{n} = 0 \\,.\n$$\n\n> *Proof.*\nFor all $n \\in \\N$ we can estimate\n$$\n-1 \\leq(-1)^{n} \\leq 1 \\,.\n$$\nTherefore\n$$\n\\frac{-1}{n} \\leq \\frac{(-1)^{n}}{n} \\leq \\frac{1}{n} \\,, \\quad \\forall \\, n \\in \\N \\,.\n$$\nMoreover\n$$\n\\lim_{n \\to \\infty} \\frac{-1}{n}= -1 \\cdot 0=0 \\,, \\quad \n\\lim_{n \\to \\infty} \\frac{1}{n}=0 \\,.\n$$\nBy the Squeeze Theorem \\ref{theorem-squeeze} we conclude\n$$\n\\lim_{n \\to \\infty} \\frac{(-1)^{n}}{n}=0 \\,.\n$$\n\n:::\n\n\n\n::: {.Example #example-squeeze}\n\nProve that \n$$\n\\lim_{n \\to \\infty} \\frac{\\cos (3 n) + 9 n^{2}}{11 n^{2}+15 \\sin (17 n)} = \\frac{9}{11} \\,.\n$$\n\n\n> *Proof.* \nWe know that \n$$\n-1 \\leq \\cos(x) \\leq 1 \\,, \\quad - 1 \\leq \\sin(x) \\leq 1 \\,, \\quad \\forall \\, x \\in \\R \\,.\n$$\nTherefore, for all $n \\in \\N$\n$$\n- 1 \\leq \\cos(3n) \\leq 1 \\,, \\quad -1 \\leq \\sin(17n) \\leq 1 \\,.\n$$\nWe can use the above to estimate the numerator in the given sequence:\n$$\n-1 + 9 n^{2} \\leq \\cos (3 n)+9 n^{2} \\leq 1+ 9 n^{2} \\,.\n$${#eq-squeeze-example-1}\nConcerning the denominator, we have\n$$\n11 n^{2}-15 \\leq 11 n^{2}+15 \\sin (17 n) \\leq  11 n^{2} + 15\n$$\nand therefore\n$$\n\\frac{1}{11 n^{2} + 15} \\leq \\frac{1}{11 n^{2}+15 \\sin (17 n)} \\leq \\frac{1}{11 n^{2}-15} \\,.\n$${#eq-squeeze-example-2}\nPutting together (@eq-squeeze-example-1)-(@eq-squeeze-example-2) we obtain \n$$\n\\frac{-1 + 9 n^{2}}{11 n^{2} + 15} \\leq \\frac{\\cos (3 n)+9 n^{2}}{11 n^{2}+15 \\sin (17 n)} \\leq \\frac{1+ 9 n^{2}}{11 n^{2}-15} \\,.\n$$\nBy the Algebra of Limits we infer\n$$\n\\frac{-1+9 n^{2}}{11 n^{2}+15}=\\frac{-\\dfrac{1}{n^{2}} + 9}{11 + \\dfrac{15}{n^{2}}} \\to \\frac{0+9}{11+0}=\\frac{9}{11}\n$$\nand\n$$\n\\frac{1+9 n^{2}}{11 n^{2} - 15}=\\frac{ \\dfrac{1}{n^{2}} + 9}{ 11 - \\dfrac{15}{n^{2}}} \\to \\frac{0+9}{11+0}=\\frac{9}{11} \\,.\n$$\nApplying the Squeeze Theorem \\ref{theorem-squeeze} we conclude\n$$\n\\lim_{n \\to \\infty} \\frac{\\cos (3 n)+ 9 n^{2}}{11 n^{2}+15 \\sin (17 n)}=\\frac{9}{11} \\,.\n$$\n\n:::\n \n::: Warning\n\nSuppose that the sequences $(a_n), (b_n), (c_n)$ satisfy\n$$\nb_n \\leq a_n \\leq c_n \\,, \\quad \\forall n \\in \\N \\,,\n$$\nand\n$$\nb_n \\to L_1 \\,, \\quad c_n \\to L_2 \\,, \\quad L_1 \\neq L_2 \\,.\n$$\nIn general, we cannot conclude that $a_n$ converges.\n\n:::\n\n\n::: Example \n\nConsider the sequence\n$$\na_n = \\left( 1 + \\frac{1}{n} \\right) (-1)^n  \\,.\n$$\nFor all $n \\in \\N$ we can bound\n$$\n- 1 - \\frac{1}{n} \\leq \\left( 1 + \\frac{1}{n} \\right) (-1)^n   \\leq 1 + \\frac{1}{n}  \\,.\n$$\nHowever \n$$\n- 1 - \\frac{1}{n} \\longrightarrow - 1 - 0 = -1 \n$$\nand\n$$\n1 + \\frac{1}{n}  \\longrightarrow  1 + 0 = 1 \\,.\n$$\nSince \n$$\n- 1 \\neq 1 \\,,\n$$\nwe cannot apply the Squeeze Theorem \\ref{theorem-squeeze} to conclude convergence of $(a_n)$. Indeed, $(a_n)$ is a divergent sequence.\n\n> *Proof.* Suppose by contradiction that $a_n \\to a$. We have\n$$\na_n = (-1)^n + \\frac{(-1)^n}{n} = b_n + c_n\n$$\nwhere\n$$\nb_n := (-1)^n \\,, \\quad c_n := \\frac{(-1)^n}{n} \\,.\n$$\nWe have seen in Example \\ref{example-squeeze} that $c_n \\to 0$. Therefore, \nby the Algebra of Limits, we have\n$$\nb_n = a_n - c_n \\longrightarrow a - 0 = a \\,.\n$$\nHowever, Theorem \\ref{theorem-1-minus-n} says that the sequence\n$b_n = (-1)^n$ diverges. Contradiction. Hence $(a_n)$ diverges.\n\n:::\n\n\n\n\n\n### Geometric sequences\n\n\n::: Definition \n\nA sequence $\\left(a_{n}\\right)$ is called a **geometric sequence** if\n$$\na_{n}=x^{n} \\,,\n$$\nfor some $x \\in \\R$.\n\n:::\n\n\nThe value of $|x|$ determines whether or not a geometric sequence converges, as shown in the following theorem.\n\n\n::: Theorem \n### Geometric Sequence Test   {#theorem-geometric-sequence}\n\nLet $x \\in \\R$ and let $\\left(a_{n}\\right)$ be the sequence defined by\n$$\na_{n}:=x^{n} \\,.\n$$\nWe have:\n\n1. If $|x|<1$, then \n$$\n\\lim_{n \\to \\infty} a_{n} = 0 \\,.\n$$\n\n2. If $|x|>1$, then sequence $\\left(a_{n}\\right)$ is unbounded, and hence divergent.\n\n:::\n\n\n\n::: Warning\n\nThe Geometric Sequence Test in Theorem \\ref{theorem-geometric-sequence} does not address the case\n$$\n|x|=1 \\,.\n$$\nThis is because, in this case, the sequence \n$$\na_n = x^n \n$$\nmight converge or diverge, depending on the value of $x$. Indeed, \n$$\n|x| = 1 \\quad \\implies \\quad x = \\pm 1 \\,.\n$$\nWe therefore have two cases:\n\n- $x = 1$: Then \n$$\na_n = 1^n = 1\n$$\nso that $a_n \\to 1$ and $(a_n)$ is convergent.\n\n- $x=-1$: Then\n$$\na_n = x^n = (-1)^n\n$$\nwhich is divergent by Theorem \\ref{theorem-1-minus-n}.\n\n:::\n\n\n\nTo prove Theorem \\ref{theorem-geometric-sequence} we need the following inequality, known as Bernoulli's inequality.\n\n\n::: Lemma \n### Bernoulli's inequality  {#lemma-bernoulli}\n\nLet $x \\in \\R$ with $x>-1$. Then\n$$\n(1+x)^{n} \\geq 1+n x \\,, \\quad \\forall \\, n \\in \\N \\,.\n$${#eq-bernoulli-inequality}\n\n:::\n\n\n::: Proof\n\nLet $x \\in \\mathbb{R}, x>-1$. We prove the statement by induction:\n\n- Base case: (@eq-bernoulli-inequality) holds with equality when $n=1$.\n\n- Induction hypothesis: Let $k \\in \\N$ and suppose that (@eq-bernoulli-inequality) holds for $n=k$, i.e.,\n$$\n(1+x)^{k} \\geq 1+k x \\,.\n$$\nThen\n\\begin{align*}\n(1+x)^{k+1} & = (1+x)^{k}(1+x) \\\\\n            & \\geq(1+k x)(1+x) \\\\\n            & =1+k x+x+k x^{2} \\\\\n            & \\geq 1+(k+1) x \\,,\n\\end{align*}\nwhere we used that $kx^2 \\geq 0$. Then (@eq-bernoulli-inequality) holds for $n=k+1$.\n\nBy induction we conclude (@eq-bernoulli-inequality).\n\n:::\n\n\nWe are ready to prove Theorem \\ref{theorem-geometric-sequence}.\n\n\n::: Proof \n### Proof of Theorem \\ref{theorem-geometric-sequence}\n\n*Part 1. The case $|x|<1$*. \n\nIf $x=0$, then\n$$\na_n = x^n = 0 \n$$\nso that $a_n \\to 0$. Hence assume $x \\neq 0$. We need to prove that\n$$\n\\forall \\, \\e> 0 \\,, \\, \\exists \\, N \\in \\N \\st \n\\forall \\, n \\geq N \\,, \\,\\, |x^n - 0| < \\e \\,.\n$$\nLet $\\e>0$. We have \n$$\n|x| < 1 \\quad \\implies \\quad \\frac{1}{|x|} > 1 \\,.\n$$\nTherefore \n$$\n|x|= \\frac{1}{1+u} \\,, \\quad u:=\\frac{1}{|x|} - 1 > 0 \\,.\n$$\nLet $N \\in \\N$ be such that \n$$\nN > \\frac{1}{\\e u} \\,,\n$$\nso that \n$$\n\\frac{1}{N u} < \\e \\,.\n$$\nLet $n \\geq N$. Then\n\\begin{align*}\n\\left|x^{n}-0\\right| & =|x|^{n} \\\\\n                     & = \\left(\\frac{1}{1+u}\\right)^{n} \\\\\n                     & =\\frac{1}{(1+u)^{n}} \\\\\n                     & \\leq \\frac{1}{1+n u} \\\\\n                     & \\leq \\frac{1}{n u} \\\\\n                     & \\leq \\frac{1}{N u} \\\\\n                     & < \\e \\,,\n\\end{align*}\nwhere we used Bernoulli's inequality (@eq-bernoulli-inequality) in the first inequality.\n\n\n*Part 2. The case $|x|>1$*. \n\nTo prove that (a_n) does not converge, we prove that it is unbounded. \nThis means showing that\n$$\n\\forall \\, M > 0 \\,, \\, \\exists n \\in \\N \\st  \\left| a_{n}\\right| >M \\,.\n$$\nLet $M > 0$. We have two cases: \n\n- $0 < M \\leq 1$: Choose $n=1$. Then\n$$\n\\left|a_{1}\\right|=|x|>1 \\geq M \\,.\n$$\n\n- $M>1$: Choose $n \\in \\N$ such that \n$$\nn> \\frac{\\log M}{\\log |x|} \\,.\n$$ \nNote that $\\log |x|>0$ since $|x|>1$. Therefore\n\\begin{align*}\nn>\\frac{\\log M}{\\log |x|} & \\iff n \\log |x|>\\log M \\\\\n                          & \\iff \\log |x|^n>\\log M \\\\\n                          & \\iff |x|^{n}>M \\,.\n\\end{align*}\nThen \n$$\n\\left|a_{n}\\right|=\\left|x^{n}\\right|=|x|^{n} > M \\,.\n$$\n\n\nHence $(a_n)$ is unbounded. By Corollary \\ref{corollary-convergent-bounded} we conclude that $(a_n)$ is divergent. \n\n\n:::\n\n\n\n\n\n\n::: Example \n\nWe can apply Theorem \\ref{theorem-geometric-sequence} to prove convergence\nor divergence for the following sequences.\n\n1. We have\n$$\n\\left(\\frac{1}{2}\\right)^{n} \\longrightarrow 0\n$$\nsince \n$$\n\\left|\\frac{1}{2}\\right|=\\frac{1}{2}<1 \\,.\n$$\n\n2. We have\n$$\n\\left(\\frac{-1}{2}\\right)^{n} \\longrightarrow 0\n$$\nsince \n$$\n\\left|\\frac{-1}{2}\\right|=\\frac{1}{2}<1 \\,.\n$$\n\n3. The sequence \n$$\na_n = \\left(\\frac{-3}{2}\\right)^{n}\n$$ \ndoes not converge, since \n$$\n\\left|\\frac{-3}{2}\\right|=\\frac{3}{2}>1 \\,.\n$$\n\n4. As $n \\rightarrow \\infty$,\n$$\n\\frac{3^{n}}{(-5)^{n}}=\\left(-\\frac{3}{5}\\right)^{n} \\longrightarrow 0\n$$\nsince \n$$\n\\left|-\\frac{3}{5}\\right|=\\frac{3}{5}<1 \\,.\n$$\n\n5. The sequence \n$$\na_{n}=\\frac{(-7)^{n}}{2^{2 n}}\n$$ \ndoes not converge, since\n$$\n\\frac{(-7)^{n}}{2^{2 n}}=\\frac{(-7)^{n}}{\\left(2^{2}\\right)^{n}}=\\left(-\\frac{7}{4}\\right)^{n}\n$$\nand \n$$\n\\left|-\\frac{7}{4}\\right|=\\frac{7}{4}>1  \\,.\n$$\n\n:::\n\n\n\n\n\n\n\n### Ratio Test\n\n\n\n::: Theorem \n### Ratio Test {#theorem-ratio-test}\n\nLet $\\left(a_{n}\\right)$ be a sequence in $\\R$ such that\n$$\na_{n} \\neq 0 \\,, \\quad \\forall \\, n \\in \\N \\,.\n$$ \n\n\n1. Suppose that the following limit exists:\n$$\nL:=\\lim_{n \\to \\infty}\\left|\\frac{a_{n+1}}{a_{n}}\\right| \\,.\n$$\nThen,\n\n    - If $L<1$ we have\n    $$\n    \\lim_{n \\to\\infty} a_{n}=0 \\,.\n    $$\n\n    - If $L>1$, the sequence $\\left(a_{n}\\right)$ is unbounded, and hence does not converge.\n\n2. Suppose that there exists $N \\in \\N$ and $L>1$ such that\n$$\n\\left|\\frac{a_{n+1}}{a_{n}}\\right| \\geq L \\,, \\quad \\forall \\, n \\geq N \\,. \n$$\nThen the sequence $\\left(a_{n}\\right)$ is unbounded, and hence does not converge.\n\n:::\n\n\n\n::: Proof\n\nDefine the sequence $b_{n}=\\left|a_{n}\\right|$. Then,\n\n$$\n\\left|\\frac{a_{n+1}}{a_{n}}\\right|=\\frac{\\left|a_{n+1}\\right|}{\\left|a_{n}\\right|}=\\frac{b_{n+1}}{b_{n}}\n$$\n\n\n*Part 1.* Suppose that there exists the limit\n$$\nL:=\\lim_{n \\to \\infty}\\left|\\frac{a_{n+1}}{a_{n}}\\right| \\,.\n$$\nTherefore \n$$\n\\lim_{n \\to \\infty} \\frac{b_{n+1}}{b_{n}} = L \\,.\n$${#eq-ratio-test-proof}\n\n- $L<1$: Choose $r>0$ such that \n$$\nL<r<1 \\,.\n$$\nSet \n$$\n\\e := r-L\n$$\nBy the convergence at (@eq-ratio-test-proof) there exists $N \\in \\N$ such that\n$$\n\\left|\\frac{b_{n+1}}{b_{n}}-L\\right| < \\e = r - L \\,, \\quad \\forall \\, n \\geq N \\,.\n$$\nIn particular \n$$\n\\frac{b_{n+1}}{b_{n}}-L < r-L \\,, \\quad \\forall \\, n \\geq N \\,,\n$$\nwhich implies \n$$\nb_{n+1} <  r \\,{b_{n}} \\,, \\quad \\forall \\, n \\geq N \\,.\n$${#eq-ratio-test-proof-1}\nLet $n \\geq N$, we can use (@eq-ratio-test-proof-1) recursively and obtain\n$$\n0 \\leq b_{n} < \\, r b_{n-1} < \\ldots < r^{n-N} \\, b_{N} = r^{n} \\, \\frac{b_{N}}{r^{N}} \\,.\n$$\nIn particular, we have proven that\n$$\n0 \\leq b_{n} < r^{n} \\, \\frac{b_{N}}{r^{N}} \\,, \\quad \\forall \\, n \\in \\N \\,.\n$${#eq-ratio-test-proof-2}\nSince $|r|<1$, by the Geometric Sequence Test Theorem \\ref{theorem-geometric-sequence} we infer\n$$\nr^{n} \\rightarrow 0 \\,.\n$$\nThe Algebra of Limits the yields\n$$\nr^{n} \\, \\frac{b_{N}}{r^{N}} \\rightarrow 0 \\,.\n$$\nBy the Squeeze Theorem \\ref{theorem-squeeze} applied to (@eq-ratio-test-proof-2), it follows that \n$$   \nb_{n} = |a_n| \\to 0 \\,.\n$$\nSince \n$$\n-\\left|a_{n}\\right| \\leq a_{n} \\leq\\left|a_{n}\\right| \\,,\n$$\nand \n$$\n-|a_n| \\to 0 \\,, \\quad |a_n| \\to 0 \\,, \n$$\nwe can again apply the Squeeze Theorem \\ref{theorem-squeeze} to infer\n$$\na_{n} \\rightarrow 0 \\,.\n$$\n\n\n- $L>1$: Choose $r>0$ such that \n$$\n1<r<L  \\,.\n$$\nDefine\n$$\n\\e := L - r > 0 \\,.\n$$\nBy the convergence (@eq-ratio-test-proof), there exists $N \\in \\N$ such that\n$$\n\\left|\\frac{b_{n+1}}{b_{n}}-L\\right| < \\e = L - r \\,, \\quad \\forall \\, n \\geq N \\,.\n$$\nIn particular,\n$$\n-(L-r) < \\frac{b_{n+1}}{b_{n}}-L \\,, \\quad \\forall \\, n \\geq N \\,,\n$$\nwhich implies\n$$\nb_{n+1} > r \\, b_{n} \\,, \\quad  \\forall \\, n \\geq N \\,.\n$${#eq-ratio-test-proof-4}\nLet $n \\geq N$. Applying (@eq-ratio-test-proof-4) recursively we get\n$$\nb_{n} > r^{n-N} \\, b_{N} = r^{n} \\, \\frac{b_{N}}{r^{N}} \\,, \\quad \\forall \\, n \\geq N \\,.\n$${#eq-ratio-test-proof-5}\nSince $|r|>1$, by the Geometric Sequence Test we have that the sequence\n$$\n(r^n)\n$$\nis unbounded. Therefore also the right hand side of (@eq-ratio-test-proof-5) is unbounded, proving that $(b_n)$ is unbounded. Since\n$$\nb_n = |a_n|\\,,\n$$\nwe conclude that $(a_n)$ is unbounded. By Corollary \\ref{corollary-convergent-bounded} we conclude that $(a_n)$ does not converge.\n\n\n*Part 2.* Suppose that there exists $N \\in \\N$ and $M>1$ such that\n$$\n\\left|\\frac{a_{n+1}}{a_{n}}\\right| \\geq M \\,, \\quad \\forall \\, n \\geq N \\,. \n$$\nSince $b_n = |a_n|$, we infer\n$$\nb_{n+1} \\geq L \\, b_n \\,, \\quad \\forall \\, n \\geq N \\,. \n$$\nArguing as above, we obtain\n$$\nb_n \\geq L^n \\, \\frac{b_N}{L^N} \\,, \\quad \\forall \\, n \\geq N \\,.\n$$\nSince $L>1$, we have that the sequence \n$$  \nL^n \\, \\frac{b_N}{L^N}\n$$\nis unbounded, by the Geometric Sequence Test. Hence also $(b_n)$ is unbounded, from which we conclude that $(a_n)$ is unbounded. By Corollary \\ref{corollary-convergent-bounded} we conclude that $(a_n)$ does not converge.\n\n\n:::\n\n\nLet us apply the Ratio Test to some concrete examples.\n\n\n\n::: Example\n\n\nLet \n$$\na_{n}=\\frac{3^{n}}{n !} \\,,\n$$\nwhere we recall that $n!$ (pronounced $n$ factorial) is defined by\n$$\nn! := n \\cdot (n-1) \\cdot (n-2) \\cdot  \\ldots \\cdot  3 \\cdot 2 \\cdot 1 \\,. \n$$\nProve that\n$$\n\\lim_{n \\to \\infty} a_n  = 0 \\,.\n$$\n\n> *Proof*. We have\n\\begin{align*}\n\\left|\\frac{a_{n+1}}{a_{n}}\\right| & = \\dfrac{\\left( \\dfrac{3^{n+1}}{(n+1) !} \\right) }{  \\left( \\dfrac{3^{n}}{n !} \\right) } \\\\\n& = \\frac{3^{n+1}}{3^{n}} \\, \\frac{n !}{(n+1) !} \\\\\n& = \\frac{3 \\cdot 3^n}{3^n} \\, \\frac{n!}{(n+1) n!} \\\\\n& =\\frac{3}{n+1} \\longrightarrow L = 0 \\,.\n\\end{align*}\nHence, $L=0<1$ so $a_{n} \\to 0$ by the Ratio Test in Theorem \\ref{theorem-ratio-test}.\n\n:::\n\n\n::: Example\n\nConsider the sequence\n$$\na_{n}=\\frac{n ! \\cdot 3^{n}}{\\sqrt{(2 n) !}} \\,.\n$$\nProve that $(a_n)$ is divergent.\n\n> *Proof.* We have\n\\begin{align*}\n\\left|\\frac{a_{n+1}}{a_{n}}\\right| & =\\frac{(n+1) ! \\cdot 3^{n+1}}{\\sqrt{(2(n+1)) !}}  \\frac{\\sqrt{(2 n) !}}{n ! \\cdot 3^{n}} \\\\\n& =\\frac{(n+1) !}{n !} \\cdot \\frac{3^{n+1}}{3^{n}} \\cdot \\frac{\\sqrt{(2 n) !}}{\\sqrt{(2(n+1)) !}}\n\\end{align*}\nFor the first two fractions we have\n$$\n\\frac{(n+1) !}{n !} \\cdot \\frac{3^{n+1}}{3^{n}} = 3(n+1) \\,,\n$$\nwhile for the third fraction\n\\begin{align*}\n\\frac{\\sqrt{(2 n) !}}{\\sqrt{(2(n+1)) !}} & =\\sqrt{\\frac{(2 n) !}{(2 n+2) !}} \\\\\n& = \\sqrt{\\frac{ (2n)! }{ (2n+2) \\cdot (2n+1) \\cdot (2n)! }} \\\\\n& = \\frac{1}{\\sqrt{(2 n+1)(2 n+2)}} \\,.\n\\end{align*}\nTherefore, using the Algebra of Limits,\n\\begin{align*}\n\\left|\\frac{a_{n+1}}{a_{n}}\\right| & = \\frac{3(n+1)}{\\sqrt{(2 n+1)(2 n+2)}}\\\\\n& = \\frac{3n \\left(1+ \\dfrac{1}{n} \\right)}{\\sqrt{n^2 \\left( 2 + \\dfrac{1}{n}\\right) \\left(2 + \\dfrac{2}{n} \\right)}} \\\\\n& = \\frac{3 \\left(1+ \\dfrac{1}{n} \\right)}{\\sqrt{ \\left( 2 + \\dfrac{1}{n}\\right) \\left(2 + \\dfrac{2}{n} \\right)}}  \\longrightarrow \\frac{3}{\\sqrt{4}} = \\frac{3}{2} > 1 \\,.\n\\end{align*}\nBy the Ratio Test we conclude that $(a_n)$ is divergent.\n\n:::\n\n\n\n::: Example \n\nLet \n$$\na_{n}=\\frac{n !}{100^{n}} \\,.\n$$\nProve that $(a_n)$ is divergent.\n\n> *Proof.*\n$$\n\\left|\\frac{a_{n+1}}{a_{n}}\\right| \n= \\frac{100^{n}}{100^{n+1}} \\frac{(n+1) !}{n !}\n=\\frac{n+1}{100} \\,.\n$$\nChoose $N=101$. Then for all $n \\geq N$,\n$$\n\\left|\\frac{a_{n+1}}{a_{n}}\\right| \\geq \\frac{101}{100}>1 \\,.\n$$\nHence $a_{n}$ is divergent by the Ratio Test.\n\n:::\n\n\n\n\n::: Warning\n\nThe Ratio Test in Theorem \\ref{theorem-ratio-test} does not address the case\n$$\nL=1 \\,.\n$$\nThis is because, in this case, the sequence $(a_n)$ \nmight converge or diverge. \n\nFor example:\n\n- Define the sequence\n$$\na_n = \\frac1n \\,.\n$$\nWe have\n$$\n\\left|\\frac{a_{n+1}}{a_{n}}\\right| = \\frac{n}{n+1} \\rightarrow L = 1 \\,. \n$$\nHence we cannot apply the Ratio Test. However we know that \n$$\n\\lim_{n \\to \\infty} \\, \\frac1n = 0 \\,.\n$$\n\n\n- Consider the sequence\n$$\na_n = n \\,.\n$$\nWe have\n$$\n\\frac{|a_{n+1}|}{|a_n|} = \\frac{n+1}{n} \\rightarrow L = 1 \\,. \n$$\nHence we cannot apply the Ratio Test. However we know that $(a_n)$ is unbounded, and thus divergent.\n\n:::\n\n\n\nIf the sequence $(a_n)$ is geometric, the Ratio Test of Theorem \\ref{theorem-ratio-test} will give the same\nanswer as the Geometric Sequence Test of Theorem \\ref{theorem-geometric-sequence}. This is the content of the following remark.\n\n\n::: Remark\n\nLet $x \\in \\R$ and define the geometric sequence \n$$\na_{n}=x^{n} \\,.\n$$\nThen\n$$\n\\left|\\frac{a_{n+1}}{a_{n}}\\right|  = \\frac{\\left|x^{n+1}\\right|}{\\left|x^{n}\\right|} \n= \\frac{|x|^{n+1}}{|x|^{n}} \n =|x| \\rightarrow |x| \\,.\n$$\nHence: \n\n- If $|x|<1$, the sequence $(a_n)$ converges by the Ratio Test\n- If $|x|>1$, the sequence $(a_n)$ diverges by the Ratio Test. \n- If $|x|=1$, the sequence $(a_n)$ might be convergent or divergent.\n\nThese results are in agreement with the Geometric Sequence Test. \n\n:::\n\n\n\n\n\n\n## Monotone sequences\n\n\nWe showed in Theorem \\ref{theorem-convergent-bounded} that convergent sequences are bounded. We noted that the converse statement is not true. \nFor example the sequence\n$$\na_n = (-1)^n\n$$\nis bounded but not convergent, as shown in Theorem \\ref{theorem-1-minus-n}.\nOn the other hand, if a bounded sequence is **monotone**, then it is convergent.\n\n\n::: Definition\n### Monotone sequence\n\nLet $(a_n)$ be a real sequence. We say that:\n\n1. $(a_n)$ is **increasing** if \n$$\na_n \\leq a_{n+1} \\,, \\quad \\forall \\, n \\geq N \\,.\n$$\n\n2. $(a_n)$ is **decreasing** if \n$$\na_n \\geq a_{n+1} \\,, \\quad \\forall \\, n \\geq N \\,.\n$$\n\n3. $(a_n)$ is **monotone** if it is either increasing or decreasing.\n\n:::\n\n\n\n::: Example\n\n- The sequence \n$$\na_n = \\frac1n\n$$\nis decreasing.\n\n  > We have\n  $$\n  a_n = \\frac1n > \\frac{1}{n+1} = a_{n+1} \\,.\n  $$\n\n- The sequence \n$$\nb_n = \\frac{n-1}{n} \n$$\nis increasing. \n\n  > We have\n  $$\n  b_{n+1} = \\frac{n}{n+1} > \\frac{n-1}{n} = b_n \\,,\n  $$\n  where the inequality holds because\n  \\begin{align*}\n  \\frac{n}{n+1} > \\frac{n-1}{n} \\quad & \\iff \\quad \n  n^2 > (n-1)(n+1) \\quad  \\\\\n  & \\iff \\quad n^2 > n^2 - 1 \\\\\n  & \\iff \\quad 0 >  - 1\n  \\end{align*}\n\n:::\n\n\nThe main result about monotone sequences is the Monotone Convergence\nTheorem.\n\n\n::: Theorem\n### Monotone Convergence Theorem  {#theorem-monotone-convergence}\n\nIf a sequence is bounded and monotone, then it converges.\n\n:::\n\n\n::: Proof\n\nAssume $(a_n)$ is bounded and monotone. Since $(a_n)$ is bounded, the set \n$$\nA:=\\{ a_n \\divider n \\in \\N \\}  \\subseteq \\R \n$$\nis bounded below and above. By the Axiom of Completeness of $\\R$ there exist $i,s \\in \\R$ such that\n$$\ni = \\inf A \\,, \\quad s = \\sup A \\,.\n$$\n\nWe have two cases:\n\n1. $(a_n)$ is increasing: We are going to prove that\n$$\n\\lim_{n \\to \\infty} a_n = s \\,.\n$$\nEquivalently, we need to prove that\n$$\n\\forall \\, \\e>0 \\,, \\, \\exists \\, N \\in \\N \\st \\forall \\, n \\geq N \\,, \\,\\,\n|a_n - s| < \\e \\,. \n$$\nLet $\\e > 0$. Since $s$ is the smallest upper bound for $A$, this means\n$$\ns - \\e\n$$\nis not an upper bound. Therefore there exists $N \\in \\N$ such that \n$$\ns - \\e < a_N \\,.\n$${#eq-monotone-convergence-proof-1}\nLet $n \\geq N$. Since $a_n$ is increasing, we have\n$$\na_N \\leq a_n \\,, \\quad \\forall \\, n \\geq N \\,.\n$${#eq-monotone-convergence-proof-2}\nMoreover $s$ is the supremum of $A$, so that\n$$\na_n \\leq s < s + \\e \\,, \\quad \\forall \\, n \\in \\N \\,.\n$${#eq-monotone-convergence-proof-3}\nPutting together estimates (@eq-monotone-convergence-proof-1)-(@eq-monotone-convergence-proof-2)-(@eq-monotone-convergence-proof-3)\nwe get\n$$\ns - \\e < a_N \\leq a_n \\leq s < s + \\e \\,, \\quad \\forall \\, n \\geq N \\,.\n$$\nTherefore, for all $n \\geq N$, we have\n$$\ns - \\e < a_n < s + \\e \\quad \\implies \\quad |a_n - s | < \\e \\,.\n$$\n\n\n\n\n2. $(a_n)$ is decreasing: With a similar proof, one can show that\n$$\n\\lim_{n \\to \\infty} a_n = i \\,.\n$$\nThis is left as an exercise.\n\n\n:::\n\n\n\n\n::: {.content-hidden}\n\n\n\n### Example: Euler's Number\n\n\nAs an application of the Monotone Convergence Theorem we can give\na formal definition for the Euler's Number\n$$\ne = 2.71828182845904523536 \\dots\n$$\n\n\n::: Theorem \n\nConsider the sequence\n$$\na_n = \\left(  1 + \\frac{1}{n}  \\right)^n  \\,.\n$$\n\nWe have that:\n\n1. $(a_n)$ is monotone increasing,\n2. $(a_n)$ is bounded.\n\nIn particular $(a_n)$ is convergent.\n\n:::\n\n\n::: Proof\n\nCopiala da Marcellini Sbordone, pagina 106\n\n:::\n\n\nThanks to Theorem \\ref{theorem-euler-number} we can define the Euler's Number $e$.\n\n\n::: Definition\n### Euler's Number\n\nThe Euler's number is defined as\n$$\ne := \\lim_{n \\to \\infty } \\, \\left(  1 + \\frac{1}{n}  \\right)^n \\,.\n$$\n\n:::\n\n\nAlready for $n=1000$ we have a good approximation of $e$:\n$$\na_{1000} = 2.7169 \\,.\n$$\n\n\n\n\n\n\n\n\n\nUse abbot, marcellini, and analisi sapienza note. For next year remove integral test and add cauchy condensation test.\n\n\n## Divergence to infinity\n\nDo this from abbott or marcellini. For the moment we have only done\ndivergent as non-convergent. Would need definition of Divergence to $\\pm \\infty$. Vedi anche forme indeterminate a pagina 96 Marcellini.\n\n\n## Some notable limits\n\nMarcellini Page 101. Also  see what is the english name for limiti notevoli.\n\n\n\n\n\n## Recurrence relations\n\nMarcellini Page 110\n\n\n\n## Example: Heron's Method\n\n\nThe first explicit algorithm for approximating \n$$\n\\sqrt{x}\n$$ \nfor $x > 0$ is known as **Heron's method**, after the first-century Greek mathematician [Heron of Alexandria](https://en.wikipedia.org/wiki/Hero_of_Alexandria) who described the method in his AD 60 work Metrica, see reference to\n[Wikipedia page](https://en.wikipedia.org/wiki/Methods_of_computing_square_roots#Babylonian_method).\n\nLet us see what is the idea of the algorithm:\n\n- Suppose that $a_1$ is an approximation of $\\sqrt{x}$ from above, that is,\n$$\n \\sqrt{x} < a_1 \\,.\n$$ {#eq-heron}\n- Multiplying (@eq-heron) by $\\sqrt{x}/a_1$ we obtain\n$$\n\\frac{x}{a_1} < \\sqrt{x} \\,,\n$${#eq-heron-1}\nobtaining an approximation of $\\sqrt{x}$ from below.\n- Therefore, putting together the above inequalities,\n$$\n\\frac{x}{a_1} < \\sqrt{x} < a_1 \n$$ {#eq-heron-2}\n- If we take the average of the points $x/a_1$ and $a_1$, it is reasonable to think that we find a better approximation of $\\sqrt{x}$. Thus our next approximation is\n$$\na_2 := \\frac{1}{2} \\left( a_1 + \\frac{x}{a_1} \\right) \\,,\n$$\nsee figure below.\n\n\n![Heron's Algorithm for approximating $\\sqrt{x}$](/images/heron.png){width=70%}\n\n\nIterating, we define by recurrence the sequence \n$$\na_{n+1} := \\frac12  \\left( a_n + \\frac{x}{a_n} \\right)\n$$\nfor all $n \\in \\N$, where the initial guess $a_1$ has to satisfy (@eq-heron). The aim of the section is to show that\n$$\n\\lim_{n \\to \\infty } \\ a_n = \\sqrt{x} \\,.\n$${#eq-heron-convergence}\nWe start by showing that (@eq-heron-2) holds for all $n \\in \\N$.\n\n::: {.Proposition  #proposition-heron}\nWe have\n$$\n\\frac{x}{a_n} < \\sqrt{x} < a_n \n$${#eq-heron-3}\nfor all $n \\in \\N$.\n:::\n\n\n::: Proof\nWe prove it by induction:\n\n1. By (@eq-heron) and (@eq-heron-1) we know that (@eq-heron-3) holds for $n=1$. \n\n2. Suppose now that (@eq-heron-3) holds for $n$. Then\n\\begin{align}\na_{n+1} - \\sqrt{x} & = \\frac12  \\left( a_n + \\frac{x}{a_n} \\right) - \\sqrt{x} \\\\\n& = \\frac{1}{2 a_n} ( a_n^2 + x - 2 a_n \\sqrt{x} ) \\\\\n& = \\frac{1}{2 a_n} ( a_n - \\sqrt{x} )^2 > 0 \\,,\n\\end{align}\nsince we are assuming that $a_n > \\sqrt{x}$. Therefore\n$$\n\\sqrt{x} < a_{n+1} \\,.\n$${#eq-heron-proof-1}\nMultiplying the above by $\\sqrt{x}/a_{n+1}$ we get\n$$\n\\frac{x}{a_{n+1}} < \\sqrt{x} \\,.\n$${#eq-heron-proof-2}\nInequalities (@eq-heron-proof-1) and (@eq-heron-proof-2) show that (@eq-heron-3) holds for $n+1$.\n\nTherefore we conclude (@eq-heron-3) by the Principle of Induction.\n:::\n\n\nWe are now ready to prove error estimates, that is, estimating how far away $a_n$ is from $\\sqrt{x}$. \n\n\n::: Proposition\n### Error estimate {#proposition-heron-error}\nFor all $n \\in \\N$ we have\n$$\na_{n+1} - \\sqrt{x} < \\frac12 (a_{n} - \\sqrt{x}) \\,.\n$${#eq-heron-half}\n:::\n\n\n::: Proof\nBy Proposition \\ref{proposition-heron} we know that \n$$\n\\frac{x}{a_n} < \\sqrt{x}\n$$\nfor all $n \\in \\N$. Therefore\n\\begin{align}\na_{n+1} & = \\frac12 \\left(  a_n +  \\frac{x}{a_n}   \\right) \\\\\n        & < \\frac12 \\left(  a_n +  \\sqrt{x}   \\right) \\,.\n\\end{align}\nSubtracting $\\sqrt{x}$ from both members in the above inequality we get the thesis.\n:::\n\n\nInequality (@eq-heron-half) is saying that the error halves at each step. Therefore we can prove that after $n$ steps the error is exponentially lower, as detailed in the following proposition.\n\n\n::: {.Proposition #proposition-heron-error-exp}\nFor all $n \\in \\N$ we have\n$$\na_{n+1} - \\sqrt{x} < \\frac{1}{2^n} (a_1 - \\sqrt{x})\n$$ {#eq-heron-error}\n:::\n\n\n\n::: Proof\nWe prove (@eq-heron-error) by induction:\n\n1. For $n=1$ we have that (@eq-heron-error) is satisfied, since it coincides with (@eq-heron-half) for $n=1$.\n2. Suppose that (@eq-heron-error) holds for $n$. By (@eq-heron-half) with $n$ replaced by $n+1$ we have\n\\begin{align}\na_{n+2} - \\sqrt{x} & < \\frac12 (a_{n+1} - \\sqrt{x}) \\\\\n& < \\frac12 \\,\\cdot \\, \\frac{1}{2^n} (a_{1} - \\sqrt{x}) \\\\\n& = \\frac{1}{2^{n+1}} (a_{1} - \\sqrt{x})\n\\end{align}\nwhere in the second inequality we used the induction hypothesis (@eq-heron-error). Hence (@eq-heron-error) holds for $n+1$.\n\nBy invoking the Induction Principle we conclude the proof.\n:::\n\nLet us comment estimate (@eq-heron-error). Denote the error at step $n$ by\n$$\ne_n := a_n - \\sqrt{x}\\,.\n$$\nThe initial error $e_1$ depends on how far the initial guess is from $\\sqrt{x}$. The estimate in (@eq-heron-error) is telling us that $e_n$ is a fraction of $e_1$, and actually\n$$\n\\lim_{n \\to \\infty} \\ e_n = 0\n$$\nexponentially fast. From this fact we are finally able to prove (@eq-heron-convergence).\n\n::: Theorem\n### Convergence of Heron's Algorithm\nWe have that \n$$\n\\lim_{n \\to \\infty} \\ a_n = \\sqrt{x} \\,.\n$$\n:::\n\n::: Proof\nBy Proposition \\ref{proposition-heron-error-exp} we have that\n$$\na_{n+1} - \\sqrt{x} < \\frac{1}{2^n} (a_1 - \\sqrt{x})\n$$\nMoreover Proposition \\ref{proposition-heron} tells us that\n$$\n\\sqrt{x} < a_{n+1} \\,.\n$$\nPutting together the two inequalities above we infer\n$$\n\\sqrt{x} < a_{n+1} < \\sqrt{x} + \\frac{1}{2^n} (a_1 - \\sqrt{x}) \\,.\n$${#eq-heron-final}\nNow note that\n$$\n\\lim_{n \\to \\infty} \\ \\frac{1}{2^n} =\n\\lim_{n \\to \\infty} \\ \\left( \\frac{1}{2} \\right)^n = 0 \\,.\n$$\nTherefore the RHS of (@eq-heron-final) converges to $\\sqrt{x}$ as $n \\to \\infty$. Applying the Squeeze Theorem to (@eq-heron-final) we conclude that $a_n \\to \\sqrt{x}$ as $n \\to \\infty$.\n:::\n\n\n\n### Coding the Algorithm\n\nHeron's Algorithm can be easily coded in Python. For example, see the function below:\n\n```python\n# x is the number for which to compute sqrt(x)\n# guess is the point a_1\n# a_1 must be strictly larger than sqrt(x)\n# n is the number of iterations\n# the function returns a_{n+1}\n\ndef herons_algorithm(x, guess, n):\n    for i in range(n):\n        guess = (guess + x / guess) / 2.0\n    return guess\n```\n\nFor example let us use the Algorithm to compute $\\sqrt{2}$ after $3$ iterations. For initial guess we take $a_1 = 2$. \n\n```python\n# Calculate sqrt(2) with 3 iterations and guess 2\nsqrt_2 = herons_algorithm(2, 2, 3)\n\nprint(f\"The sqrt(2) is approximately {sqrt_2}\")\n```\n\n::: {.cell execution_count=1}\n\n::: {.cell-output .cell-output-stdout}\n```\nThe sqrt(2) is approximately 1.4142156862745097\n```\n:::\n:::\n\n\nThat is a pretty good approximation in just $3$ iterations!\n\n\n\n\n\n## Fibonacci Sequence\n\n\n\n\n:::\n\n",
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-    "text": "6.6 Algebra of limits\nProving convergence using Definition 5 can be a tedious task. In this section we discuss how to prove convergence, starting from known convergence results.\n\nTheorem 26: Algebra of limits\nLet \\(\\left(a_{n}\\right)_{n \\in \\mathbb{N}}\\) and \\(\\left(b_{n}\\right)_{n \\in \\mathbb{N}}\\) be sequences in \\(\\mathbb{R}\\). Suppose that \\[\n\\lim_{n \\rightarrow \\infty} a_{n}=a \\,, \\quad  \\lim_{n \\rightarrow \\infty} b_{n}=b \\,,\n\\] for some \\(a,b \\in \\mathbb{R}\\). Then,\n\nLimit of sum is the sum of limits: \\[\n\\lim_{n \\rightarrow \\infty}\\left(a_{n} \\pm b_{n}\\right)=a \\pm b\n\\]\nLimit of product is the product of limits: \\[\n\\lim _{n \\rightarrow \\infty}\\left(a_{n}  b_{n}\\right) = a  b\n\\]\nIf \\(b_{n} \\neq 0\\) for all \\(n \\in \\mathbb{N}\\) and \\(b \\neq 0\\), then \\[\n\\lim_{n \\rightarrow \\infty} \\left(\\frac{a_{n}}{b_{n}}\\right)=\\frac{a}{b}\n\\]\n\n\n\n\nProofLet \\(\\left(a_{n}\\right)_{n \\in \\mathbb{N}}\\) and \\(\\left(b_{n}\\right)_{n \\in \\mathbb{N}}\\) be sequences in \\(\\mathbb{R}\\) and let \\(c \\in \\mathbb{R}\\). Suppose that, for some \\(a, b \\in \\mathbb{R}\\) \\[\n\\lim _{n \\rightarrow \\infty} a_{n}=a \\,, \\quad  \\lim _{n \\rightarrow \\infty} b_{n}=b \\,.\n\\]\nProof of Point 1.\nWe need to show that \\[\n\\lim_{n \\to \\infty} (a_n \\pm  b_n) = a \\pm b \\,.\n\\] We only give a proof of the formula with \\(+\\), since the case with \\(-\\) follows with a very similar proof. Hence, we need to show that \\[\n\\forall \\, \\varepsilon&gt; 0 \\,, \\, \\exists  \\, N \\in \\mathbb{N}\\, \\text{ s.t. } \\, \\forall \\, n \\geq N \\,, \\, \\left|(a_{n} + b_n) - (a+b)  \\right| &lt; \\varepsilon\\,.\n\\] Let \\(\\varepsilon&gt;0\\) and set \\[\n\\widetilde{\\varepsilon} := \\frac{\\varepsilon}{2} \\,.\n\\] Since \\(a_{n} \\rightarrow a\\) and \\(\\widetilde{\\varepsilon}&gt;0\\), there exists \\(N_1 \\in \\mathbb{N}\\) such that \\[\n|a_n - a|&lt; \\widetilde{\\varepsilon} \\,, \\quad \\forall \\, n \\geq N_1 \\,.\n\\] Since \\(b_{n} \\rightarrow b\\) and \\(\\widetilde{\\varepsilon}&gt;0\\), there exists \\(N_2 \\in \\mathbb{N}\\) such that \\[\n|b_n - b|&lt; \\widetilde{\\varepsilon} \\,, \\quad \\forall \\, n \\geq N_2 \\,.\n\\] Define \\[\nN : = \\max  \\{ N_1, N_2 \\} \\,.\n\\] For all \\(n \\geq N\\) we have, by the triangle inequality, \\[\\begin{align*}\n\\left|\\left(a_{n}+b_{n}\\right)-(a+b)\\right|\n& = \\left|\\left(a_{n}-a\\right) + \\left(b_{n}-b\\right)\\right| \\\\\n& \\leq \\left|a_{n}-a\\right|+\\left|b_{n}-b\\right| \\\\\n& &lt; \\widetilde{\\varepsilon}+\\widetilde{\\varepsilon} \\\\\n& =\\varepsilon\\,.\n\\end{align*}\\]\nProof of Point 2.\nWe need to show that \\[\n\\lim_{n \\to \\infty} (a_n  b_n) = a b  \\,,\n\\] which is equivalent to \\[\n\\forall \\, \\varepsilon&gt; 0 \\,, \\, \\exists \\, N \\in \\mathbb{N}\\, \\text{ s.t. } \\, \\forall \\, n \\geq N \\,, \\, \\left|a_{n} b_{n} - a b \\right| &lt; \\varepsilon\\,.\n\\] Let \\(\\varepsilon&gt;0\\). The sequence \\(\\left(a_{n}\\right)\\) converges, and hence is bounded, by Theorem 19. This means there exists some \\(M&gt;0\\) such that \\[\n\\left|a_{n}\\right| \\leq M \\,, \\quad \\forall \\, n \\in \\mathbb{N}\\,.\n\\] Define \\[\n\\widetilde{\\varepsilon} = \\frac{\\varepsilon}{ M + |b| } \\,.\n\\] Since \\(a_{n} \\rightarrow a\\) and \\(\\widetilde{\\varepsilon} &gt; 0\\), there exists \\(N_1 \\in \\mathbb{N}\\) such that \\[\n| a_n  - a | &lt; \\widetilde{\\varepsilon} \\,, \\quad \\forall \\, n \\geq N_1 \\,.\n\\] Since \\(b_{n} \\rightarrow b\\) and \\(\\widetilde{\\varepsilon} &gt; 0\\), there exists \\(N_2 \\in \\mathbb{N}\\) such that \\[\n| b_n  - b | &lt; \\widetilde{\\varepsilon} \\,, \\quad \\forall \\, n \\geq N_2 \\,.\n\\] Let \\[\nN:= \\max\\{ N_1, N_2 \\} \\,.\n\\] For all \\(n \\geq N\\) we have \\[\\begin{align*}\n\\left|a_{n}b_{n}-a b\\right|\n& = \\left|a_{n} b_{n} - a_n  b + a_n  b - a  b \\right| \\\\\n& \\leq \\left|a_{n}  b_{n} - a_n  b \\right| + \\left|a_n  b - a b \\right| \\\\\n& =\\left|a_{n}\\right| \\left|b_{n}-b\\right|+ |b|  \\left|a_{n}-a\\right| \\\\\n& &lt; M \\, \\widetilde{\\varepsilon} +|b| \\, \\widetilde{\\varepsilon} \\\\\n& = (M + |b|) \\, \\widetilde{\\varepsilon} \\\\\n& = \\varepsilon\\,.\n\\end{align*}\\]\nProof of Point 3.\nSuppose in addition that \\(b_n \\neq 0\\) and \\(b \\neq 0\\). We need to show that \\[\n\\lim_{n \\to \\infty} \\, \\frac{a_n}{b_n} = \\frac{a}{b} \\,,\n\\] which is equivalent to \\[\n\\forall \\, \\varepsilon&gt; 0 \\,, \\, \\exists \\, N \\in \\mathbb{N}\\, \\text{ s.t. } \\, \\forall \\, n \\geq N \\,, \\, \\left|\\frac{a_{n}}{ b_{n}} - \\frac{a}{b} \\right| &lt; \\varepsilon\\,.\n\\] We suppose in addition that \\(b&gt;0\\). The proof is very similar for the case \\(b &lt;0\\). Let \\(\\varepsilon&gt; 0\\). Set \\[\n\\delta := \\frac{b}{2} \\,.\n\\] Since \\(b_n \\to b\\) and \\(\\delta&gt;0\\), there exists \\(N_1 \\in \\mathbb{N}\\) such that \\[\n|b_n - b| &lt; \\delta \\, \\quad \\forall \\, n \\geq N_1 \\,.\n\\] In particular we have \\[\nb_n &gt; b - \\delta = b - \\frac{b}{2} = \\frac{b}{2}   \\, \\quad \\forall \\, n \\geq N_1 \\,.\n\\] Define \\[\n\\widetilde{\\varepsilon} := \\frac{ b^2 }{ 2 (b + |a|)} \\, \\varepsilon\\,.\n\\] Since \\(\\widetilde{\\varepsilon}&gt;0\\) and \\(a_n \\to a\\), there exists \\(N_2 \\in \\mathbb{N}\\) such that \\[\n|a_n - a |&lt; \\widetilde{\\varepsilon} \\,, \\quad \\forall \\, n \\geq N_2 \\,.\n\\] Since \\(\\widetilde{\\varepsilon}&gt;0\\) and \\(b_n \\to b\\), there exists \\(N_3 \\in \\mathbb{N}\\) such that \\[\n|b_n - b |&lt; \\widetilde{\\varepsilon} \\,, \\quad \\forall \\, n \\geq N_3 \\,.\n\\] Define \\[\nN:= \\max\\{ N_1, N_2, N_3 \\} \\,.\n\\] For all \\(n \\geq N\\) we have \\[\\begin{align*}\n\\left|\\frac{a_{n}}{ b_{n}} - \\frac{a}{b} \\right| & =\n\\left|\\frac{a_{n} b - a b_n}{ b_{n} b }  \\right| \\\\\n& = \\frac{1}{ \\left|b_{n} b \\right|} \\,  \n\\left| a_{n}b - a b + a b - a b_n    \\right| \\\\\n& = \\frac{1}{ \\left|b_{n} b \\right|} \\,  \n\\left| (a_{n} - a) b + a(b-b_n)   \\right| \\\\\n& \\leq \\frac{1}{ \\left|b_{n} b \\right|} \\,  \n\\left( |a_{n} - a| |b| + |a| |b-b_n| \\right) \\\\\n& &lt; \\frac{1}{ \\dfrac{b}{2} \\, b } \\, \\left(  \\widetilde{\\varepsilon} \\, b + \\widetilde{\\varepsilon} |a|   \\right) \\\\\n& = \\frac{2 (b + |a|) }{b^2} \\, \\widetilde{\\varepsilon} \\\\\n& = \\varepsilon\\,.\n\\end{align*}\\]\n\n\nIn the future we will refer to Theorem 26 as the Algebra of Limits. We now show how to use Theorem 26 for computing certain limits.\n\nExample 27\nProve that \\[\n\\lim_{n \\to \\infty} \\, \\frac{3 n}{7 n+4} = \\frac{3}{7} \\,.\n\\]\n\nProof. We can rewrite \\[\n\\frac{3 n}{7 n+4}=\\frac{3}{7+\\dfrac{4}{n}}\n\\] By Theorem 12 we know that \\[\n3 \\rightarrow 3\\,,  \\quad 4 \\to 4 \\,, \\quad 7 \\to 7 \\,.\n\\] From Theorem 8 we know that \\[\n\\frac{1}{n} \\rightarrow 0 \\,.\n\\] Hence, it follows from Theorem 26 Point 2 that \\[\n\\frac{4}{n} = 4 \\cdot \\frac1n \\rightarrow 4 \\cdot 0 = 0 \\,.\n\\] By Theorem 26 Point 1 we have \\[\n7 + \\frac{4}{n} \\rightarrow  7 + 0 = 7 \\,.\n\\] Finally we can use Theorem 26 Point 3 to infer \\[\n\\frac{3 n}{7 n+4}=\\frac{3}{7+\\dfrac{4}{n}} \\rightarrow \\frac{3}{7} \\,.\n\\]\n\n\n\n\nImportantThe technique shown in Example 27 is useful to compute limits of fractions of polynomials. To identify the possible limit, if it exists, it is often best to divide by the largest power of \\(n\\) in the denominator.\n\n\n\nExample 28\nProve that \\[\n\\lim_{n \\rightarrow \\infty} \\frac{n^{2}-1}{2 n^{2}-3} = \\frac{1}{2} \\,.\n\\]\n\nProof. Factor \\(n^2\\) to obtain \\[\n\\frac{n^{2}-1}{2 n^{2}-3} = \\frac{1-\\dfrac{1}{n^{2}}}{2-\\dfrac{3}{n^{2}}} \\,.\n\\] By Theorem 9 we have \\[\n\\frac{1}{n^2} \\to 0 \\,.\n\\] We can then use the Algebra of Limits Theorem 26 Point 2 to infer \\[\n\\frac{3}{n^2} \\to 3 \\cdot 0 = 0\n\\] and Theorem 26 Point 1 to get \\[\n1 - \\frac{1}{n^2} \\to 1 - 0 = 1 \\,, \\quad\n2 - \\frac{3}{n^2} \\to 2 - 0 = 2 \\,.\n\\] Finally we use Theorem 26 Point 3 and conclude \\[\n\\frac{1-\\dfrac{1}{n^{2}}}{2-\\dfrac{3}{n^{2}}} \\to \\frac{1}{2} \\,.\n\\] Therefore \\[\n\\lim_{n \\to \\infty } \\, \\frac{n^{2}-1}{2 n^{2}-3} = \\lim_{n \\to \\infty} \\, \\frac{1-\\dfrac{1}{n^{2}}}{2-\\dfrac{3}{n^{2}}} = \\frac{1}{2} \\,.\n\\]\n\n\n\nWe can also use the Algebra of Limits to prove that certain limits do not exist.\n\nExample 29\nProve that the sequence \\[\na_n = \\frac{4 n^{3}+8 n+1}{7 n^{2}+2 n+1}\n\\] does not converge.\n\nProof. To show that the sequence \\(\\left(a_n\\right)\\) does not converge, we divide by the largest power in the denominator, which in this case is \\(n^2\\) \\[\\begin{align*}\na_n & = \\frac{4 n^{3}+8 n+1}{7 n^{2}+2 n+1} \\\\\n& =\\frac{4 n+ \\dfrac{8}{n}  +  \\dfrac{1}{n^{2}} }{7+  \\dfrac{2}{n}  +  \\dfrac{1}{n^{2}} } \\\\\n& = \\frac{b_n}{c_n}\n\\end{align*}\\] where we set \\[\nb_n := 4 n+ \\dfrac{8}{n}  +  \\dfrac{1}{n^{2}}\\,, \\quad\nc_n := 7 + \\dfrac{2}{n}  +  \\dfrac{1}{n^{2}} \\,.\n\\] Using the Algebra of Limits Theorem 26 we see that \\[\nc_n = 7+  \\dfrac{2}{n}  +  \\dfrac{1}{n^{2}}  \\to 7 \\,.\n\\] Suppose by contradiction that \\[\na_n \\to a\n\\] for some \\(a \\in \\mathbb{R}\\). Then, by the Algebra of Limits Theorem 26 we would infer \\[\nb_n = c_n \\cdot a_n \\to 7 a \\,,\n\\] concluding that \\(b_n\\) is convergent to \\(7a\\). We have that \\[\nb_n = 4n + d_n \\,, \\quad d_n := \\dfrac{8}{n}  +  \\dfrac{1}{n^{2}} \\,.\n\\] Again by the Algebra of Limits Theorem 26 we get that \\[\nd_n =  \\dfrac{8}{n}  +  \\dfrac{1}{n^{2}} \\to 0 \\,,\n\\] and hence \\[\n4n = b_n - d_n \\to 7a - 0 = 7a \\,.\n\\] This is a contradiction, since the sequence \\((4n)\\) is unbounded, and hence cannot be convergent. Hence \\((a_n)\\) is not convergent.\n\n\n\n\nWarningConsider the sequence \\[\na_n = \\frac{4 n^{3}+8 n+1}{7 n^{2}+2 n+1}\n\\] from the previous example. We have proven that \\((a_n)\\) is not convergent, by making use of the Algebra of Limits.\nLet us review a faulty argument to conclude that \\((a_n)\\) is not convergent. Write \\[\na_n = \\frac{b_n}{c_n} \\,, \\quad b_n:= 4 n^{3}+8 n+1\\,,\n\\quad c_n : =7 n^{2}+2 n+1 \\,.\n\\] The numerator \\[\nb_n = 4 n^{3}+8 n+1\n\\] and denominator \\[\nc_n  =7 n^{2}+2 n+1\n\\] are both unbounded, and hence \\((b_n)\\) and \\((c_n)\\) do not converge. One might be tempted to conclude that \\((a_n)\\) does not converge. However this is false in general: as seen in Example 28, we have \\[\n\\lim_{n \\rightarrow \\infty} \\frac{n^{2}-1}{2 n^{2}-3} = \\frac{1}{2} \\,,\n\\] while numerator and denominator are unbounded.\n\n\nSometimes it is useful to rearrange the terms of a sequence, before applying the Algebra of Limits.\n\nExample 30\nDefine \\[\na_n := \\frac{2 n^{3}+7 n+1}{5 n+9} \\cdot \\frac{8 n+9}{6 n^{3}+8 n^{2}+3} \\,.\n\\] Prove that \\[\n\\lim_{n \\to \\infty}  a_n = \\frac{8}{15} \\,.\n\\]\n\nProof. The first fraction in \\((a_n)\\) does not converge, as it is unbounded. Therefore we cannot use Point 2 in Theorem 26 directly. However, we note that \\[\\begin{align*}\na_{n} & =\\frac{2 n^{3}+7 n+1}{5 n+9} \\cdot \\frac{8 n+9}{6 n^{3}+8 n^{2}+3} \\\\\n& = \\frac{8 n+9}{5 n+9} \\cdot \\frac{2 n^{3}+7 n+1}{6 n^{3}+8 n^{2}+3} \\,.\n\\end{align*}\\] Factoring out \\(n\\) and \\(n^3\\), respectively, and using the Algebra of Limits, we see that \\[\n\\frac{8 n+9}{5 n+9}=\\frac{8+9 / n}{5+9 / n} \\to \\frac{8+0}{5+0}=\\frac{8}{5}\n\\] and \\[\n\\frac{2+7 / n^{2}+1 / n^{3}}{6+8 / n+3 / n^{3}} \\to \\frac{2+0+0}{6+0+0}=\\frac{1}{3}\n\\] Therefore Theorem 26 Point 2 ensures that \\[\na_{n} \\to \\frac{8}{5} \\cdot \\frac{1}{3}=\\frac{8}{15} \\,.\n\\]"
+    "text": "6.6 Algebra of limits\nProving convergence using Definition 5 can be a tedious task. In this section we discuss how to prove convergence, starting from known convergence results.\n\nTheorem 26: Algebra of limits\nLet \\(\\left(a_{n}\\right)_{n \\in \\mathbb{N}}\\) and \\(\\left(b_{n}\\right)_{n \\in \\mathbb{N}}\\) be sequences in \\(\\mathbb{R}\\). Suppose that \\[\n\\lim_{n \\rightarrow \\infty} a_{n}=a \\,, \\quad  \\lim_{n \\rightarrow \\infty} b_{n}=b \\,,\n\\] for some \\(a,b \\in \\mathbb{R}\\). Then,\n\nLimit of sum is the sum of limits: \\[\n\\lim_{n \\rightarrow \\infty}\\left(a_{n} \\pm b_{n}\\right)=a \\pm b\n\\]\nLimit of product is the product of limits: \\[\n\\lim _{n \\rightarrow \\infty}\\left(a_{n}  b_{n}\\right) = a  b\n\\]\nIf \\(b_{n} \\neq 0\\) for all \\(n \\in \\mathbb{N}\\) and \\(b \\neq 0\\), then \\[\n\\lim_{n \\rightarrow \\infty} \\left(\\frac{a_{n}}{b_{n}}\\right)=\\frac{a}{b}\n\\]\n\n\n\n\nProofLet \\(\\left(a_{n}\\right)_{n \\in \\mathbb{N}}\\) and \\(\\left(b_{n}\\right)_{n \\in \\mathbb{N}}\\) be sequences in \\(\\mathbb{R}\\) and let \\(c \\in \\mathbb{R}\\). Suppose that, for some \\(a, b \\in \\mathbb{R}\\) \\[\n\\lim _{n \\rightarrow \\infty} a_{n}=a \\,, \\quad  \\lim _{n \\rightarrow \\infty} b_{n}=b \\,.\n\\]\nProof of Point 1.\nWe need to show that \\[\n\\lim_{n \\to \\infty} (a_n \\pm  b_n) = a \\pm b \\,.\n\\] We only give a proof of the formula with \\(+\\), since the case with \\(-\\) follows with a very similar proof. Hence, we need to show that \\[\n\\forall \\, \\varepsilon&gt; 0 \\,, \\, \\exists  \\, N \\in \\mathbb{N}\\, \\text{ s.t. } \\, \\forall \\, n \\geq N \\,, \\, \\left|(a_{n} + b_n) - (a+b)  \\right| &lt; \\varepsilon\\,.\n\\] Let \\(\\varepsilon&gt;0\\) and set \\[\n\\widetilde{\\varepsilon} := \\frac{\\varepsilon}{2} \\,.\n\\] Since \\(a_{n} \\rightarrow a, b_n \\to b\\), and \\(\\widetilde{\\varepsilon}&gt;0\\), there exist \\(N_1, N_2 \\in \\mathbb{N}\\) such that \\[\\begin{align*}\n|a_n - a| & &lt; \\widetilde{\\varepsilon} \\,, \\quad \\forall \\, n \\geq N_1 \\,, \\\\\n|b_n - b| & &lt; \\widetilde{\\varepsilon} \\,, \\quad \\forall \\, n \\geq N_2 \\,.\n\\end{align*}\\] Define \\[\nN : = \\max  \\{ N_1, N_2 \\} \\,.\n\\] For all \\(n \\geq N\\) we have, by the triangle inequality, \\[\\begin{align*}\n\\left|\\left(a_{n}+b_{n}\\right)-(a+b)\\right|\n& = \\left|\\left(a_{n}-a\\right) + \\left(b_{n}-b\\right)\\right| \\\\\n& \\leq \\left|a_{n}-a\\right|+\\left|b_{n}-b\\right| \\\\\n& &lt; \\widetilde{\\varepsilon}+\\widetilde{\\varepsilon} \\\\\n& =\\varepsilon\\,.\n\\end{align*}\\]\nProof of Point 2.\nWe need to show that \\[\n\\lim_{n \\to \\infty} (a_n  b_n) = a b  \\,,\n\\] which is equivalent to \\[\n\\forall \\, \\varepsilon&gt; 0 \\,, \\, \\exists \\, N \\in \\mathbb{N}\\, \\text{ s.t. } \\, \\forall \\, n \\geq N \\,, \\, \\left|a_{n} b_{n} - a b \\right| &lt; \\varepsilon\\,.\n\\] Let \\(\\varepsilon&gt;0\\). The sequence \\(\\left(a_{n}\\right)\\) converges, and hence is bounded, by Theorem 19. This means there exists some \\(M&gt;0\\) such that \\[\n\\left|a_{n}\\right| \\leq M \\,, \\quad \\forall \\, n \\in \\mathbb{N}\\,.\n\\] Define \\[\n\\widetilde{\\varepsilon} = \\frac{\\varepsilon}{ M + |b| } \\,.\n\\] Since \\(a_{n} \\rightarrow a, b_n \\to b\\), and \\(\\widetilde{\\varepsilon} &gt; 0\\), there exist \\(N_1, N_2 \\in \\mathbb{N}\\) such that \\[\\begin{align*}\n| a_n  - a | & &lt; \\widetilde{\\varepsilon} \\,, \\quad \\forall \\, n \\geq N_1 \\,. \\\\\n| b_n  - b | & &lt; \\widetilde{\\varepsilon} \\,, \\quad \\forall \\, n \\geq N_2 \\,.\n\\end{align*}\\] Let \\[\nN:= \\max\\{ N_1, N_2 \\} \\,.\n\\] For all \\(n \\geq N\\) we have \\[\\begin{align*}\n\\left|a_{n}b_{n}-a b\\right|\n& = \\left|a_{n} b_{n} - a_n  b + a_n  b - a  b \\right| \\\\\n& \\leq \\left|a_{n}  b_{n} - a_n  b \\right| + \\left|a_n  b - a b \\right| \\\\\n& =\\left|a_{n}\\right| \\left|b_{n}-b\\right|+ |b|  \\left|a_{n}-a\\right| \\\\\n& &lt; M \\, \\widetilde{\\varepsilon} +|b| \\, \\widetilde{\\varepsilon} \\\\\n& = (M + |b|) \\, \\widetilde{\\varepsilon} \\\\\n& = \\varepsilon\\,.\n\\end{align*}\\]\nProof of Point 3.\nSuppose in addition that \\(b_n \\neq 0\\) and \\(b \\neq 0\\). We need to show that \\[\n\\lim_{n \\to \\infty} \\, \\frac{a_n}{b_n} = \\frac{a}{b} \\,,\n\\] which is equivalent to \\[\n\\forall \\, \\varepsilon&gt; 0 \\,, \\, \\exists \\, N \\in \\mathbb{N}\\, \\text{ s.t. } \\, \\forall \\, n \\geq N \\,, \\, \\left|\\frac{a_{n}}{ b_{n}} - \\frac{a}{b} \\right| &lt; \\varepsilon\\,.\n\\] We suppose in addition that \\(b&gt;0\\). The proof is very similar for the case \\(b &lt;0\\), and is hence omitted. Let \\(\\varepsilon&gt; 0\\). Set \\[\n\\delta := \\frac{b}{2} \\,.\n\\] Since \\(b_n \\to b\\) and \\(\\delta&gt;0\\), there exists \\(N_1 \\in \\mathbb{N}\\) such that \\[\n|b_n - b| &lt; \\delta \\, \\quad \\forall \\, n \\geq N_1 \\,.\n\\] In particular we have \\[\nb_n &gt; b - \\delta = b - \\frac{b}{2} = \\frac{b}{2}   \\, \\quad \\forall \\, n \\geq N_1 \\,.\n\\] Define \\[\n\\widetilde{\\varepsilon} := \\frac{ b^2 }{ 2 (b + |a|)} \\, \\varepsilon\\,.\n\\] Since \\(\\widetilde{\\varepsilon}&gt;0\\) and \\(a_n \\to a, b_n \\to b\\), there exist \\(N_2, N_3 \\in \\mathbb{N}\\) such that \\[\\begin{align*}\n|a_n - a | & &lt; \\widetilde{\\varepsilon} \\,, \\quad \\forall \\, n \\geq N_2 \\,, \\\\\n|b_n - b | & &lt; \\widetilde{\\varepsilon} \\,, \\quad \\forall \\, n \\geq N_3 \\,.\n\\end{align*}\\] Define \\[\nN:= \\max\\{ N_1, N_2, N_3 \\} \\,.\n\\] For all \\(n \\geq N\\) we have \\[\\begin{align*}\n\\left|\\frac{a_{n}}{ b_{n}} - \\frac{a}{b} \\right| & =\n\\left|\\frac{a_{n} b - a b_n}{ b_{n} b }  \\right| \\\\\n& = \\frac{1}{ \\left|b_{n} b \\right|} \\,  \n\\left| a_{n}b - a b + a b - a b_n    \\right| \\\\\n& = \\frac{1}{ \\left|b_{n} b \\right|} \\,  \n\\left| (a_{n} - a) b + a(b-b_n)   \\right| \\\\\n& \\leq \\frac{1}{ \\left|b_{n} b \\right|} \\,  \n\\left( |a_{n} - a| |b| + |a| |b-b_n| \\right) \\\\\n& &lt; \\frac{1}{ \\dfrac{b}{2} \\, b } \\, \\left(  \\widetilde{\\varepsilon} \\, b + \\widetilde{\\varepsilon} |a|   \\right) \\\\\n& = \\frac{2 (b + |a|) }{b^2} \\, \\widetilde{\\varepsilon} \\\\\n& = \\varepsilon\\,.\n\\end{align*}\\]\n\n\nIn the future we will refer to Theorem 26 as the Algebra of Limits. We now show how to use Theorem 26 for computing certain limits.\n\nExample 27\nProve that \\[\n\\lim_{n \\to \\infty} \\, \\frac{3 n}{7 n+4} = \\frac{3}{7} \\,.\n\\]\n\nProof. We can rewrite \\[\n\\frac{3 n}{7 n+4}=\\frac{3}{7+\\dfrac{4}{n}}\n\\] By Theorem 12 we know that \\[\n3 \\rightarrow 3\\,,  \\quad 4 \\to 4 \\,, \\quad 7 \\to 7 \\,.\n\\] From Theorem 8 we know that \\[\n\\frac{1}{n} \\rightarrow 0 \\,.\n\\] Hence, it follows from Theorem 26 Point 2 that \\[\n\\frac{4}{n} = 4 \\cdot \\frac1n \\rightarrow 4 \\cdot 0 = 0 \\,.\n\\] By Theorem 26 Point 1 we have \\[\n7 + \\frac{4}{n} \\rightarrow  7 + 0 = 7 \\,.\n\\] Finally we can use Theorem 26 Point 3 to infer \\[\n\\frac{3 n}{7 n+4}=\\frac{3}{7+\\dfrac{4}{n}} \\rightarrow \\frac{3}{7} \\,.\n\\]\n\n\n\n\nImportantThe technique shown in Example 27 is useful to compute limits of fractions of polynomials. To identify the possible limit, if it exists, it is often best to divide by the largest power of \\(n\\) in the denominator.\n\n\n\nExample 28\nProve that \\[\n\\lim_{n \\rightarrow \\infty} \\frac{n^{2}-1}{2 n^{2}-3} = \\frac{1}{2} \\,.\n\\]\n\nProof. Factor \\(n^2\\) to obtain \\[\n\\frac{n^{2}-1}{2 n^{2}-3} = \\frac{1-\\dfrac{1}{n^{2}}}{2-\\dfrac{3}{n^{2}}} \\,.\n\\] By Theorem 9 we have \\[\n\\frac{1}{n^2} \\to 0 \\,.\n\\] We can then use the Algebra of Limits Theorem 26 Point 2 to infer \\[\n\\frac{3}{n^2} \\to 3 \\cdot 0 = 0\n\\] and Theorem 26 Point 1 to get \\[\n1 - \\frac{1}{n^2} \\to 1 - 0 = 1 \\,, \\quad\n2 - \\frac{3}{n^2} \\to 2 - 0 = 2 \\,.\n\\] Finally we use Theorem 26 Point 3 and conclude \\[\n\\frac{1-\\dfrac{1}{n^{2}}}{2-\\dfrac{3}{n^{2}}} \\to \\frac{1}{2} \\,.\n\\] Therefore \\[\n\\lim_{n \\to \\infty } \\, \\frac{n^{2}-1}{2 n^{2}-3} = \\lim_{n \\to \\infty} \\, \\frac{1-\\dfrac{1}{n^{2}}}{2-\\dfrac{3}{n^{2}}} = \\frac{1}{2} \\,.\n\\]\n\n\n\nWe can also use the Algebra of Limits to prove that certain limits do not exist.\n\nExample 29\nProve that the sequence \\[\na_n = \\frac{4 n^{3}+8 n+1}{7 n^{2}+2 n+1}\n\\] does not converge.\n\nProof. To show that the sequence \\(\\left(a_n\\right)\\) does not converge, we divide by the largest power in the denominator, which in this case is \\(n^2\\) \\[\\begin{align*}\na_n & = \\frac{4 n^{3}+8 n+1}{7 n^{2}+2 n+1} \\\\\n& =\\frac{4 n+ \\dfrac{8}{n}  +  \\dfrac{1}{n^{2}} }{7+  \\dfrac{2}{n}  +  \\dfrac{1}{n^{2}} } \\\\\n& = \\frac{b_n}{c_n}\n\\end{align*}\\] where we set \\[\nb_n := 4 n+ \\dfrac{8}{n}  +  \\dfrac{1}{n^{2}}\\,, \\quad\nc_n := 7 + \\dfrac{2}{n}  +  \\dfrac{1}{n^{2}} \\,.\n\\] Using the Algebra of Limits Theorem 26 we see that \\[\nc_n = 7+  \\dfrac{2}{n}  +  \\dfrac{1}{n^{2}}  \\to 7 \\,.\n\\] Suppose by contradiction that \\[\na_n \\to a\n\\] for some \\(a \\in \\mathbb{R}\\). Then, by the Algebra of Limits Theorem 26 we would infer \\[\nb_n = c_n \\cdot a_n \\to 7 a \\,,\n\\] concluding that \\(b_n\\) is convergent to \\(7a\\). We have that \\[\nb_n = 4n + d_n \\,, \\quad d_n := \\dfrac{8}{n}  +  \\dfrac{1}{n^{2}} \\,.\n\\] Again by the Algebra of Limits Theorem 26 we get that \\[\nd_n =  \\dfrac{8}{n}  +  \\dfrac{1}{n^{2}} \\to 0 \\,,\n\\] and hence \\[\n4n = b_n - d_n \\to 7a - 0 = 7a \\,.\n\\] This is a contradiction, since the sequence \\((4n)\\) is unbounded, and hence cannot be convergent. Hence \\((a_n)\\) is not convergent.\n\n\n\n\nWarningConsider the sequence \\[\na_n = \\frac{4 n^{3}+8 n+1}{7 n^{2}+2 n+1}\n\\] from the previous example. We have proven that \\((a_n)\\) is not convergent, by making use of the Algebra of Limits.\nLet us review a faulty argument to conclude that \\((a_n)\\) is not convergent. Write \\[\na_n = \\frac{b_n}{c_n} \\,, \\quad b_n:= 4 n^{3}+8 n+1\\,,\n\\quad c_n : =7 n^{2}+2 n+1 \\,.\n\\] The numerator \\[\nb_n = 4 n^{3}+8 n+1\n\\] and denominator \\[\nc_n  =7 n^{2}+2 n+1\n\\] are both unbounded, and hence \\((b_n)\\) and \\((c_n)\\) do not converge. One might be tempted to conclude that \\((a_n)\\) does not converge. However this is false in general: as seen in Example 28, we have \\[\n\\lim_{n \\rightarrow \\infty} \\frac{n^{2}-1}{2 n^{2}-3} = \\frac{1}{2} \\,,\n\\] while numerator and denominator are unbounded.\n\n\nSometimes it is useful to rearrange the terms of a sequence, before applying the Algebra of Limits.\n\nExample 30\nDefine \\[\na_n := \\frac{2 n^{3}+7 n+1}{5 n+9} \\cdot \\frac{8 n+9}{6 n^{3}+8 n^{2}+3} \\,.\n\\] Prove that \\[\n\\lim_{n \\to \\infty}  a_n = \\frac{8}{15} \\,.\n\\]\n\nProof. The first fraction in \\((a_n)\\) does not converge, as it is unbounded. Therefore we cannot use Point 2 in Theorem 26 directly. However, we note that \\[\\begin{align*}\na_{n} & =\\frac{2 n^{3}+7 n+1}{5 n+9} \\cdot \\frac{8 n+9}{6 n^{3}+8 n^{2}+3} \\\\\n& = \\frac{8 n+9}{5 n+9} \\cdot \\frac{2 n^{3}+7 n+1}{6 n^{3}+8 n^{2}+3} \\,.\n\\end{align*}\\] Factoring out \\(n\\) and \\(n^3\\), respectively, and using the Algebra of Limits, we see that \\[\n\\frac{8 n+9}{5 n+9}=\\frac{8+9 / n}{5+9 / n} \\to \\frac{8+0}{5+0}=\\frac{8}{5}\n\\] and \\[\n\\frac{2+7 / n^{2}+1 / n^{3}}{6+8 / n+3 / n^{3}} \\to \\frac{2+0+0}{6+0+0}=\\frac{1}{3}\n\\] Therefore Theorem 26 Point 2 ensures that \\[\na_{n} \\to \\frac{8}{5} \\cdot \\frac{1}{3}=\\frac{8}{15} \\,.\n\\]"
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diff --git a/docs/sections/chap_4.html b/docs/sections/chap_4.html
index 2a642f8..a07e637 100644
--- a/docs/sections/chap_4.html
+++ b/docs/sections/chap_4.html
@@ -66,6 +66,10 @@
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@@ -122,15 +126,11 @@
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diff --git a/docs/sections/chap_5.html b/docs/sections/chap_5.html
index 6f98d10..726af06 100644
--- a/docs/sections/chap_5.html
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@@ -66,74 +66,74 @@
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diff --git a/docs/sections/chap_6.html b/docs/sections/chap_6.html
index 199c0e8..2d8b38d 100644
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@@ -66,67 +66,51 @@
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@@ -867,11 +867,10 @@ <h2 data-number="6.6" class="anchored" data-anchor-id="algebra-of-limits"><span
 \forall \, \varepsilon&gt; 0 \,, \, \exists  \, N \in \mathbb{N}\, \text{ s.t. } \, \forall \, n \geq N \,, \, \left|(a_{n} + b_n) - (a+b)  \right| &lt; \varepsilon\,.
 \]</span> Let <span class="math inline">\(\varepsilon&gt;0\)</span> and set <span class="math display">\[
 \widetilde{\varepsilon} := \frac{\varepsilon}{2} \,.
-\]</span> Since <span class="math inline">\(a_{n} \rightarrow a\)</span> and <span class="math inline">\(\widetilde{\varepsilon}&gt;0\)</span>, there exists <span class="math inline">\(N_1 \in \mathbb{N}\)</span> such that <span class="math display">\[
-|a_n - a|&lt; \widetilde{\varepsilon} \,, \quad \forall \, n \geq N_1 \,.
-\]</span> Since <span class="math inline">\(b_{n} \rightarrow b\)</span> and <span class="math inline">\(\widetilde{\varepsilon}&gt;0\)</span>, there exists <span class="math inline">\(N_2 \in \mathbb{N}\)</span> such that <span class="math display">\[
-|b_n - b|&lt; \widetilde{\varepsilon} \,, \quad \forall \, n \geq N_2 \,.
-\]</span> Define <span class="math display">\[
+\]</span> Since <span class="math inline">\(a_{n} \rightarrow a, b_n \to b\)</span>, and <span class="math inline">\(\widetilde{\varepsilon}&gt;0\)</span>, there exist <span class="math inline">\(N_1, N_2 \in \mathbb{N}\)</span> such that <span class="math display">\[\begin{align*}
+|a_n - a| &amp; &lt; \widetilde{\varepsilon} \,, \quad \forall \, n \geq N_1 \,, \\
+|b_n - b| &amp; &lt; \widetilde{\varepsilon} \,, \quad \forall \, n \geq N_2 \,.
+\end{align*}\]</span> Define <span class="math display">\[
 N : = \max  \{ N_1, N_2 \} \,.
 \]</span> For all <span class="math inline">\(n \geq N\)</span> we have, by the triangle inequality, <span class="math display">\[\begin{align*}
 \left|\left(a_{n}+b_{n}\right)-(a+b)\right|
@@ -889,11 +888,10 @@ <h2 data-number="6.6" class="anchored" data-anchor-id="algebra-of-limits"><span
 \left|a_{n}\right| \leq M \,, \quad \forall \, n \in \mathbb{N}\,.
 \]</span> Define <span class="math display">\[
 \widetilde{\varepsilon} = \frac{\varepsilon}{ M + |b| } \,.
-\]</span> Since <span class="math inline">\(a_{n} \rightarrow a\)</span> and <span class="math inline">\(\widetilde{\varepsilon} &gt; 0\)</span>, there exists <span class="math inline">\(N_1 \in \mathbb{N}\)</span> such that <span class="math display">\[
-| a_n  - a | &lt; \widetilde{\varepsilon} \,, \quad \forall \, n \geq N_1 \,.
-\]</span> Since <span class="math inline">\(b_{n} \rightarrow b\)</span> and <span class="math inline">\(\widetilde{\varepsilon} &gt; 0\)</span>, there exists <span class="math inline">\(N_2 \in \mathbb{N}\)</span> such that <span class="math display">\[
-| b_n  - b | &lt; \widetilde{\varepsilon} \,, \quad \forall \, n \geq N_2 \,.
-\]</span> Let <span class="math display">\[
+\]</span> Since <span class="math inline">\(a_{n} \rightarrow a, b_n \to b\)</span>, and <span class="math inline">\(\widetilde{\varepsilon} &gt; 0\)</span>, there exist <span class="math inline">\(N_1, N_2 \in \mathbb{N}\)</span> such that <span class="math display">\[\begin{align*}
+| a_n  - a | &amp; &lt; \widetilde{\varepsilon} \,, \quad \forall \, n \geq N_1 \,. \\
+| b_n  - b | &amp; &lt; \widetilde{\varepsilon} \,, \quad \forall \, n \geq N_2 \,.
+\end{align*}\]</span> Let <span class="math display">\[
 N:= \max\{ N_1, N_2 \} \,.
 \]</span> For all <span class="math inline">\(n \geq N\)</span> we have <span class="math display">\[\begin{align*}
 \left|a_{n}b_{n}-a b\right|
@@ -909,7 +907,7 @@ <h2 data-number="6.6" class="anchored" data-anchor-id="algebra-of-limits"><span
 \lim_{n \to \infty} \, \frac{a_n}{b_n} = \frac{a}{b} \,,
 \]</span> which is equivalent to <span class="math display">\[
 \forall \, \varepsilon&gt; 0 \,, \, \exists \, N \in \mathbb{N}\, \text{ s.t. } \, \forall \, n \geq N \,, \, \left|\frac{a_{n}}{ b_{n}} - \frac{a}{b} \right| &lt; \varepsilon\,.
-\]</span> We suppose in addition that <span class="math inline">\(b&gt;0\)</span>. The proof is very similar for the case <span class="math inline">\(b &lt;0\)</span>. Let <span class="math inline">\(\varepsilon&gt; 0\)</span>. Set <span class="math display">\[
+\]</span> We suppose in addition that <span class="math inline">\(b&gt;0\)</span>. The proof is very similar for the case <span class="math inline">\(b &lt;0\)</span>, and is hence omitted. Let <span class="math inline">\(\varepsilon&gt; 0\)</span>. Set <span class="math display">\[
 \delta := \frac{b}{2} \,.
 \]</span> Since <span class="math inline">\(b_n \to b\)</span> and <span class="math inline">\(\delta&gt;0\)</span>, there exists <span class="math inline">\(N_1 \in \mathbb{N}\)</span> such that <span class="math display">\[
 |b_n - b| &lt; \delta \, \quad \forall \, n \geq N_1 \,.
@@ -917,11 +915,10 @@ <h2 data-number="6.6" class="anchored" data-anchor-id="algebra-of-limits"><span
 b_n &gt; b - \delta = b - \frac{b}{2} = \frac{b}{2}   \, \quad \forall \, n \geq N_1 \,.
 \]</span> Define <span class="math display">\[
 \widetilde{\varepsilon} := \frac{ b^2 }{ 2 (b + |a|)} \, \varepsilon\,.
-\]</span> Since <span class="math inline">\(\widetilde{\varepsilon}&gt;0\)</span> and <span class="math inline">\(a_n \to a\)</span>, there exists <span class="math inline">\(N_2 \in \mathbb{N}\)</span> such that <span class="math display">\[
-|a_n - a |&lt; \widetilde{\varepsilon} \,, \quad \forall \, n \geq N_2 \,.
-\]</span> Since <span class="math inline">\(\widetilde{\varepsilon}&gt;0\)</span> and <span class="math inline">\(b_n \to b\)</span>, there exists <span class="math inline">\(N_3 \in \mathbb{N}\)</span> such that <span class="math display">\[
-|b_n - b |&lt; \widetilde{\varepsilon} \,, \quad \forall \, n \geq N_3 \,.
-\]</span> Define <span class="math display">\[
+\]</span> Since <span class="math inline">\(\widetilde{\varepsilon}&gt;0\)</span> and <span class="math inline">\(a_n \to a, b_n \to b\)</span>, there exist <span class="math inline">\(N_2, N_3 \in \mathbb{N}\)</span> such that <span class="math display">\[\begin{align*}
+|a_n - a | &amp; &lt; \widetilde{\varepsilon} \,, \quad \forall \, n \geq N_2 \,, \\
+|b_n - b | &amp; &lt; \widetilde{\varepsilon} \,, \quad \forall \, n \geq N_3 \,.
+\end{align*}\]</span> Define <span class="math display">\[
 N:= \max\{ N_1, N_2, N_3 \} \,.
 \]</span> For all <span class="math inline">\(n \geq N\)</span> we have <span class="math display">\[\begin{align*}
 \left|\frac{a_{n}}{ b_{n}} - \frac{a}{b} \right| &amp; =
diff --git a/docs/sections/chap_7.html b/docs/sections/chap_7.html
index 021eb11..ca0a082 100644
--- a/docs/sections/chap_7.html
+++ b/docs/sections/chap_7.html
@@ -66,43 +66,27 @@
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@@ -130,10 +106,34 @@
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diff --git a/docs/sections/chap_8.html b/docs/sections/chap_8.html
index b536551..4dd7ca4 100644
--- a/docs/sections/chap_8.html
+++ b/docs/sections/chap_8.html
@@ -66,23 +66,47 @@
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diff --git a/docs/sections/license.html b/docs/sections/license.html
index 0ca7d1d..a072c84 100644
--- a/docs/sections/license.html
+++ b/docs/sections/license.html
@@ -100,31 +100,39 @@
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 }
-.Proof {
-  --color1: #f1f1f2;
-  --color2: #c0c0c1;
-}
-.Axiom {
+.Question {
   --color1: #f4cce0;
   --color2: #db4d92;
 }
diff --git a/docs/sections/references.html b/docs/sections/references.html
index e62be71..d0e4bc7 100644
--- a/docs/sections/references.html
+++ b/docs/sections/references.html
@@ -84,74 +84,74 @@
   }
 }</script>
 <style>
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-  --color1: #bce5de;
-  --color2: #21aa93;
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-}
-.TODO {
-  --color1: #e7b1b4;
-  --color2: #8c3236;
-}
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+.Theorem {
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 }
-.Definition {
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diff --git a/sections/chap_6.qmd b/sections/chap_6.qmd
index bd6fb28..eacde3a 100644
--- a/sections/chap_6.qmd
+++ b/sections/chap_6.qmd
@@ -974,14 +974,11 @@ Let $\e>0$ and set
 $$
 \widetilde{\e} := \frac{\e}{2} \,.
 $$
-Since $a_{n} \rightarrow a$ and $\widetilde{\e}>0$, there exists $N_1 \in \N$ such that
-$$
-|a_n - a|< \widetilde{\e} \,, \quad \forall \, n \geq N_1 \,.
-$$
-Since $b_{n} \rightarrow b$ and $\widetilde{\e}>0$, there exists $N_2 \in \N$ such that
-$$
-|b_n - b|< \widetilde{\e} \,, \quad \forall \, n \geq N_2 \,.
-$$
+Since $a_{n} \rightarrow a, b_n \to b$, and $\widetilde{\e}>0$, there exist $N_1, N_2 \in \N$ such that
+\begin{align*}
+|a_n - a| & < \widetilde{\e} \,, \quad \forall \, n \geq N_1 \,, \\
+|b_n - b| & < \widetilde{\e} \,, \quad \forall \, n \geq N_2 \,.
+\end{align*}
 Define
 $$
 N : = \max  \{ N_1, N_2 \} \,.
@@ -1016,16 +1013,12 @@ Define
 $$
 \widetilde{\e} = \frac{\e}{ M + |b| } \,. 
 $$
-Since $a_{n} \rightarrow a$ and $\widetilde{\e} > 0$, there exists 
-$N_1 \in \N$ such that
-$$
-| a_n  - a | < \widetilde{\e} \,, \quad \forall \, n \geq N_1 \,.
-$$
-Since $b_{n} \rightarrow b$ and $\widetilde{\e} > 0$, there exists 
-$N_2 \in \N$ such that
-$$
-| b_n  - b | < \widetilde{\e} \,, \quad \forall \, n \geq N_2 \,.
-$$
+Since $a_{n} \rightarrow a, b_n \to b$, and $\widetilde{\e} > 0$, there exist 
+$N_1, N_2 \in \N$ such that
+\begin{align*}
+| a_n  - a | & < \widetilde{\e} \,, \quad \forall \, n \geq N_1 \,. \\
+| b_n  - b | & < \widetilde{\e} \,, \quad \forall \, n \geq N_2 \,.
+\end{align*}
 Let 
 $$
 N:= \max\{ N_1, N_2 \} \,.
@@ -1052,7 +1045,7 @@ which is equivalent to
 $$
 \forall \, \e > 0 \,, \, \exists \, N \in \N \st  \forall \, n \geq N \,, \, \left|\frac{a_{n}}{ b_{n}} - \frac{a}{b} \right| < \e \,.
 $$
-We suppose in addition that $b>0$. The proof is very similar for the case $b <0$. Let $\e > 0$. Set
+We suppose in addition that $b>0$. The proof is very similar for the case $b <0$, and is hence omitted. Let $\e > 0$. Set
 $$
 \delta := \frac{b}{2} \,.
 $$
@@ -1068,14 +1061,11 @@ Define
 $$
 \widetilde{\e} := \frac{ b^2 }{ 2 (b + |a|)} \, \e \,.
 $$
-Since $\widetilde{\e}>0$ and $a_n \to a$, there exists $N_2 \in \N$ such that
-$$
-|a_n - a |< \widetilde{\e} \,, \quad \forall \, n \geq N_2 \,.
-$$
-Since $\widetilde{\e}>0$ and $b_n \to b$, there exists $N_3 \in \N$ such that
-$$
-|b_n - b |< \widetilde{\e} \,, \quad \forall \, n \geq N_3 \,.
-$$
+Since $\widetilde{\e}>0$ and $a_n \to a, b_n \to b$, there exist $N_2, N_3 \in \N$ such that
+\begin{align*}
+|a_n - a | & < \widetilde{\e} \,, \quad \forall \, n \geq N_2 \,, \\
+|b_n - b | & < \widetilde{\e} \,, \quad \forall \, n \geq N_3 \,.
+\end{align*}
 Define
 $$
 N:= \max\{ N_1, N_2, N_3 \} \,.
diff --git a/sections/list-of-mathstuff.qmd b/sections/list-of-mathstuff.qmd
index 51731b6..89d7772 100644
--- a/sections/list-of-mathstuff.qmd
+++ b/sections/list-of-mathstuff.qmd
@@ -249124,3 +249124,1593 @@
  [**Remark 11**](chap_8.qmd#Remark-11)\Pageref{Remark-11}
 
  [**Example 12**](chap_8.qmd#Example-12)\Pageref{Example-12}
+
+ [**Definition 1**](chap_8.qmd#partial-sums): Partial sums\Pageref{partial-sums}
+
+ [**Definition 2**](chap_8.qmd#convergent-series-1): Convergent series\Pageref{convergent-series-1}
+
+ [**Definition 3**](chap_8.qmd#divergent-series): Divergent series\Pageref{divergent-series}
+
+ [**Example 4**](chap_8.qmd#Example-4)\Pageref{Example-4}
+
+ [**Example 5**](chap_8.qmd#Example-5)\Pageref{Example-5}
+
+ [**Theorem 6**](chap_8.qmd#theorem-series-necessary)\Pageref{theorem-series-necessary}
+
+ [**Proof**](chap_8.qmd#Proof*-1)\Pageref{Proof*-1}
+
+ [**Theorem 7**](chap_8.qmd#theorem-series-necessary-1)\Pageref{theorem-series-necessary-1}
+
+ [**Example 8**](chap_8.qmd#Example-8)\Pageref{Example-8}
+
+ [**Example 9**](chap_8.qmd#Example-9)\Pageref{Example-9}
+
+ [**Important**](chap_8.qmd#Important*-2)\Pageref{Important*-2}
+
+ [**Example 10**](chap_8.qmd#Example-10)\Pageref{Example-10}
+
+ [**Remark 11**](chap_8.qmd#Remark-11)\Pageref{Remark-11}
+
+ [**Example 12**](chap_8.qmd#Example-12)\Pageref{Example-12}
+
+ [**Remark 1**](chap_6.qmd#Remark-1)\Pageref{Remark-1}
+
+ [**Definition 2**](chap_6.qmd#sequence-of-real-numbers): Sequence of Real numbers\Pageref{sequence-of-real-numbers}
+
+ [**Notation 3**](chap_6.qmd#Notation-3)\Pageref{Notation-3}
+
+ [**Example 4**](chap_6.qmd#Example-4)\Pageref{Example-4}
+
+ [**Definition 5**](chap_6.qmd#definition-convergent-sequence): Convergent sequence\Pageref{definition-convergent-sequence}
+
+ [**Notation 6**](chap_6.qmd#Notation-6)\Pageref{Notation-6}
+
+ [**Remark 7**](chap_6.qmd#Remark-7)\Pageref{Remark-7}
+
+ [**Theorem 8**](chap_6.qmd#theorem-one-over-n)\Pageref{theorem-one-over-n}
+
+ [**Proof**](chap_6.qmd#proof-of-theorem-long-version): Proof of Theorem 8 (Long version)\Pageref{proof-of-theorem-long-version}
+
+ [**Proof**](chap_6.qmd#proof-of-theorem-short-version): Proof of Theorem 8 (Short version)\Pageref{proof-of-theorem-short-version}
+
+ [**Theorem 9**](chap_6.qmd#theorem-one-over-np)\Pageref{theorem-one-over-np}
+
+ [**Proof**](chap_6.qmd#Proof*-3)\Pageref{Proof*-3}
+
+ [**Question 10**](chap_6.qmd#Question-10)\Pageref{Question-10}
+
+ [**Important**](chap_6.qmd#Important*-4)\Pageref{Important*-4}
+
+ [**Example 11**](chap_6.qmd#Example-11)\Pageref{Example-11}
+
+ [**Theorem 12**](chap_6.qmd#theorem-constant-sequence)\Pageref{theorem-constant-sequence}
+
+ [**Proof**](chap_6.qmd#Proof*-5)\Pageref{Proof*-5}
+
+ [**Definition 13**](chap_6.qmd#divergent-sequence): Divergent sequence\Pageref{divergent-sequence}
+
+ [**Remark 14**](chap_6.qmd#Remark-14)\Pageref{Remark-14}
+
+ [**Theorem 15**](chap_6.qmd#theorem-1-minus-n)\Pageref{theorem-1-minus-n}
+
+ [**Proof**](chap_6.qmd#Proof*-6)\Pageref{Proof*-6}
+
+ [**Theorem 16**](chap_6.qmd#theorem-uniqueness-limit): Uniqueness of limit\Pageref{theorem-uniqueness-limit}
+
+ [**Proof**](chap_6.qmd#Proof*-7)\Pageref{Proof*-7}
+
+ [**Example 17**](chap_6.qmd#Example-17)\Pageref{Example-17}
+
+ [**Definition 18**](chap_6.qmd#definition-bounded-sequence): Bounded sequence\Pageref{definition-bounded-sequence}
+
+ [**Theorem 19**](chap_6.qmd#theorem-convergent-bounded)\Pageref{theorem-convergent-bounded}
+
+ [**Proof**](chap_6.qmd#Proof*-8)\Pageref{Proof*-8}
+
+ [**Example 20**](chap_6.qmd#Example-20)\Pageref{Example-20}
+
+ [**Warning**](chap_6.qmd#Warning*-9)\Pageref{Warning*-9}
+
+ [**Example 21**](chap_6.qmd#Example-21)\Pageref{Example-21}
+
+ [**Corollary 22**](chap_6.qmd#corollary-convergent-bounded)\Pageref{corollary-convergent-bounded}
+
+ [**Remark 23**](chap_6.qmd#Remark-23)\Pageref{Remark-23}
+
+ [**Theorem 24**](chap_6.qmd#Theorem-24)\Pageref{Theorem-24}
+
+ [**Proof**](chap_6.qmd#Proof*-10)\Pageref{Proof*-10}
+
+ [**Theorem 25**](chap_6.qmd#Theorem-25)\Pageref{Theorem-25}
+
+ [**Proof**](chap_6.qmd#Proof*-11)\Pageref{Proof*-11}
+
+ [**Theorem 26**](chap_6.qmd#theorem-algebra-limits): Algebra of limits\Pageref{theorem-algebra-limits}
+
+ [**Proof**](chap_6.qmd#Proof*-12)\Pageref{Proof*-12}
+
+ [**Example 27**](chap_6.qmd#example-limit-ploynomial)\Pageref{example-limit-ploynomial}
+
+ [**Important**](chap_6.qmd#Important*-13)\Pageref{Important*-13}
+
+ [**Example 28**](chap_6.qmd#example-algebra-of-limits-2)\Pageref{example-algebra-of-limits-2}
+
+ [**Example 29**](chap_6.qmd#Example-29)\Pageref{Example-29}
+
+ [**Warning**](chap_6.qmd#Warning*-14)\Pageref{Warning*-14}
+
+ [**Example 30**](chap_6.qmd#Example-30)\Pageref{Example-30}
+
+ [**Example 31**](chap_6.qmd#Example-31)\Pageref{Example-31}
+
+ [**Theorem 32**](chap_6.qmd#theorem-square-root-limit)\Pageref{theorem-square-root-limit}
+
+ [**Proof**](chap_6.qmd#Proof*-15)\Pageref{Proof*-15}
+
+ [**Example 33**](chap_6.qmd#Example-33)\Pageref{Example-33}
+
+ [**Example 34**](chap_6.qmd#Example-34)\Pageref{Example-34}
+
+ [**Theorem 35**](chap_6.qmd#theorem-squeeze): Squeeze theorem\Pageref{theorem-squeeze}
+
+ [**Proof**](chap_6.qmd#Proof*-16)\Pageref{Proof*-16}
+
+ [**Example 36**](chap_6.qmd#Example-36)\Pageref{Example-36}
+
+ [**Example 37**](chap_6.qmd#example-squeeze)\Pageref{example-squeeze}
+
+ [**Warning**](chap_6.qmd#Warning*-17)\Pageref{Warning*-17}
+
+ [**Example 38**](chap_6.qmd#Example-38)\Pageref{Example-38}
+
+ [**Definition 39**](chap_6.qmd#Definition-39)\Pageref{Definition-39}
+
+ [**Theorem 40**](chap_6.qmd#theorem-geometric-sequence): Geometric Sequence Test\Pageref{theorem-geometric-sequence}
+
+ [**Warning**](chap_6.qmd#Warning*-18)\Pageref{Warning*-18}
+
+ [**Lemma 41**](chap_6.qmd#lemma-bernoulli): Bernoulli’s inequality\Pageref{lemma-bernoulli}
+
+ [**Proof**](chap_6.qmd#Proof*-19)\Pageref{Proof*-19}
+
+ [**Proof**](chap_6.qmd#proof-of-theorem): Proof of Theorem 40\Pageref{proof-of-theorem}
+
+ [**Example 42**](chap_6.qmd#Example-42)\Pageref{Example-42}
+
+ [**Theorem 43**](chap_6.qmd#theorem-ratio-test): Ratio Test\Pageref{theorem-ratio-test}
+
+ [**Proof**](chap_6.qmd#Proof*-21)\Pageref{Proof*-21}
+
+ [**Example 44**](chap_6.qmd#Example-44)\Pageref{Example-44}
+
+ [**Example 45**](chap_6.qmd#Example-45)\Pageref{Example-45}
+
+ [**Example 46**](chap_6.qmd#Example-46)\Pageref{Example-46}
+
+ [**Warning**](chap_6.qmd#Warning*-22)\Pageref{Warning*-22}
+
+ [**Remark 47**](chap_6.qmd#Remark-47)\Pageref{Remark-47}
+
+ [**Definition 48**](chap_6.qmd#monotone-sequence): Monotone sequence\Pageref{monotone-sequence}
+
+ [**Example 49**](chap_6.qmd#Example-49)\Pageref{Example-49}
+
+ [**Theorem 50**](chap_6.qmd#theorem-monotone-convergence): Monotone Convergence Theorem\Pageref{theorem-monotone-convergence}
+
+ [**Proof**](chap_6.qmd#Proof*-23)\Pageref{Proof*-23}
+
+ [**Remark 1**](chap_6.qmd#Remark-1)\Pageref{Remark-1}
+
+ [**Definition 2**](chap_6.qmd#sequence-of-real-numbers): Sequence of Real numbers\Pageref{sequence-of-real-numbers}
+
+ [**Notation 3**](chap_6.qmd#Notation-3)\Pageref{Notation-3}
+
+ [**Example 4**](chap_6.qmd#Example-4)\Pageref{Example-4}
+
+ [**Definition 5**](chap_6.qmd#definition-convergent-sequence): Convergent sequence\Pageref{definition-convergent-sequence}
+
+ [**Notation 6**](chap_6.qmd#Notation-6)\Pageref{Notation-6}
+
+ [**Remark 7**](chap_6.qmd#Remark-7)\Pageref{Remark-7}
+
+ [**Theorem 8**](chap_6.qmd#theorem-one-over-n)\Pageref{theorem-one-over-n}
+
+ [**Proof**](chap_6.qmd#proof-of-theorem-long-version): Proof of Theorem 8 (Long version)\Pageref{proof-of-theorem-long-version}
+
+ [**Proof**](chap_6.qmd#proof-of-theorem-short-version): Proof of Theorem 8 (Short version)\Pageref{proof-of-theorem-short-version}
+
+ [**Theorem 9**](chap_6.qmd#theorem-one-over-np)\Pageref{theorem-one-over-np}
+
+ [**Proof**](chap_6.qmd#Proof*-3)\Pageref{Proof*-3}
+
+ [**Question 10**](chap_6.qmd#Question-10)\Pageref{Question-10}
+
+ [**Important**](chap_6.qmd#Important*-4)\Pageref{Important*-4}
+
+ [**Example 11**](chap_6.qmd#Example-11)\Pageref{Example-11}
+
+ [**Theorem 12**](chap_6.qmd#theorem-constant-sequence)\Pageref{theorem-constant-sequence}
+
+ [**Proof**](chap_6.qmd#Proof*-5)\Pageref{Proof*-5}
+
+ [**Definition 13**](chap_6.qmd#divergent-sequence): Divergent sequence\Pageref{divergent-sequence}
+
+ [**Remark 14**](chap_6.qmd#Remark-14)\Pageref{Remark-14}
+
+ [**Theorem 15**](chap_6.qmd#theorem-1-minus-n)\Pageref{theorem-1-minus-n}
+
+ [**Proof**](chap_6.qmd#Proof*-6)\Pageref{Proof*-6}
+
+ [**Theorem 16**](chap_6.qmd#theorem-uniqueness-limit): Uniqueness of limit\Pageref{theorem-uniqueness-limit}
+
+ [**Proof**](chap_6.qmd#Proof*-7)\Pageref{Proof*-7}
+
+ [**Example 17**](chap_6.qmd#Example-17)\Pageref{Example-17}
+
+ [**Definition 18**](chap_6.qmd#definition-bounded-sequence): Bounded sequence\Pageref{definition-bounded-sequence}
+
+ [**Theorem 19**](chap_6.qmd#theorem-convergent-bounded)\Pageref{theorem-convergent-bounded}
+
+ [**Proof**](chap_6.qmd#Proof*-8)\Pageref{Proof*-8}
+
+ [**Example 20**](chap_6.qmd#Example-20)\Pageref{Example-20}
+
+ [**Warning**](chap_6.qmd#Warning*-9)\Pageref{Warning*-9}
+
+ [**Example 21**](chap_6.qmd#Example-21)\Pageref{Example-21}
+
+ [**Corollary 22**](chap_6.qmd#corollary-convergent-bounded)\Pageref{corollary-convergent-bounded}
+
+ [**Remark 23**](chap_6.qmd#Remark-23)\Pageref{Remark-23}
+
+ [**Theorem 24**](chap_6.qmd#Theorem-24)\Pageref{Theorem-24}
+
+ [**Proof**](chap_6.qmd#Proof*-10)\Pageref{Proof*-10}
+
+ [**Theorem 25**](chap_6.qmd#Theorem-25)\Pageref{Theorem-25}
+
+ [**Proof**](chap_6.qmd#Proof*-11)\Pageref{Proof*-11}
+
+ [**Theorem 26**](chap_6.qmd#theorem-algebra-limits): Algebra of limits\Pageref{theorem-algebra-limits}
+
+ [**Proof**](chap_6.qmd#Proof*-12)\Pageref{Proof*-12}
+
+ [**Example 27**](chap_6.qmd#example-limit-ploynomial)\Pageref{example-limit-ploynomial}
+
+ [**Important**](chap_6.qmd#Important*-13)\Pageref{Important*-13}
+
+ [**Example 28**](chap_6.qmd#example-algebra-of-limits-2)\Pageref{example-algebra-of-limits-2}
+
+ [**Example 29**](chap_6.qmd#Example-29)\Pageref{Example-29}
+
+ [**Warning**](chap_6.qmd#Warning*-14)\Pageref{Warning*-14}
+
+ [**Example 30**](chap_6.qmd#Example-30)\Pageref{Example-30}
+
+ [**Example 31**](chap_6.qmd#Example-31)\Pageref{Example-31}
+
+ [**Theorem 32**](chap_6.qmd#theorem-square-root-limit)\Pageref{theorem-square-root-limit}
+
+ [**Proof**](chap_6.qmd#Proof*-15)\Pageref{Proof*-15}
+
+ [**Example 33**](chap_6.qmd#Example-33)\Pageref{Example-33}
+
+ [**Example 34**](chap_6.qmd#Example-34)\Pageref{Example-34}
+
+ [**Theorem 35**](chap_6.qmd#theorem-squeeze): Squeeze theorem\Pageref{theorem-squeeze}
+
+ [**Proof**](chap_6.qmd#Proof*-16)\Pageref{Proof*-16}
+
+ [**Example 36**](chap_6.qmd#Example-36)\Pageref{Example-36}
+
+ [**Example 37**](chap_6.qmd#example-squeeze)\Pageref{example-squeeze}
+
+ [**Warning**](chap_6.qmd#Warning*-17)\Pageref{Warning*-17}
+
+ [**Example 38**](chap_6.qmd#Example-38)\Pageref{Example-38}
+
+ [**Definition 39**](chap_6.qmd#Definition-39)\Pageref{Definition-39}
+
+ [**Theorem 40**](chap_6.qmd#theorem-geometric-sequence): Geometric Sequence Test\Pageref{theorem-geometric-sequence}
+
+ [**Warning**](chap_6.qmd#Warning*-18)\Pageref{Warning*-18}
+
+ [**Lemma 41**](chap_6.qmd#lemma-bernoulli): Bernoulli’s inequality\Pageref{lemma-bernoulli}
+
+ [**Proof**](chap_6.qmd#Proof*-19)\Pageref{Proof*-19}
+
+ [**Proof**](chap_6.qmd#proof-of-theorem): Proof of Theorem 40\Pageref{proof-of-theorem}
+
+ [**Example 42**](chap_6.qmd#Example-42)\Pageref{Example-42}
+
+ [**Theorem 43**](chap_6.qmd#theorem-ratio-test): Ratio Test\Pageref{theorem-ratio-test}
+
+ [**Proof**](chap_6.qmd#Proof*-21)\Pageref{Proof*-21}
+
+ [**Example 44**](chap_6.qmd#Example-44)\Pageref{Example-44}
+
+ [**Example 45**](chap_6.qmd#Example-45)\Pageref{Example-45}
+
+ [**Example 46**](chap_6.qmd#Example-46)\Pageref{Example-46}
+
+ [**Warning**](chap_6.qmd#Warning*-22)\Pageref{Warning*-22}
+
+ [**Remark 47**](chap_6.qmd#Remark-47)\Pageref{Remark-47}
+
+ [**Definition 48**](chap_6.qmd#monotone-sequence): Monotone sequence\Pageref{monotone-sequence}
+
+ [**Example 49**](chap_6.qmd#Example-49)\Pageref{Example-49}
+
+ [**Theorem 50**](chap_6.qmd#theorem-monotone-convergence): Monotone Convergence Theorem\Pageref{theorem-monotone-convergence}
+
+ [**Proof**](chap_6.qmd#Proof*-23)\Pageref{Proof*-23}
+
+ [**Remark 1**](chap_6.qmd#Remark-1)\Pageref{Remark-1}
+
+ [**Definition 2**](chap_6.qmd#sequence-of-real-numbers): Sequence of Real numbers\Pageref{sequence-of-real-numbers}
+
+ [**Notation 3**](chap_6.qmd#Notation-3)\Pageref{Notation-3}
+
+ [**Example 4**](chap_6.qmd#Example-4)\Pageref{Example-4}
+
+ [**Definition 5**](chap_6.qmd#definition-convergent-sequence): Convergent sequence\Pageref{definition-convergent-sequence}
+
+ [**Notation 6**](chap_6.qmd#Notation-6)\Pageref{Notation-6}
+
+ [**Remark 7**](chap_6.qmd#Remark-7)\Pageref{Remark-7}
+
+ [**Theorem 8**](chap_6.qmd#theorem-one-over-n)\Pageref{theorem-one-over-n}
+
+ [**Proof**](chap_6.qmd#proof-of-theorem-long-version): Proof of Theorem 8 (Long version)\Pageref{proof-of-theorem-long-version}
+
+ [**Proof**](chap_6.qmd#proof-of-theorem-short-version): Proof of Theorem 8 (Short version)\Pageref{proof-of-theorem-short-version}
+
+ [**Theorem 9**](chap_6.qmd#theorem-one-over-np)\Pageref{theorem-one-over-np}
+
+ [**Proof**](chap_6.qmd#Proof*-3)\Pageref{Proof*-3}
+
+ [**Question 10**](chap_6.qmd#Question-10)\Pageref{Question-10}
+
+ [**Important**](chap_6.qmd#Important*-4)\Pageref{Important*-4}
+
+ [**Example 11**](chap_6.qmd#Example-11)\Pageref{Example-11}
+
+ [**Theorem 12**](chap_6.qmd#theorem-constant-sequence)\Pageref{theorem-constant-sequence}
+
+ [**Proof**](chap_6.qmd#Proof*-5)\Pageref{Proof*-5}
+
+ [**Definition 13**](chap_6.qmd#divergent-sequence): Divergent sequence\Pageref{divergent-sequence}
+
+ [**Remark 14**](chap_6.qmd#Remark-14)\Pageref{Remark-14}
+
+ [**Theorem 15**](chap_6.qmd#theorem-1-minus-n)\Pageref{theorem-1-minus-n}
+
+ [**Proof**](chap_6.qmd#Proof*-6)\Pageref{Proof*-6}
+
+ [**Theorem 16**](chap_6.qmd#theorem-uniqueness-limit): Uniqueness of limit\Pageref{theorem-uniqueness-limit}
+
+ [**Proof**](chap_6.qmd#Proof*-7)\Pageref{Proof*-7}
+
+ [**Example 17**](chap_6.qmd#Example-17)\Pageref{Example-17}
+
+ [**Definition 18**](chap_6.qmd#definition-bounded-sequence): Bounded sequence\Pageref{definition-bounded-sequence}
+
+ [**Theorem 19**](chap_6.qmd#theorem-convergent-bounded)\Pageref{theorem-convergent-bounded}
+
+ [**Proof**](chap_6.qmd#Proof*-8)\Pageref{Proof*-8}
+
+ [**Example 20**](chap_6.qmd#Example-20)\Pageref{Example-20}
+
+ [**Warning**](chap_6.qmd#Warning*-9)\Pageref{Warning*-9}
+
+ [**Example 21**](chap_6.qmd#Example-21)\Pageref{Example-21}
+
+ [**Corollary 22**](chap_6.qmd#corollary-convergent-bounded)\Pageref{corollary-convergent-bounded}
+
+ [**Remark 23**](chap_6.qmd#Remark-23)\Pageref{Remark-23}
+
+ [**Theorem 24**](chap_6.qmd#Theorem-24)\Pageref{Theorem-24}
+
+ [**Proof**](chap_6.qmd#Proof*-10)\Pageref{Proof*-10}
+
+ [**Theorem 25**](chap_6.qmd#Theorem-25)\Pageref{Theorem-25}
+
+ [**Proof**](chap_6.qmd#Proof*-11)\Pageref{Proof*-11}
+
+ [**Theorem 26**](chap_6.qmd#theorem-algebra-limits): Algebra of limits\Pageref{theorem-algebra-limits}
+
+ [**Proof**](chap_6.qmd#Proof*-12)\Pageref{Proof*-12}
+
+ [**Example 27**](chap_6.qmd#example-limit-ploynomial)\Pageref{example-limit-ploynomial}
+
+ [**Important**](chap_6.qmd#Important*-13)\Pageref{Important*-13}
+
+ [**Example 28**](chap_6.qmd#example-algebra-of-limits-2)\Pageref{example-algebra-of-limits-2}
+
+ [**Example 29**](chap_6.qmd#Example-29)\Pageref{Example-29}
+
+ [**Warning**](chap_6.qmd#Warning*-14)\Pageref{Warning*-14}
+
+ [**Example 30**](chap_6.qmd#Example-30)\Pageref{Example-30}
+
+ [**Example 31**](chap_6.qmd#Example-31)\Pageref{Example-31}
+
+ [**Theorem 32**](chap_6.qmd#theorem-square-root-limit)\Pageref{theorem-square-root-limit}
+
+ [**Proof**](chap_6.qmd#Proof*-15)\Pageref{Proof*-15}
+
+ [**Example 33**](chap_6.qmd#Example-33)\Pageref{Example-33}
+
+ [**Example 34**](chap_6.qmd#Example-34)\Pageref{Example-34}
+
+ [**Theorem 35**](chap_6.qmd#theorem-squeeze): Squeeze theorem\Pageref{theorem-squeeze}
+
+ [**Proof**](chap_6.qmd#Proof*-16)\Pageref{Proof*-16}
+
+ [**Example 36**](chap_6.qmd#Example-36)\Pageref{Example-36}
+
+ [**Example 37**](chap_6.qmd#example-squeeze)\Pageref{example-squeeze}
+
+ [**Warning**](chap_6.qmd#Warning*-17)\Pageref{Warning*-17}
+
+ [**Example 38**](chap_6.qmd#Example-38)\Pageref{Example-38}
+
+ [**Definition 39**](chap_6.qmd#Definition-39)\Pageref{Definition-39}
+
+ [**Theorem 40**](chap_6.qmd#theorem-geometric-sequence): Geometric Sequence Test\Pageref{theorem-geometric-sequence}
+
+ [**Warning**](chap_6.qmd#Warning*-18)\Pageref{Warning*-18}
+
+ [**Lemma 41**](chap_6.qmd#lemma-bernoulli): Bernoulli’s inequality\Pageref{lemma-bernoulli}
+
+ [**Proof**](chap_6.qmd#Proof*-19)\Pageref{Proof*-19}
+
+ [**Proof**](chap_6.qmd#proof-of-theorem): Proof of Theorem 40\Pageref{proof-of-theorem}
+
+ [**Example 42**](chap_6.qmd#Example-42)\Pageref{Example-42}
+
+ [**Theorem 43**](chap_6.qmd#theorem-ratio-test): Ratio Test\Pageref{theorem-ratio-test}
+
+ [**Proof**](chap_6.qmd#Proof*-21)\Pageref{Proof*-21}
+
+ [**Example 44**](chap_6.qmd#Example-44)\Pageref{Example-44}
+
+ [**Example 45**](chap_6.qmd#Example-45)\Pageref{Example-45}
+
+ [**Example 46**](chap_6.qmd#Example-46)\Pageref{Example-46}
+
+ [**Warning**](chap_6.qmd#Warning*-22)\Pageref{Warning*-22}
+
+ [**Remark 47**](chap_6.qmd#Remark-47)\Pageref{Remark-47}
+
+ [**Definition 48**](chap_6.qmd#monotone-sequence): Monotone sequence\Pageref{monotone-sequence}
+
+ [**Example 49**](chap_6.qmd#Example-49)\Pageref{Example-49}
+
+ [**Theorem 50**](chap_6.qmd#theorem-monotone-convergence): Monotone Convergence Theorem\Pageref{theorem-monotone-convergence}
+
+ [**Proof**](chap_6.qmd#Proof*-23)\Pageref{Proof*-23}
+
+ [**Remark 1**](chap_6.qmd#Remark-1)\Pageref{Remark-1}
+
+ [**Definition 2**](chap_6.qmd#sequence-of-real-numbers): Sequence of Real numbers\Pageref{sequence-of-real-numbers}
+
+ [**Notation 3**](chap_6.qmd#Notation-3)\Pageref{Notation-3}
+
+ [**Example 4**](chap_6.qmd#Example-4)\Pageref{Example-4}
+
+ [**Definition 5**](chap_6.qmd#definition-convergent-sequence): Convergent sequence\Pageref{definition-convergent-sequence}
+
+ [**Notation 6**](chap_6.qmd#Notation-6)\Pageref{Notation-6}
+
+ [**Remark 7**](chap_6.qmd#Remark-7)\Pageref{Remark-7}
+
+ [**Theorem 8**](chap_6.qmd#theorem-one-over-n)\Pageref{theorem-one-over-n}
+
+ [**Proof**](chap_6.qmd#proof-of-theorem-long-version): Proof of Theorem 8 (Long version)\Pageref{proof-of-theorem-long-version}
+
+ [**Proof**](chap_6.qmd#proof-of-theorem-short-version): Proof of Theorem 8 (Short version)\Pageref{proof-of-theorem-short-version}
+
+ [**Theorem 9**](chap_6.qmd#theorem-one-over-np)\Pageref{theorem-one-over-np}
+
+ [**Proof**](chap_6.qmd#Proof*-3)\Pageref{Proof*-3}
+
+ [**Question 10**](chap_6.qmd#Question-10)\Pageref{Question-10}
+
+ [**Important**](chap_6.qmd#Important*-4)\Pageref{Important*-4}
+
+ [**Example 11**](chap_6.qmd#Example-11)\Pageref{Example-11}
+
+ [**Theorem 12**](chap_6.qmd#theorem-constant-sequence)\Pageref{theorem-constant-sequence}
+
+ [**Proof**](chap_6.qmd#Proof*-5)\Pageref{Proof*-5}
+
+ [**Definition 13**](chap_6.qmd#divergent-sequence): Divergent sequence\Pageref{divergent-sequence}
+
+ [**Remark 14**](chap_6.qmd#Remark-14)\Pageref{Remark-14}
+
+ [**Theorem 15**](chap_6.qmd#theorem-1-minus-n)\Pageref{theorem-1-minus-n}
+
+ [**Proof**](chap_6.qmd#Proof*-6)\Pageref{Proof*-6}
+
+ [**Theorem 16**](chap_6.qmd#theorem-uniqueness-limit): Uniqueness of limit\Pageref{theorem-uniqueness-limit}
+
+ [**Proof**](chap_6.qmd#Proof*-7)\Pageref{Proof*-7}
+
+ [**Example 17**](chap_6.qmd#Example-17)\Pageref{Example-17}
+
+ [**Definition 18**](chap_6.qmd#definition-bounded-sequence): Bounded sequence\Pageref{definition-bounded-sequence}
+
+ [**Theorem 19**](chap_6.qmd#theorem-convergent-bounded)\Pageref{theorem-convergent-bounded}
+
+ [**Proof**](chap_6.qmd#Proof*-8)\Pageref{Proof*-8}
+
+ [**Example 20**](chap_6.qmd#Example-20)\Pageref{Example-20}
+
+ [**Warning**](chap_6.qmd#Warning*-9)\Pageref{Warning*-9}
+
+ [**Example 21**](chap_6.qmd#Example-21)\Pageref{Example-21}
+
+ [**Corollary 22**](chap_6.qmd#corollary-convergent-bounded)\Pageref{corollary-convergent-bounded}
+
+ [**Remark 23**](chap_6.qmd#Remark-23)\Pageref{Remark-23}
+
+ [**Theorem 24**](chap_6.qmd#Theorem-24)\Pageref{Theorem-24}
+
+ [**Proof**](chap_6.qmd#Proof*-10)\Pageref{Proof*-10}
+
+ [**Theorem 25**](chap_6.qmd#Theorem-25)\Pageref{Theorem-25}
+
+ [**Proof**](chap_6.qmd#Proof*-11)\Pageref{Proof*-11}
+
+ [**Theorem 26**](chap_6.qmd#theorem-algebra-limits): Algebra of limits\Pageref{theorem-algebra-limits}
+
+ [**Proof**](chap_6.qmd#Proof*-12)\Pageref{Proof*-12}
+
+ [**Example 27**](chap_6.qmd#example-limit-ploynomial)\Pageref{example-limit-ploynomial}
+
+ [**Important**](chap_6.qmd#Important*-13)\Pageref{Important*-13}
+
+ [**Example 28**](chap_6.qmd#example-algebra-of-limits-2)\Pageref{example-algebra-of-limits-2}
+
+ [**Example 29**](chap_6.qmd#Example-29)\Pageref{Example-29}
+
+ [**Warning**](chap_6.qmd#Warning*-14)\Pageref{Warning*-14}
+
+ [**Example 30**](chap_6.qmd#Example-30)\Pageref{Example-30}
+
+ [**Example 31**](chap_6.qmd#Example-31)\Pageref{Example-31}
+
+ [**Theorem 32**](chap_6.qmd#theorem-square-root-limit)\Pageref{theorem-square-root-limit}
+
+ [**Proof**](chap_6.qmd#Proof*-15)\Pageref{Proof*-15}
+
+ [**Example 33**](chap_6.qmd#Example-33)\Pageref{Example-33}
+
+ [**Example 34**](chap_6.qmd#Example-34)\Pageref{Example-34}
+
+ [**Theorem 35**](chap_6.qmd#theorem-squeeze): Squeeze theorem\Pageref{theorem-squeeze}
+
+ [**Proof**](chap_6.qmd#Proof*-16)\Pageref{Proof*-16}
+
+ [**Example 36**](chap_6.qmd#Example-36)\Pageref{Example-36}
+
+ [**Example 37**](chap_6.qmd#example-squeeze)\Pageref{example-squeeze}
+
+ [**Warning**](chap_6.qmd#Warning*-17)\Pageref{Warning*-17}
+
+ [**Example 38**](chap_6.qmd#Example-38)\Pageref{Example-38}
+
+ [**Definition 39**](chap_6.qmd#Definition-39)\Pageref{Definition-39}
+
+ [**Theorem 40**](chap_6.qmd#theorem-geometric-sequence): Geometric Sequence Test\Pageref{theorem-geometric-sequence}
+
+ [**Warning**](chap_6.qmd#Warning*-18)\Pageref{Warning*-18}
+
+ [**Lemma 41**](chap_6.qmd#lemma-bernoulli): Bernoulli’s inequality\Pageref{lemma-bernoulli}
+
+ [**Proof**](chap_6.qmd#Proof*-19)\Pageref{Proof*-19}
+
+ [**Proof**](chap_6.qmd#proof-of-theorem): Proof of Theorem 40\Pageref{proof-of-theorem}
+
+ [**Example 42**](chap_6.qmd#Example-42)\Pageref{Example-42}
+
+ [**Theorem 43**](chap_6.qmd#theorem-ratio-test): Ratio Test\Pageref{theorem-ratio-test}
+
+ [**Proof**](chap_6.qmd#Proof*-21)\Pageref{Proof*-21}
+
+ [**Example 44**](chap_6.qmd#Example-44)\Pageref{Example-44}
+
+ [**Example 45**](chap_6.qmd#Example-45)\Pageref{Example-45}
+
+ [**Example 46**](chap_6.qmd#Example-46)\Pageref{Example-46}
+
+ [**Warning**](chap_6.qmd#Warning*-22)\Pageref{Warning*-22}
+
+ [**Remark 47**](chap_6.qmd#Remark-47)\Pageref{Remark-47}
+
+ [**Definition 48**](chap_6.qmd#monotone-sequence): Monotone sequence\Pageref{monotone-sequence}
+
+ [**Example 49**](chap_6.qmd#Example-49)\Pageref{Example-49}
+
+ [**Theorem 50**](chap_6.qmd#theorem-monotone-convergence): Monotone Convergence Theorem\Pageref{theorem-monotone-convergence}
+
+ [**Proof**](chap_6.qmd#Proof*-23)\Pageref{Proof*-23}
+
+ [**Remark 1**](chap_6.qmd#Remark-1)\Pageref{Remark-1}
+
+ [**Definition 2**](chap_6.qmd#sequence-of-real-numbers): Sequence of Real numbers\Pageref{sequence-of-real-numbers}
+
+ [**Notation 3**](chap_6.qmd#Notation-3)\Pageref{Notation-3}
+
+ [**Example 4**](chap_6.qmd#Example-4)\Pageref{Example-4}
+
+ [**Definition 5**](chap_6.qmd#definition-convergent-sequence): Convergent sequence\Pageref{definition-convergent-sequence}
+
+ [**Notation 6**](chap_6.qmd#Notation-6)\Pageref{Notation-6}
+
+ [**Remark 7**](chap_6.qmd#Remark-7)\Pageref{Remark-7}
+
+ [**Theorem 8**](chap_6.qmd#theorem-one-over-n)\Pageref{theorem-one-over-n}
+
+ [**Proof**](chap_6.qmd#proof-of-theorem-long-version): Proof of Theorem 8 (Long version)\Pageref{proof-of-theorem-long-version}
+
+ [**Proof**](chap_6.qmd#proof-of-theorem-short-version): Proof of Theorem 8 (Short version)\Pageref{proof-of-theorem-short-version}
+
+ [**Theorem 9**](chap_6.qmd#theorem-one-over-np)\Pageref{theorem-one-over-np}
+
+ [**Proof**](chap_6.qmd#Proof*-3)\Pageref{Proof*-3}
+
+ [**Question 10**](chap_6.qmd#Question-10)\Pageref{Question-10}
+
+ [**Important**](chap_6.qmd#Important*-4)\Pageref{Important*-4}
+
+ [**Example 11**](chap_6.qmd#Example-11)\Pageref{Example-11}
+
+ [**Theorem 12**](chap_6.qmd#theorem-constant-sequence)\Pageref{theorem-constant-sequence}
+
+ [**Proof**](chap_6.qmd#Proof*-5)\Pageref{Proof*-5}
+
+ [**Definition 13**](chap_6.qmd#divergent-sequence): Divergent sequence\Pageref{divergent-sequence}
+
+ [**Remark 14**](chap_6.qmd#Remark-14)\Pageref{Remark-14}
+
+ [**Theorem 15**](chap_6.qmd#theorem-1-minus-n)\Pageref{theorem-1-minus-n}
+
+ [**Proof**](chap_6.qmd#Proof*-6)\Pageref{Proof*-6}
+
+ [**Theorem 16**](chap_6.qmd#theorem-uniqueness-limit): Uniqueness of limit\Pageref{theorem-uniqueness-limit}
+
+ [**Proof**](chap_6.qmd#Proof*-7)\Pageref{Proof*-7}
+
+ [**Example 17**](chap_6.qmd#Example-17)\Pageref{Example-17}
+
+ [**Definition 18**](chap_6.qmd#definition-bounded-sequence): Bounded sequence\Pageref{definition-bounded-sequence}
+
+ [**Theorem 19**](chap_6.qmd#theorem-convergent-bounded)\Pageref{theorem-convergent-bounded}
+
+ [**Proof**](chap_6.qmd#Proof*-8)\Pageref{Proof*-8}
+
+ [**Example 20**](chap_6.qmd#Example-20)\Pageref{Example-20}
+
+ [**Warning**](chap_6.qmd#Warning*-9)\Pageref{Warning*-9}
+
+ [**Example 21**](chap_6.qmd#Example-21)\Pageref{Example-21}
+
+ [**Corollary 22**](chap_6.qmd#corollary-convergent-bounded)\Pageref{corollary-convergent-bounded}
+
+ [**Remark 23**](chap_6.qmd#Remark-23)\Pageref{Remark-23}
+
+ [**Theorem 24**](chap_6.qmd#Theorem-24)\Pageref{Theorem-24}
+
+ [**Proof**](chap_6.qmd#Proof*-10)\Pageref{Proof*-10}
+
+ [**Theorem 25**](chap_6.qmd#Theorem-25)\Pageref{Theorem-25}
+
+ [**Proof**](chap_6.qmd#Proof*-11)\Pageref{Proof*-11}
+
+ [**Theorem 26**](chap_6.qmd#theorem-algebra-limits): Algebra of limits\Pageref{theorem-algebra-limits}
+
+ [**Proof**](chap_6.qmd#Proof*-12)\Pageref{Proof*-12}
+
+ [**Example 27**](chap_6.qmd#example-limit-ploynomial)\Pageref{example-limit-ploynomial}
+
+ [**Important**](chap_6.qmd#Important*-13)\Pageref{Important*-13}
+
+ [**Example 28**](chap_6.qmd#example-algebra-of-limits-2)\Pageref{example-algebra-of-limits-2}
+
+ [**Example 29**](chap_6.qmd#Example-29)\Pageref{Example-29}
+
+ [**Warning**](chap_6.qmd#Warning*-14)\Pageref{Warning*-14}
+
+ [**Example 30**](chap_6.qmd#Example-30)\Pageref{Example-30}
+
+ [**Example 31**](chap_6.qmd#Example-31)\Pageref{Example-31}
+
+ [**Theorem 32**](chap_6.qmd#theorem-square-root-limit)\Pageref{theorem-square-root-limit}
+
+ [**Proof**](chap_6.qmd#Proof*-15)\Pageref{Proof*-15}
+
+ [**Example 33**](chap_6.qmd#Example-33)\Pageref{Example-33}
+
+ [**Example 34**](chap_6.qmd#Example-34)\Pageref{Example-34}
+
+ [**Theorem 35**](chap_6.qmd#theorem-squeeze): Squeeze theorem\Pageref{theorem-squeeze}
+
+ [**Proof**](chap_6.qmd#Proof*-16)\Pageref{Proof*-16}
+
+ [**Example 36**](chap_6.qmd#Example-36)\Pageref{Example-36}
+
+ [**Example 37**](chap_6.qmd#example-squeeze)\Pageref{example-squeeze}
+
+ [**Warning**](chap_6.qmd#Warning*-17)\Pageref{Warning*-17}
+
+ [**Example 38**](chap_6.qmd#Example-38)\Pageref{Example-38}
+
+ [**Definition 39**](chap_6.qmd#Definition-39)\Pageref{Definition-39}
+
+ [**Theorem 40**](chap_6.qmd#theorem-geometric-sequence): Geometric Sequence Test\Pageref{theorem-geometric-sequence}
+
+ [**Warning**](chap_6.qmd#Warning*-18)\Pageref{Warning*-18}
+
+ [**Lemma 41**](chap_6.qmd#lemma-bernoulli): Bernoulli’s inequality\Pageref{lemma-bernoulli}
+
+ [**Proof**](chap_6.qmd#Proof*-19)\Pageref{Proof*-19}
+
+ [**Proof**](chap_6.qmd#proof-of-theorem): Proof of Theorem 40\Pageref{proof-of-theorem}
+
+ [**Example 42**](chap_6.qmd#Example-42)\Pageref{Example-42}
+
+ [**Theorem 43**](chap_6.qmd#theorem-ratio-test): Ratio Test\Pageref{theorem-ratio-test}
+
+ [**Proof**](chap_6.qmd#Proof*-21)\Pageref{Proof*-21}
+
+ [**Example 44**](chap_6.qmd#Example-44)\Pageref{Example-44}
+
+ [**Example 45**](chap_6.qmd#Example-45)\Pageref{Example-45}
+
+ [**Example 46**](chap_6.qmd#Example-46)\Pageref{Example-46}
+
+ [**Warning**](chap_6.qmd#Warning*-22)\Pageref{Warning*-22}
+
+ [**Remark 47**](chap_6.qmd#Remark-47)\Pageref{Remark-47}
+
+ [**Definition 48**](chap_6.qmd#monotone-sequence): Monotone sequence\Pageref{monotone-sequence}
+
+ [**Example 49**](chap_6.qmd#Example-49)\Pageref{Example-49}
+
+ [**Theorem 50**](chap_6.qmd#theorem-monotone-convergence): Monotone Convergence Theorem\Pageref{theorem-monotone-convergence}
+
+ [**Proof**](chap_6.qmd#Proof*-23)\Pageref{Proof*-23}
+
+ [**Definition 1**](chap_7.qmd#sequence-of-complex-numbers): Sequence of Complex numbers\Pageref{sequence-of-complex-numbers}
+
+ [**Definition 2**](chap_7.qmd#definition-convergent-sequence-C): Convergent sequence in $\mathbb{C}$\Pageref{definition-convergent-sequence-C}
+
+ [**Important**](chap_7.qmd#Important*-1)\Pageref{Important*-1}
+
+ [**Example 3**](chap_7.qmd#Example-3)\Pageref{Example-3}
+
+ [**Definition 4**](chap_7.qmd#bounded-sequence-in-mathbbc): Bounded sequence in $\mathbb{C}$\Pageref{bounded-sequence-in-mathbbc}
+
+ [**Theorem 5**](chap_7.qmd#theorem-convergent-bounded-C)\Pageref{theorem-convergent-bounded-C}
+
+ [**Definition 6**](chap_7.qmd#divergent-sequences-in-mathbbc): Divergent sequences in $\mathbb{C}$\Pageref{divergent-sequences-in-mathbbc}
+
+ [**Corollary 7**](chap_7.qmd#Corollary-7)\Pageref{Corollary-7}
+
+ [**Theorem 8**](chap_7.qmd#theorem-algebra-limits-C): Algebra of limits in $\mathbb{C}$\Pageref{theorem-algebra-limits-C}
+
+ [**Example 9**](chap_7.qmd#Example-9)\Pageref{Example-9}
+
+ [**Theorem 10**](chap_7.qmd#theorem-convergence-zero-C)\Pageref{theorem-convergence-zero-C}
+
+ [**Proof**](chap_7.qmd#Proof*-2)\Pageref{Proof*-2}
+
+ [**Example 11**](chap_7.qmd#Example-11)\Pageref{Example-11}
+
+ [**Example 12**](chap_7.qmd#Example-12)\Pageref{Example-12}
+
+ [**Theorem 13**](chap_7.qmd#geometric-sequence-test-in-mathbbc): Geometric sequence Test in $\mathbb{C}$\Pageref{geometric-sequence-test-in-mathbbc}
+
+ [**Example 14**](chap_7.qmd#Example-14)\Pageref{Example-14}
+
+ [**Theorem 15**](chap_7.qmd#theorem-ratop-test-C): Ratio Test in \Pageref{theorem-ratop-test-C}
+
+ [**Example 16**](chap_7.qmd#Example-16)\Pageref{Example-16}
+
+ [**Theorem 17**](chap_7.qmd#theorem-convergence-real-imaginary)\Pageref{theorem-convergence-real-imaginary}
+
+ [**Proof**](chap_7.qmd#Proof*-3)\Pageref{Proof*-3}
+
+ [**Example 18**](chap_7.qmd#Example-18)\Pageref{Example-18}
+
+ [**Question 1**](chap_1.qmd#Question-1)\Pageref{Question-1}
+
+ [**Question 2**](chap_1.qmd#Question-2)\Pageref{Question-2}
+
+ [**Question 3**](chap_1.qmd#Question-3)\Pageref{Question-3}
+
+ [**Question 4**](chap_1.qmd#Question-4)\Pageref{Question-4}
+
+ [**Conjecture 5**](chap_1.qmd#conj-line)\Pageref{conj-line}
+
+ [**Theorem 6**](chap_1.qmd#theorem-root-2)\Pageref{theorem-root-2}
+
+ [**Proof**](chap_1.qmd#proof-of-theorem): Proof of Theorem 6\Pageref{proof-of-theorem}
+
+ [**Remark 7**](chap_1.qmd#Remark-7)\Pageref{Remark-7}
+
+ [**Theorem 8**](chap_1.qmd#we-will-prove-this-in-the-future): We will prove this in the future\Pageref{we-will-prove-this-in-the-future}
+
+ [**Theorem 9**](chap_1.qmd#we-will-prove-this-in-the-future-1): We will prove this in the future\Pageref{we-will-prove-this-in-the-future-1}
+
+ [**Remark 1**](chap_2.qmd#Remark-1)\Pageref{Remark-1}
+
+ [**Example 2**](chap_2.qmd#Example-2)\Pageref{Example-2}
+
+ [**Example 3**](chap_2.qmd#Example-3)\Pageref{Example-3}
+
+ [**Example 4**](chap_2.qmd#Example-4)\Pageref{Example-4}
+
+ [**Example 5**](chap_2.qmd#Example-5)\Pageref{Example-5}
+
+ [**Example 6**](chap_2.qmd#Example-6)\Pageref{Example-6}
+
+ [**Proposition 7**](chap_2.qmd#Proposition-Equality-Sets)\Pageref{Proposition-Equality-Sets}
+
+ [**Proof**](chap_2.qmd#Proof*-1)\Pageref{Proof*-1}
+
+ [**Example 8**](chap_2.qmd#Example-8)\Pageref{Example-8}
+
+ [**Remark 9**](chap_2.qmd#Remark-9)\Pageref{Remark-9}
+
+ [**Example 10**](chap_2.qmd#Example-10)\Pageref{Example-10}
+
+ [**Proposition 11**](chap_2.qmd#de-morgans-laws): De Morgan’s Laws\Pageref{de-morgans-laws}
+
+ [**Remark 12**](chap_2.qmd#Remark-12)\Pageref{Remark-12}
+
+ [**Example 13**](chap_2.qmd#Example-13)\Pageref{Example-13}
+
+ [**Definition 14**](chap_2.qmd#equivalence-relation-1): Equivalence relation\Pageref{equivalence-relation-1}
+
+ [**Definition 15**](chap_2.qmd#equivalence-classes): Equivalence classes\Pageref{equivalence-classes}
+
+ [**Example 16**](chap_2.qmd#equality-is-an-equivalence-relation): Equality is an equivalence relation\Pageref{equality-is-an-equivalence-relation}
+
+ [**Example 17**](chap_2.qmd#Example-17)\Pageref{Example-17}
+
+ [**Definition 18**](chap_2.qmd#partial-order): Partial order\Pageref{partial-order}
+
+ [**Definition 19**](chap_2.qmd#total-order): Total order\Pageref{total-order}
+
+ [**Example 20**](chap_2.qmd#set-inclusion-is-a-partial-order): Set inclusion is a partial order\Pageref{set-inclusion-is-a-partial-order}
+
+ [**Example 21**](chap_2.qmd#inequality-is-a-total-order): Inequality is a total order\Pageref{inequality-is-a-total-order}
+
+ [**Notation 22**](chap_2.qmd#Notation-22)\Pageref{Notation-22}
+
+ [**Definition 23**](chap_2.qmd#Definition-23)\Pageref{Definition-23}
+
+ [**Definition 24**](chap_2.qmd#definition-function): Functions\Pageref{definition-function}
+
+ [**Warning**](chap_2.qmd#Warning*-2)\Pageref{Warning*-2}
+
+ [**Example 25**](chap_2.qmd#Example-25)\Pageref{Example-25}
+
+ [**Example 26**](chap_2.qmd#Example-26)\Pageref{Example-26}
+
+ [**Definition 27**](chap_2.qmd#absolute-value): Absolute value\Pageref{absolute-value}
+
+ [**Example 28**](chap_2.qmd#Example-28)\Pageref{Example-28}
+
+ [**Remark 29**](chap_2.qmd#Remark-29)\Pageref{Remark-29}
+
+ [**Remark 30**](chap_2.qmd#rmk-absolute-value-sym)\Pageref{rmk-absolute-value-sym}
+
+ [**Remark 31**](chap_2.qmd#geometric-interpretation-of-x): Geometric interpretation of $|x|$\Pageref{geometric-interpretation-of-x}
+
+ [**Remark 32**](chap_2.qmd#geometric-interpretation-of-x-y): Geometric interpretation of $|x-y|$\Pageref{geometric-interpretation-of-x-y}
+
+ [**Lemma 33**](chap_2.qmd#lemma-absolute-value)\Pageref{lemma-absolute-value}
+
+ [**Proof**](chap_2.qmd#proof-of-lemma): Proof of Lemma 33\Pageref{proof-of-lemma}
+
+ [**Lemma 34**](chap_2.qmd#Lemma-34)\Pageref{Lemma-34}
+
+ [**Theorem 35**](chap_2.qmd#theorem-triangle-inequality): Triangle inequality\Pageref{theorem-triangle-inequality}
+
+ [**Remark 36**](chap_2.qmd#geometric-meaning-of-triangle-inequality): Geometric meaning of triangle inequality\Pageref{geometric-meaning-of-triangle-inequality}
+
+ [**Proof**](chap_2.qmd#proof-of-theorem): Proof of Theorem 35\Pageref{proof-of-theorem}
+
+ [**Remark 37**](chap_2.qmd#Remark-37)\Pageref{Remark-37}
+
+ [**Proposition 38**](chap_2.qmd#proposition-easy)\Pageref{proposition-easy}
+
+ [**Proof**](chap_2.qmd#of-proposition): of Proposition 38\Pageref{of-proposition}
+
+ [**Axiom 39**](chap_2.qmd#principle-of-induction): Principle of Induction\Pageref{principle-of-induction}
+
+ [**Important**](chap_2.qmd#Important*-6)\Pageref{Important*-6}
+
+ [**Remark 40**](chap_2.qmd#Remark-40)\Pageref{Remark-40}
+
+ [**Corollary 41**](chap_2.qmd#principle-of-inducion---alternative-formulation): Principle of Inducion - Alternative formulation\Pageref{principle-of-inducion---alternative-formulation}
+
+ [**Proof**](chap_2.qmd#Proof*-7)\Pageref{Proof*-7}
+
+ [**Example 42**](chap_2.qmd#formula-for-summing-first-n-natural-numbers): Formula for summing first $n$ natural numbers\Pageref{formula-for-summing-first-n-natural-numbers}
+
+ [**Example 43**](chap_2.qmd#statements-about-sequences-of-numbers): Statements about sequences of numbers\Pageref{statements-about-sequences-of-numbers}
+
+ [**Definition 1**](chap_3.qmd#binary-operation): Binary operation\Pageref{binary-operation}
+
+ [**Notation 2**](chap_3.qmd#Notation-2)\Pageref{Notation-2}
+
+ [**Example 3**](chap_3.qmd#example-binary-operation): of binary operation\Pageref{example-binary-operation}
+
+ [**Definition 4**](chap_3.qmd#Definition-4)\Pageref{Definition-4}
+
+ [**Example 5**](chap_3.qmd#example-binary-operation-2)\Pageref{example-binary-operation-2}
+
+ [**Example 6**](chap_3.qmd#Example-6)\Pageref{Example-6}
+
+ [**Definition 7**](chap_3.qmd#field): Field\Pageref{field}
+
+ [**Example 8**](chap_3.qmd#Example-8)\Pageref{Example-8}
+
+ [**Definition 9**](chap_3.qmd#subtraction-and-division): Subtraction and division\Pageref{subtraction-and-division}
+
+ [**Proposition 10**](chap_3.qmd#uniqueness-of-neutral-elements-and-inverses): Uniqueness of neutral elements and inverses\Pageref{uniqueness-of-neutral-elements-and-inverses}
+
+ [**Proof**](chap_3.qmd#Proof*-1)\Pageref{Proof*-1}
+
+ [**Proposition 11**](chap_3.qmd#properties-of-field-operations): Properties of field operations\Pageref{properties-of-field-operations}
+
+ [**Theorem 12**](chap_3.qmd#Theorem-12)\Pageref{Theorem-12}
+
+ [**Definition 13**](chap_3.qmd#Definition-13)\Pageref{Definition-13}
+
+ [**Example 14**](chap_3.qmd#Example-14)\Pageref{Example-14}
+
+ [**Definition 15**](chap_3.qmd#partition-of-a-set): Partition of a set\Pageref{partition-of-a-set}
+
+ [**Definition 16**](chap_3.qmd#cut-of-a-set): Cut of a set\Pageref{cut-of-a-set}
+
+ [**Definition 17**](chap_3.qmd#cut-property-1): Cut property\Pageref{cut-property-1}
+
+ [**Example 18**](chap_3.qmd#Example-18)\Pageref{Example-18}
+
+ [**Question 19**](chap_3.qmd#Question-19)\Pageref{Question-19}
+
+ [**Theorem 20**](chap_3.qmd#theorem-Q-cut): $\mathbb{Q}$ does not have the cut property.\Pageref{theorem-Q-cut}
+
+ [**Remark 21**](chap_3.qmd#remark-idea-proof-cut): Ideas for the proof of Theorem 20\Pageref{remark-idea-proof-cut}
+
+ [**Proof**](chap_3.qmd#proof-of-theorem): Proof of Theorem 20\Pageref{proof-of-theorem}
+
+ [**Remark 22**](chap_3.qmd#Remark-22)\Pageref{Remark-22}
+
+ [**Example 23**](chap_3.qmd#intuition-about-supremum-and-infimum): Intuition about supremum and infimum\Pageref{intuition-about-supremum-and-infimum}
+
+ [**Definition 24**](chap_3.qmd#upper-bound-and-bounded-above): Upper bound and bounded above\Pageref{upper-bound-and-bounded-above}
+
+ [**Definition 25**](chap_3.qmd#supremum): Supremum\Pageref{supremum}
+
+ [**Notation 26**](chap_3.qmd#Notation-26)\Pageref{Notation-26}
+
+ [**Remark 27**](chap_3.qmd#Remark-27)\Pageref{Remark-27}
+
+ [**Proposition 28**](chap_3.qmd#Proposition-28)\Pageref{Proposition-28}
+
+ [**Proof**](chap_3.qmd#Proof*-3)\Pageref{Proof*-3}
+
+ [**Warning**](chap_3.qmd#Warning*-4)\Pageref{Warning*-4}
+
+ [**Warning**](chap_3.qmd#Warning*-5)\Pageref{Warning*-5}
+
+ [**Definition 29**](chap_3.qmd#maximum): Maximum\Pageref{maximum}
+
+ [**Proposition 30**](chap_3.qmd#Proposition-30)\Pageref{Proposition-30}
+
+ [**Proof**](chap_3.qmd#Proof*-6)\Pageref{Proof*-6}
+
+ [**Warning**](chap_3.qmd#Warning*-7)\Pageref{Warning*-7}
+
+ [**Definition 31**](chap_3.qmd#upper-bound-bounded-below-infimum-minimum): Upper bound, bounded below, infimum, minimum\Pageref{upper-bound-bounded-below-infimum-minimum}
+
+ [**Proposition 32**](chap_3.qmd#Proposition-32)\Pageref{Proposition-32}
+
+ [**Warning**](chap_3.qmd#Warning*-8)\Pageref{Warning*-8}
+
+ [**Warning**](chap_3.qmd#Warning*-9)\Pageref{Warning*-9}
+
+ [**Warning**](chap_3.qmd#Warning*-10)\Pageref{Warning*-10}
+
+ [**Proposition 33**](chap_3.qmd#Proposition-33)\Pageref{Proposition-33}
+
+ [**Proposition 34**](chap_3.qmd#proposition-inf-sup): Relationship between sup and inf\Pageref{proposition-inf-sup}
+
+ [**Question 35**](chap_3.qmd#Question-35)\Pageref{Question-35}
+
+ [**Theorem 36**](chap_3.qmd#theorem-Q-not-complete)\Pageref{theorem-Q-not-complete}
+
+ [**Proof**](chap_3.qmd#Proof*-11)\Pageref{Proof*-11}
+
+ [**Question 37**](chap_3.qmd#Question-37)\Pageref{Question-37}
+
+ [**Definition 38**](chap_3.qmd#completeness-1): Completeness\Pageref{completeness-1}
+
+ [**Notation 39**](chap_3.qmd#Notation-39)\Pageref{Notation-39}
+
+ [**Proposition 40**](chap_3.qmd#Proposition-40)\Pageref{Proposition-40}
+
+ [**Proof**](chap_3.qmd#Proof*-12)\Pageref{Proof*-12}
+
+ [**Theorem 41**](chap_3.qmd#theorem-cut-complete): Equivalence of Cut Property and Completeness\Pageref{theorem-cut-complete}
+
+ [**Remark 42**](chap_3.qmd#ideas-for-proving-theorem): Ideas for proving Theorem 41\Pageref{ideas-for-proving-theorem}
+
+ [**Proof**](chap_3.qmd#proof-of-theorem-1): Proof of Theorem 41\Pageref{proof-of-theorem-1}
+
+ [**Definition 43**](chap_3.qmd#definition-R): System of Real Numbers $\mathbb{R}$\Pageref{definition-R}
+
+ [**Remark 44**](chap_3.qmd#Remark-44)\Pageref{Remark-44}
+
+ [**Notation 45**](chap_3.qmd#Notation-45)\Pageref{Notation-45}
+
+ [**Remark 46**](chap_3.qmd#Remark-46)\Pageref{Remark-46}
+
+ [**Important**](chap_3.qmd#Important*-14)\Pageref{Important*-14}
+
+ [**Question 47**](chap_3.qmd#Question-47)\Pageref{Question-47}
+
+ [**Definition 48**](chap_3.qmd#inductive-set): Inductive set\Pageref{inductive-set}
+
+ [**Example 49**](chap_3.qmd#Example-49)\Pageref{Example-49}
+
+ [**Proposition 50**](chap_3.qmd#Proposition-50)\Pageref{Proposition-50}
+
+ [**Proof**](chap_3.qmd#Proof*-15)\Pageref{Proof*-15}
+
+ [**Definition 51**](chap_3.qmd#definition-natural-numbers): Set of Natural Numbers\Pageref{definition-natural-numbers}
+
+ [**Proposition 52**](chap_3.qmd#proposition-smallest-inductive): ${\mathbb{N}}_{\mathbb{R}}$ is the smallest inductive subset of $\mathbb{R}$\Pageref{proposition-smallest-inductive}
+
+ [**Proof**](chap_3.qmd#Proof*-16)\Pageref{Proof*-16}
+
+ [**Theorem 53**](chap_3.qmd#theorem-n-1)\Pageref{theorem-n-1}
+
+ [**Proof**](chap_3.qmd#Proof*-17)\Pageref{Proof*-17}
+
+ [**Notation 54**](chap_3.qmd#Notation-54)\Pageref{Notation-54}
+
+ [**Theorem 55**](chap_3.qmd#theorem-induction-new): Principle of Induction\Pageref{theorem-induction-new}
+
+ [**Proof**](chap_3.qmd#Proof*-18)\Pageref{Proof*-18}
+
+ [**Theorem 56**](chap_3.qmd#theorem-N-closed)\Pageref{theorem-N-closed}
+
+ [**Proof**](chap_3.qmd#Proof*-19)\Pageref{Proof*-19}
+
+ [**Theorem 57**](chap_3.qmd#Theorem-57)\Pageref{Theorem-57}
+
+ [**Definition 58**](chap_3.qmd#set-of-integers): Set of Integers\Pageref{set-of-integers}
+
+ [**Theorem 59**](chap_3.qmd#theorem-equivalent-Z)\Pageref{theorem-equivalent-Z}
+
+ [**Proof**](chap_3.qmd#Proof*-20)\Pageref{Proof*-20}
+
+ [**Theorem 60**](chap_3.qmd#theorem-Z-closed)\Pageref{theorem-Z-closed}
+
+ [**Theorem 61**](chap_3.qmd#theorem-Z-axioms)\Pageref{theorem-Z-axioms}
+
+ [**Proof**](chap_3.qmd#Proof*-21)\Pageref{Proof*-21}
+
+ [**Remark 62**](chap_3.qmd#remark-Z-axioms)\Pageref{remark-Z-axioms}
+
+ [**Definition 63**](chap_3.qmd#set-of-rational-numbers): Set of Rational Numbers\Pageref{set-of-rational-numbers}
+
+ [**Theorem 64**](chap_3.qmd#Theorem-64)\Pageref{Theorem-64}
+
+ [**Proof**](chap_3.qmd#Proof*-22)\Pageref{Proof*-22}
+
+ [**Theorem 65**](chap_3.qmd#Theorem-65)\Pageref{Theorem-65}
+
+ [**Notation 66**](chap_3.qmd#Notation-66)\Pageref{Notation-66}
+
+ [**Remark 1**](chap_4.qmd#Remark-1)\Pageref{Remark-1}
+
+ [**Theorem 2**](chap_4.qmd#theorem-archimedean): Archimedean Property\Pageref{theorem-archimedean}
+
+ [**Proof**](chap_4.qmd#Proof*-1)\Pageref{Proof*-1}
+
+ [**Theorem 3**](chap_4.qmd#theorem-archimedean-alternative): Archimedean Property (Alternative formulation)\Pageref{theorem-archimedean-alternative}
+
+ [**Proof**](chap_4.qmd#Proof*-2)\Pageref{Proof*-2}
+
+ [**Question 4**](chap_4.qmd#Question-4)\Pageref{Question-4}
+
+ [**Theorem 5**](chap_4.qmd#theorem-nested): Nested Interval Property\Pageref{theorem-nested}
+
+ [**Proof**](chap_4.qmd#Proof*-3)\Pageref{Proof*-3}
+
+ [**Important**](chap_4.qmd#Important*-4)\Pageref{Important*-4}
+
+ [**Example 6**](chap_4.qmd#example-nested-open)\Pageref{example-nested-open}
+
+ [**Proposition 7**](chap_4.qmd#proposition-supremum-characterization): Characterization of Supremum\Pageref{proposition-supremum-characterization}
+
+ [**Proof**](chap_4.qmd#proof-of-proposition): Proof of Proposition 7\Pageref{proof-of-proposition}
+
+ [**Proposition 8**](chap_4.qmd#prop-supremum-infimum): Characterization of Infimum\Pageref{prop-supremum-infimum}
+
+ [**Proposition 9**](chap_4.qmd#proposition-interval-inf-sup)\Pageref{proposition-interval-inf-sup}
+
+ [**Proof**](chap_4.qmd#Proof*-6)\Pageref{Proof*-6}
+
+ [**Corollary 10**](chap_4.qmd#Corollary-10)\Pageref{Corollary-10}
+
+ [**Proof**](chap_4.qmd#Proof*-7)\Pageref{Proof*-7}
+
+ [**Corollary 11**](chap_4.qmd#Corollary-11)\Pageref{Corollary-11}
+
+ [**Proposition 12**](chap_4.qmd#Proposition-12)\Pageref{Proposition-12}
+
+ [**Proof**](chap_4.qmd#Proof*-8)\Pageref{Proof*-8}
+
+ [**Definition 13**](chap_4.qmd#dense-set): Dense set\Pageref{dense-set}
+
+ [**Remark 14**](chap_4.qmd#Remark-14)\Pageref{Remark-14}
+
+ [**Theorem 15**](chap_4.qmd#theorem-density): Density of $\mathbb{Q}$ in $\mathbb{R}$\Pageref{theorem-density}
+
+ [**Proof**](chap_4.qmd#Proof*-9)\Pageref{Proof*-9}
+
+ [**Definition 16**](chap_4.qmd#irrational-numbers): Irrational numbers\Pageref{irrational-numbers}
+
+ [**Question 17**](chap_4.qmd#Question-17)\Pageref{Question-17}
+
+ [**Corollary 18**](chap_4.qmd#Corollary-18)\Pageref{Corollary-18}
+
+ [**Proof**](chap_4.qmd#Proof*-10)\Pageref{Proof*-10}
+
+ [**Theorem 19**](chap_4.qmd#theorem_existence_root): Existence of $k$-th roots\Pageref{theorem_existence_root}
+
+ [**Proof**](chap_4.qmd#proof-of-theorem): Proof of Theorem 19\Pageref{proof-of-theorem}
+
+ [**Definition 20**](chap_4.qmd#k-th-root-of-a-number): $k$-th root of a number\Pageref{k-th-root-of-a-number}
+
+ [**Definition 21**](chap_4.qmd#bijective-function): Bijective function\Pageref{bijective-function}
+
+ [**Example 22**](chap_4.qmd#injectivity): Injectivity\Pageref{injectivity}
+
+ [**Example 23**](chap_4.qmd#surjectivity): Surjectivity\Pageref{surjectivity}
+
+ [**Example 24**](chap_4.qmd#bijectivity): Bijectivity\Pageref{bijectivity}
+
+ [**Definition 25**](chap_4.qmd#cardinality-finite-countable-uncountable): Cardinality, finite, countable, uncountable\Pageref{cardinality-finite-countable-uncountable}
+
+ [**Question 26**](chap_4.qmd#Question-26)\Pageref{Question-26}
+
+ [**Proposition 27**](chap_4.qmd#proposition-countable-subset)\Pageref{proposition-countable-subset}
+
+ [**Proof**](chap_4.qmd#Proof*-12)\Pageref{Proof*-12}
+
+ [**Example 28**](chap_4.qmd#Example-28)\Pageref{Example-28}
+
+ [**Proposition 29**](chap_4.qmd#proposition-countable-union)\Pageref{proposition-countable-union}
+
+ [**Proof**](chap_4.qmd#Proof*-13)\Pageref{Proof*-13}
+
+ [**Theorem 30**](chap_4.qmd#theorem-Q-countable): $\mathbb{Q}$ is countable\Pageref{theorem-Q-countable}
+
+ [**Proof**](chap_4.qmd#Proof*-14)\Pageref{Proof*-14}
+
+ [**Theorem 31**](chap_4.qmd#theorem-R-uncountable): $\mathbb{R}$ is uncountable\Pageref{theorem-R-uncountable}
+
+ [**Proof**](chap_4.qmd#Proof*-15)\Pageref{Proof*-15}
+
+ [**Theorem 32**](chap_4.qmd#Theorem-32)\Pageref{Theorem-32}
+
+ [**Proof**](chap_4.qmd#Proof*-16)\Pageref{Proof*-16}
+
+ [**Question 1**](chap_5.qmd#Question-1)\Pageref{Question-1}
+
+ [**Definition 2**](chap_5.qmd#complex-numbers): Complex Numbers\Pageref{complex-numbers}
+
+ [**Definition 3**](chap_5.qmd#Definition-3)\Pageref{Definition-3}
+
+ [**Definition 4**](chap_5.qmd#definition-addition-C): Addition in $\mathbb{C}$\Pageref{definition-addition-C}
+
+ [**Notation 5**](chap_5.qmd#Notation-5)\Pageref{Notation-5}
+
+ [**Remark 6**](chap_5.qmd#remark-formal-multiplication): Formal calculation for multiplication in $\mathbb{C}$\Pageref{remark-formal-multiplication}
+
+ [**Definition 7**](chap_5.qmd#definition-multiplication-C): Multiplication in $\mathbb{C}$\Pageref{definition-multiplication-C}
+
+ [**Remark 8**](chap_5.qmd#remark-complex-i)\Pageref{remark-complex-i}
+
+ [**Important**](chap_5.qmd#Important*-1)\Pageref{Important*-1}
+
+ [**Example 9**](chap_5.qmd#Example-9)\Pageref{Example-9}
+
+ [**Proposition 10**](chap_5.qmd#proposition-C-additive-inverse): Additive inverse in $\mathbb{C}$\Pageref{proposition-C-additive-inverse}
+
+ [**Remark 11**](chap_5.qmd#remark-formal-inverse): Formal calculation for multiplicative inverse\Pageref{remark-formal-inverse}
+
+ [**Proposition 12**](chap_5.qmd#proposition-C-multiplicative-inverse): Multiplicative inverse in $\mathbb{C}$\Pageref{proposition-C-multiplicative-inverse}
+
+ [**Proof**](chap_5.qmd#Proof*-2)\Pageref{Proof*-2}
+
+ [**Important**](chap_5.qmd#Important*-3)\Pageref{Important*-3}
+
+ [**Example 13**](chap_5.qmd#Example-13)\Pageref{Example-13}
+
+ [**Theorem 14**](chap_5.qmd#Theorem-14)\Pageref{Theorem-14}
+
+ [**Proof**](chap_5.qmd#Proof*-4)\Pageref{Proof*-4}
+
+ [**Example 15**](chap_5.qmd#Example-15)\Pageref{Example-15}
+
+ [**Theorem 16**](chap_5.qmd#theorem-C-order)\Pageref{theorem-C-order}
+
+ [**Proof**](chap_5.qmd#Proof*-5)\Pageref{Proof*-5}
+
+ [**Definition 17**](chap_5.qmd#complex-conjugate): Complex conjugate\Pageref{complex-conjugate}
+
+ [**Example 18**](chap_5.qmd#Example-18)\Pageref{Example-18}
+
+ [**Theorem 19**](chap_5.qmd#Theorem-19)\Pageref{Theorem-19}
+
+ [**Proof**](chap_5.qmd#Proof*-6)\Pageref{Proof*-6}
+
+ [**Example 20**](chap_5.qmd#Example-20)\Pageref{Example-20}
+
+ [**Definition 21**](chap_5.qmd#modulus): Modulus\Pageref{modulus}
+
+ [**Remark 22**](chap_5.qmd#modulus-of-real-numbers): Modulus of Real numbers\Pageref{modulus-of-real-numbers}
+
+ [**Definition 23**](chap_5.qmd#distance-in-mathbbc): Distance in $\mathbb{C}$\Pageref{distance-in-mathbbc}
+
+ [**Theorem 24**](chap_5.qmd#Theorem-24)\Pageref{Theorem-24}
+
+ [**Proof**](chap_5.qmd#Proof*-7)\Pageref{Proof*-7}
+
+ [**Example 25**](chap_5.qmd#Example-25)\Pageref{Example-25}
+
+ [**Theorem 26**](chap_5.qmd#Theorem-26)\Pageref{Theorem-26}
+
+ [**Proof**](chap_5.qmd#Proof*-8)\Pageref{Proof*-8}
+
+ [**Theorem 27**](chap_5.qmd#triangle-inequality-in-mathbbc): Triangle inequality in $\mathbb{C}$\Pageref{triangle-inequality-in-mathbbc}
+
+ [**Proof**](chap_5.qmd#Proof*-9)\Pageref{Proof*-9}
+
+ [**Remark 28**](chap_5.qmd#geometric-interpretation-of-triangle-inequality): Geometric interpretation of triangle inequality\Pageref{geometric-interpretation-of-triangle-inequality}
+
+ [**Definition 29**](chap_5.qmd#argument): Argument\Pageref{argument}
+
+ [**Warning**](chap_5.qmd#Warning*-10)\Pageref{Warning*-10}
+
+ [**Remark 30**](chap_5.qmd#principal-value): Principal Value\Pageref{principal-value}
+
+ [**Example 31**](chap_5.qmd#Example-31)\Pageref{Example-31}
+
+ [**Theorem 32**](chap_5.qmd#theorem-polar-coordinates): Polar coordinates\Pageref{theorem-polar-coordinates}
+
+ [**Definition 33**](chap_5.qmd#trigonometric-form): Trigonometric form\Pageref{trigonometric-form}
+
+ [**Example 34**](chap_5.qmd#example-trigonometric-form)\Pageref{example-trigonometric-form}
+
+ [**Corollary 35**](chap_5.qmd#corollary-arg): Computing $\arg(z)$\Pageref{corollary-arg}
+
+ [**Proof**](chap_5.qmd#Proof*-11)\Pageref{Proof*-11}
+
+ [**Example 36**](chap_5.qmd#Example-36)\Pageref{Example-36}
+
+ [**Theorem 37**](chap_5.qmd#theorem-euler-identity): Euler’s identity\Pageref{theorem-euler-identity}
+
+ [**Proof**](chap_5.qmd#Proof*-12)\Pageref{Proof*-12}
+
+ [**Theorem 38**](chap_5.qmd#Theorem-38)\Pageref{Theorem-38}
+
+ [**Proof**](chap_5.qmd#Proof*-13)\Pageref{Proof*-13}
+
+ [**Theorem 39**](chap_5.qmd#Theorem-39)\Pageref{Theorem-39}
+
+ [**Proof**](chap_5.qmd#Proof*-14)\Pageref{Proof*-14}
+
+ [**Definition 40**](chap_5.qmd#exponential-form-1): Exponential form\Pageref{exponential-form-1}
+
+ [**Example 41**](chap_5.qmd#example-exponential-form)\Pageref{example-exponential-form}
+
+ [**Remark 42**](chap_5.qmd#periodicity-of-exponential): Periodicity of exponential\Pageref{periodicity-of-exponential}
+
+ [**Proposition 43**](chap_5.qmd#proposition-complex-exponential)\Pageref{proposition-complex-exponential}
+
+ [**Example 44**](chap_5.qmd#Example-44)\Pageref{Example-44}
+
+ [**Example 45**](chap_5.qmd#Example-45)\Pageref{Example-45}
+
+ [**Question 46**](chap_5.qmd#Question-46)\Pageref{Question-46}
+
+ [**Theorem 47**](chap_5.qmd#theorem-fta): Fundamental theorem of algebra\Pageref{theorem-fta}
+
+ [**Example 48**](chap_5.qmd#Example-48)\Pageref{Example-48}
+
+ [**Example 49**](chap_5.qmd#example-degree-4)\Pageref{example-degree-4}
+
+ [**Definition 50**](chap_5.qmd#Definition-50)\Pageref{Definition-50}
+
+ [**Example 51**](chap_5.qmd#example-multiplicity)\Pageref{example-multiplicity}
+
+ [**Question 52**](chap_5.qmd#Question-52)\Pageref{Question-52}
+
+ [**Theorem 53**](chap_5.qmd#theorem-abel-ruffini): Abel-Ruffini\Pageref{theorem-abel-ruffini}
+
+ [**Proposition 54**](chap_5.qmd#proposition-quadratic-formula): Quadratic formula\Pageref{proposition-quadratic-formula}
+
+ [**Example 55**](chap_5.qmd#Example-55)\Pageref{Example-55}
+
+ [**Example 56**](chap_5.qmd#Example-56)\Pageref{Example-56}
+
+ [**Example 57**](chap_5.qmd#Example-57)\Pageref{Example-57}
+
+ [**Question 58**](chap_5.qmd#Question-58)\Pageref{Question-58}
+
+ [**Proposition 59**](chap_5.qmd#proposition-quadratic-formula-C): Generalization of quadratci formula\Pageref{proposition-quadratic-formula-C}
+
+ [**Remark 60**](chap_5.qmd#Remark-60)\Pageref{Remark-60}
+
+ [**Example 61**](chap_5.qmd#Example-61)\Pageref{Example-61}
+
+ [**Remark 62**](chap_5.qmd#polynomial-equations-of-order-n34): Polynomial equations of order $n=3,4$\Pageref{polynomial-equations-of-order-n34}
+
+ [**Example 63**](chap_5.qmd#Example-63)\Pageref{Example-63}
+
+ [**Example 64**](chap_5.qmd#Example-64)\Pageref{Example-64}
+
+ [**Problem**](chap_5.qmd#Problem*-15)\Pageref{Problem*-15}
+
+ [**Question 65**](chap_5.qmd#Question-65)\Pageref{Question-65}
+
+ [**Example 66**](chap_5.qmd#Example-66)\Pageref{Example-66}
+
+ [**Theorem 67**](chap_5.qmd#Theorem-67)\Pageref{Theorem-67}
+
+ [**Proof**](chap_5.qmd#Proof*-16)\Pageref{Proof*-16}
+
+ [**Definition 68**](chap_5.qmd#Definition-68)\Pageref{Definition-68}
+
+ [**Example 69**](chap_5.qmd#Example-69)\Pageref{Example-69}
+
+ [**Example 70**](chap_5.qmd#Example-70)\Pageref{Example-70}
+
+ [**Problem**](chap_5.qmd#Problem*-17)\Pageref{Problem*-17}
+
+ [**Theorem 71**](chap_5.qmd#Theorem-71)\Pageref{Theorem-71}
+
+ [**Proof**](chap_5.qmd#Proof*-18)\Pageref{Proof*-18}
+
+ [**Example 72**](chap_5.qmd#Example-72)\Pageref{Example-72}
+
+ [**Example 73**](chap_5.qmd#Example-73)\Pageref{Example-73}
+
+ [**Remark 1**](chap_6.qmd#Remark-1)\Pageref{Remark-1}
+
+ [**Definition 2**](chap_6.qmd#sequence-of-real-numbers): Sequence of Real numbers\Pageref{sequence-of-real-numbers}
+
+ [**Notation 3**](chap_6.qmd#Notation-3)\Pageref{Notation-3}
+
+ [**Example 4**](chap_6.qmd#Example-4)\Pageref{Example-4}
+
+ [**Definition 5**](chap_6.qmd#definition-convergent-sequence): Convergent sequence\Pageref{definition-convergent-sequence}
+
+ [**Notation 6**](chap_6.qmd#Notation-6)\Pageref{Notation-6}
+
+ [**Remark 7**](chap_6.qmd#Remark-7)\Pageref{Remark-7}
+
+ [**Theorem 8**](chap_6.qmd#theorem-one-over-n)\Pageref{theorem-one-over-n}
+
+ [**Proof**](chap_6.qmd#proof-of-theorem-long-version): Proof of Theorem 8 (Long version)\Pageref{proof-of-theorem-long-version}
+
+ [**Proof**](chap_6.qmd#proof-of-theorem-short-version): Proof of Theorem 8 (Short version)\Pageref{proof-of-theorem-short-version}
+
+ [**Theorem 9**](chap_6.qmd#theorem-one-over-np)\Pageref{theorem-one-over-np}
+
+ [**Proof**](chap_6.qmd#Proof*-3)\Pageref{Proof*-3}
+
+ [**Question 10**](chap_6.qmd#Question-10)\Pageref{Question-10}
+
+ [**Important**](chap_6.qmd#Important*-4)\Pageref{Important*-4}
+
+ [**Example 11**](chap_6.qmd#Example-11)\Pageref{Example-11}
+
+ [**Theorem 12**](chap_6.qmd#theorem-constant-sequence)\Pageref{theorem-constant-sequence}
+
+ [**Proof**](chap_6.qmd#Proof*-5)\Pageref{Proof*-5}
+
+ [**Definition 13**](chap_6.qmd#divergent-sequence): Divergent sequence\Pageref{divergent-sequence}
+
+ [**Remark 14**](chap_6.qmd#Remark-14)\Pageref{Remark-14}
+
+ [**Theorem 15**](chap_6.qmd#theorem-1-minus-n)\Pageref{theorem-1-minus-n}
+
+ [**Proof**](chap_6.qmd#Proof*-6)\Pageref{Proof*-6}
+
+ [**Theorem 16**](chap_6.qmd#theorem-uniqueness-limit): Uniqueness of limit\Pageref{theorem-uniqueness-limit}
+
+ [**Proof**](chap_6.qmd#Proof*-7)\Pageref{Proof*-7}
+
+ [**Example 17**](chap_6.qmd#Example-17)\Pageref{Example-17}
+
+ [**Definition 18**](chap_6.qmd#definition-bounded-sequence): Bounded sequence\Pageref{definition-bounded-sequence}
+
+ [**Theorem 19**](chap_6.qmd#theorem-convergent-bounded)\Pageref{theorem-convergent-bounded}
+
+ [**Proof**](chap_6.qmd#Proof*-8)\Pageref{Proof*-8}
+
+ [**Example 20**](chap_6.qmd#Example-20)\Pageref{Example-20}
+
+ [**Warning**](chap_6.qmd#Warning*-9)\Pageref{Warning*-9}
+
+ [**Example 21**](chap_6.qmd#Example-21)\Pageref{Example-21}
+
+ [**Corollary 22**](chap_6.qmd#corollary-convergent-bounded)\Pageref{corollary-convergent-bounded}
+
+ [**Remark 23**](chap_6.qmd#Remark-23)\Pageref{Remark-23}
+
+ [**Theorem 24**](chap_6.qmd#Theorem-24)\Pageref{Theorem-24}
+
+ [**Proof**](chap_6.qmd#Proof*-10)\Pageref{Proof*-10}
+
+ [**Theorem 25**](chap_6.qmd#Theorem-25)\Pageref{Theorem-25}
+
+ [**Proof**](chap_6.qmd#Proof*-11)\Pageref{Proof*-11}
+
+ [**Theorem 26**](chap_6.qmd#theorem-algebra-limits): Algebra of limits\Pageref{theorem-algebra-limits}
+
+ [**Proof**](chap_6.qmd#Proof*-12)\Pageref{Proof*-12}
+
+ [**Example 27**](chap_6.qmd#example-limit-ploynomial)\Pageref{example-limit-ploynomial}
+
+ [**Important**](chap_6.qmd#Important*-13)\Pageref{Important*-13}
+
+ [**Example 28**](chap_6.qmd#example-algebra-of-limits-2)\Pageref{example-algebra-of-limits-2}
+
+ [**Example 29**](chap_6.qmd#Example-29)\Pageref{Example-29}
+
+ [**Warning**](chap_6.qmd#Warning*-14)\Pageref{Warning*-14}
+
+ [**Example 30**](chap_6.qmd#Example-30)\Pageref{Example-30}
+
+ [**Example 31**](chap_6.qmd#Example-31)\Pageref{Example-31}
+
+ [**Theorem 32**](chap_6.qmd#theorem-square-root-limit)\Pageref{theorem-square-root-limit}
+
+ [**Proof**](chap_6.qmd#Proof*-15)\Pageref{Proof*-15}
+
+ [**Example 33**](chap_6.qmd#Example-33)\Pageref{Example-33}
+
+ [**Example 34**](chap_6.qmd#Example-34)\Pageref{Example-34}
+
+ [**Theorem 35**](chap_6.qmd#theorem-squeeze): Squeeze theorem\Pageref{theorem-squeeze}
+
+ [**Proof**](chap_6.qmd#Proof*-16)\Pageref{Proof*-16}
+
+ [**Example 36**](chap_6.qmd#Example-36)\Pageref{Example-36}
+
+ [**Example 37**](chap_6.qmd#example-squeeze)\Pageref{example-squeeze}
+
+ [**Warning**](chap_6.qmd#Warning*-17)\Pageref{Warning*-17}
+
+ [**Example 38**](chap_6.qmd#Example-38)\Pageref{Example-38}
+
+ [**Definition 39**](chap_6.qmd#Definition-39)\Pageref{Definition-39}
+
+ [**Theorem 40**](chap_6.qmd#theorem-geometric-sequence): Geometric Sequence Test\Pageref{theorem-geometric-sequence}
+
+ [**Warning**](chap_6.qmd#Warning*-18)\Pageref{Warning*-18}
+
+ [**Lemma 41**](chap_6.qmd#lemma-bernoulli): Bernoulli’s inequality\Pageref{lemma-bernoulli}
+
+ [**Proof**](chap_6.qmd#Proof*-19)\Pageref{Proof*-19}
+
+ [**Proof**](chap_6.qmd#proof-of-theorem): Proof of Theorem 40\Pageref{proof-of-theorem}
+
+ [**Example 42**](chap_6.qmd#Example-42)\Pageref{Example-42}
+
+ [**Theorem 43**](chap_6.qmd#theorem-ratio-test): Ratio Test\Pageref{theorem-ratio-test}
+
+ [**Proof**](chap_6.qmd#Proof*-21)\Pageref{Proof*-21}
+
+ [**Example 44**](chap_6.qmd#Example-44)\Pageref{Example-44}
+
+ [**Example 45**](chap_6.qmd#Example-45)\Pageref{Example-45}
+
+ [**Example 46**](chap_6.qmd#Example-46)\Pageref{Example-46}
+
+ [**Warning**](chap_6.qmd#Warning*-22)\Pageref{Warning*-22}
+
+ [**Remark 47**](chap_6.qmd#Remark-47)\Pageref{Remark-47}
+
+ [**Definition 48**](chap_6.qmd#monotone-sequence): Monotone sequence\Pageref{monotone-sequence}
+
+ [**Example 49**](chap_6.qmd#Example-49)\Pageref{Example-49}
+
+ [**Theorem 50**](chap_6.qmd#theorem-monotone-convergence): Monotone Convergence Theorem\Pageref{theorem-monotone-convergence}
+
+ [**Proof**](chap_6.qmd#Proof*-23)\Pageref{Proof*-23}
+
+ [**Definition 1**](chap_7.qmd#sequence-of-complex-numbers): Sequence of Complex numbers\Pageref{sequence-of-complex-numbers}
+
+ [**Definition 2**](chap_7.qmd#definition-convergent-sequence-C): Convergent sequence in $\mathbb{C}$\Pageref{definition-convergent-sequence-C}
+
+ [**Important**](chap_7.qmd#Important*-1)\Pageref{Important*-1}
+
+ [**Example 3**](chap_7.qmd#Example-3)\Pageref{Example-3}
+
+ [**Definition 4**](chap_7.qmd#bounded-sequence-in-mathbbc): Bounded sequence in $\mathbb{C}$\Pageref{bounded-sequence-in-mathbbc}
+
+ [**Theorem 5**](chap_7.qmd#theorem-convergent-bounded-C)\Pageref{theorem-convergent-bounded-C}
+
+ [**Definition 6**](chap_7.qmd#divergent-sequences-in-mathbbc): Divergent sequences in $\mathbb{C}$\Pageref{divergent-sequences-in-mathbbc}
+
+ [**Corollary 7**](chap_7.qmd#Corollary-7)\Pageref{Corollary-7}
+
+ [**Theorem 8**](chap_7.qmd#theorem-algebra-limits-C): Algebra of limits in $\mathbb{C}$\Pageref{theorem-algebra-limits-C}
+
+ [**Example 9**](chap_7.qmd#Example-9)\Pageref{Example-9}
+
+ [**Theorem 10**](chap_7.qmd#theorem-convergence-zero-C)\Pageref{theorem-convergence-zero-C}
+
+ [**Proof**](chap_7.qmd#Proof*-2)\Pageref{Proof*-2}
+
+ [**Example 11**](chap_7.qmd#Example-11)\Pageref{Example-11}
+
+ [**Example 12**](chap_7.qmd#Example-12)\Pageref{Example-12}
+
+ [**Theorem 13**](chap_7.qmd#geometric-sequence-test-in-mathbbc): Geometric sequence Test in $\mathbb{C}$\Pageref{geometric-sequence-test-in-mathbbc}
+
+ [**Example 14**](chap_7.qmd#Example-14)\Pageref{Example-14}
+
+ [**Theorem 15**](chap_7.qmd#theorem-ratop-test-C): Ratio Test in \Pageref{theorem-ratop-test-C}
+
+ [**Example 16**](chap_7.qmd#Example-16)\Pageref{Example-16}
+
+ [**Theorem 17**](chap_7.qmd#theorem-convergence-real-imaginary)\Pageref{theorem-convergence-real-imaginary}
+
+ [**Proof**](chap_7.qmd#Proof*-3)\Pageref{Proof*-3}
+
+ [**Example 18**](chap_7.qmd#Example-18)\Pageref{Example-18}
+
+ [**Definition 1**](chap_8.qmd#partial-sums): Partial sums\Pageref{partial-sums}
+
+ [**Definition 2**](chap_8.qmd#convergent-series-1): Convergent series\Pageref{convergent-series-1}
+
+ [**Definition 3**](chap_8.qmd#divergent-series): Divergent series\Pageref{divergent-series}
+
+ [**Example 4**](chap_8.qmd#Example-4)\Pageref{Example-4}
+
+ [**Example 5**](chap_8.qmd#Example-5)\Pageref{Example-5}
+
+ [**Theorem 6**](chap_8.qmd#theorem-series-necessary)\Pageref{theorem-series-necessary}
+
+ [**Proof**](chap_8.qmd#Proof*-1)\Pageref{Proof*-1}
+
+ [**Theorem 7**](chap_8.qmd#theorem-series-necessary-1)\Pageref{theorem-series-necessary-1}
+
+ [**Example 8**](chap_8.qmd#Example-8)\Pageref{Example-8}
+
+ [**Example 9**](chap_8.qmd#Example-9)\Pageref{Example-9}
+
+ [**Important**](chap_8.qmd#Important*-2)\Pageref{Important*-2}
+
+ [**Example 10**](chap_8.qmd#Example-10)\Pageref{Example-10}
+
+ [**Remark 11**](chap_8.qmd#Remark-11)\Pageref{Remark-11}
+
+ [**Example 12**](chap_8.qmd#Example-12)\Pageref{Example-12}