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6.25"},{"file":"index.tex","autoid":"Example-6.29","reftag":"6.29","label":"Example","id":"Example-6.29","reflabel":"Example","refnum":"6.29","neu":true,"cls":"Example","title":""},{"file":"index.tex","autoid":"Example-6.30","reftag":"6.30","label":"Example","id":"Example-6.30","reflabel":"Example","refnum":"6.30","neu":true,"cls":"Example","title":""},{"file":"index.tex","autoid":"Theorem-6.31","reftag":"6.31","reflabel":"Theorem","neu":true,"id":"Theorem-6.31","refnum":"6.31","label":"Theorem","cls":"Theorem","title":""},{"file":"index.tex","autoid":"Proof*-6.15","reftag":"","label":"Proof","id":"Proof*-6.15","reflabel":"Proof","refnum":"??","neu":true,"cls":"Proof","title":""},{"file":"index.tex","autoid":"Example-6.32","reftag":"6.32","label":"Example","id":"Example-6.32","reflabel":"Example","refnum":"6.32","neu":true,"cls":"Example","title":""},{"file":"index.tex","autoid":"Example-6.33","reftag":"6.33","label":"Example","id":"Example-6.33","reflabel":"Example","refnum":"6.33","neu":true,"cls":"Example","title":""},{"file":"index.tex","autoid":"Proof*-6.12","reftag":"","label":"Proof","id":"Proof*-6.12","reflabel":"Proof","refnum":"??","neu":true,"cls":"Proof","title":""},{"file":"index.tex","autoid":"Important*-6.13","reftag":"","label":"Important","id":"Important*-6.13","reflabel":"Important","refnum":"??","neu":true,"cls":"Important","title":""},{"file":"index.tex","autoid":"Warning*-6.14","reftag":"","label":"Warning","id":"Warning*-6.14","reflabel":"Warning","refnum":"??","neu":true,"cls":"Warning","title":""},{"file":"index.tex","autoid":"Theorem-id-theorem-square-root-limit","reftag":"6.31","label":"Theorem","id":"theorem-square-root-limit","reflabel":"Theorem","refnum":"6.31","neu":true,"cls":"Theorem","title":""},{"file":"index.tex","autoid":"Theorem-6.34","reftag":"6.34","label":"Theorem","id":"theorem-squeeze","reflabel":"Theorem","refnum":"6.34","neu":true,"cls":"Theorem","title":"Squeeze 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6.39"},{"file":"index.tex","autoid":"Example-6.41","reftag":"6.41","label":"Example","id":"Example-6.41","reflabel":"Example","refnum":"6.41","neu":true,"cls":"Example","title":""},{"file":"index.tex","autoid":"Theorem-6.42","reftag":"6.42","label":"Theorem","id":"theorem-ratio-test","reflabel":"Theorem","refnum":"6.42","neu":true,"cls":"Theorem","title":"Ratio 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%norma\n\n\n\n\n\n\n\n\n\n$\n:::\n\n\n\n\n\n# Sequences\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## Sequences in $\\R$\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$$\n\\lim _{n \\rightarrow \\infty} a_{n}=a \\text {, }\n$$\n\nif for all $\\e \\in \\mathbb{R}, \\e>0$, there exists an $N \\in \\N$ 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::: 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\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 $M>1$ such that\n$$\n\\left|\\frac{a_{n+1}}{a_{n}}\\right| \\geq M \\,, \\quad \\forall \\, n \\geq N \\,. \\,.\n$$\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$$\nL1$: Choose $r>0$ such that $10$. If $\\frac{b_{n+1}}{b_{n}} \\rightarrow L$, there exists $n_{0} \\in \\mathbb{N}$ such that for all $n \\geq n_{0}$\n\n$$\n\\left|\\frac{b_{n+1}}{b_{n}}-L\\right| \\leq \\varepsilon=L-r\n$$\n\nIn particular,\n\n$$\n-(L-r) \\leq \\frac{b_{n+1}}{b_{n}}-L \\Longleftrightarrow \\frac{b_{n+1}}{b_{n}} \\geq r \\Longleftrightarrow b_{n+1} \\geq r b_{n}\n$$\n\n(In the other case we assumed this is the case, so this statement also holds there.) Hence, for all $n \\geq n_{0}$\n\n$$\nb_{n} \\geq r^{n-n_{0}} b_{n_{0}}=r^{n} \\frac{b_{n_{0}}}{r^{n_{0}}}\n$$\n\nSince $|r|>1$, it follows from the geometric sequence test that the right hand side is unbounded and hence also $b_{n}$ is unbounded.\n\n\n:::\n\n\n\n\n::: Example\n\nLet $a_{n}=\\frac{3^{n}}{n !}$. Then,\n\n$$\n\\left|\\frac{a_{n+1}}{a_{n}}\\right|=\\frac{3^{n+1}}{(n+1) !} / \\frac{3^{n}}{n !}=\\frac{3^{n+1}}{3^{n}} \\frac{n !}{(n+1) !}=3 \\frac{1 \\cdot 2 \\cdots n}{1 \\cdot 2 \\cdots n \\cdot(n+1)}=\\frac{3}{n+1} \\longrightarrow 0,\n$$\n\nas $n \\rightarrow \\infty$. Hence, $L=0<1$ so $\\lim _{n \\rightarrow \\infty} a_{n}=0$ by the ratio test.\n\n- Let $b_{n}=\\frac{n !}{100^{n}}$. Then\n\n$$\n\\left|\\frac{b_{n+1}}{b_{n}}\\right|=\\frac{(n+1) !}{100^{n+1}} / \\frac{n !}{100^{n}}=\\frac{100^{n}}{100^{n+1}} \\frac{(n+1) !}{(n !)}=\\frac{n+1}{100} .\n$$\n\nChoose $n_{0}=101$. Then for all $n \\geq n_{0}$,\n\n$$\n\\left|\\frac{b_{n+1}}{b_{n}}\\right| \\geq \\frac{101}{100}>1\n$$\n\nHence, $b_{n}$ is an unbounded sequence and does not converge.\n\n:::\n\n\n::: Warning\n\nFor part (c) of the theorem we give some examples. If you try to apply the ratio test for a fraction of two polynomials, the limit will always be equal to 1 , so the ratio test is inconclusive for this type of sequence.\n\n:::\n\nExample 2.4.13. $\\quad$ Let $a_{n}=\\frac{1}{n}$. Then,\n\n$$\n\\left|\\frac{a_{n+1}}{a_{n}}\\right|=\\frac{1}{n+1} / \\frac{1}{n}=\\frac{n}{n+1} \\rightarrow 1\n$$\n\nso the ratio test is inconclusive. But we know that $\\frac{1}{n} \\rightarrow 0$.\n\n- Let $b_{n}=\\frac{n^{2}}{n+1}$. Then,\n\n$$\n\\left|\\frac{b_{n+1}}{b_{n}}\\right|=\\frac{(n+1)^{2}}{(n+1)+1} / \\frac{n^{2}}{n+1}=\\frac{(n+1)^{3}}{(n+2) n^{2}}=\\frac{(1+1 / n)^{3}}{(1+2 / n) 1} \\rightarrow 1\n$$\n\nso the ratio test is inconclusive. But we can compute $\\frac{n^{2}}{n+1}=\\frac{n}{1+1 / n}$, which is unbounded so $b_{n}$ does not converge.\n\n- Let $c_{n}=\\frac{n}{2 n+1}$. Then,\n\n$$\n\\left|\\frac{c_{n+1}}{c_{n}}\\right|=\\frac{n+1}{2(n+1)+1} / \\frac{n}{2 n+1}=\\frac{(n+1)(2 n+1)}{(2 n+3) n}=\\frac{(1+1 / n)(2+1 / n)}{(2+3 / n) 1} \\rightarrow 1,\n$$\n\nso the ratio test is inconclusive. But we can compute $\\frac{n}{2 n+1}=\\frac{1}{2+1 / n} \\rightarrow \\frac{1}{2}$.\n\nAs mentioned, the ratio test is especially useful if factorials and geometric terms are involved. It is important to make sure you use brackets correctly, it is very easy to make mistakes.\n\nExample 2.4.14. Let\n\n$$\na_{n}=\\frac{n ! \\cdot 3^{n}}{\\sqrt{(2 n) !}}\n$$\n\nThen,\n\n$$\n\\left|\\frac{a_{n+1}}{a_{n}}\\right|=\\frac{(n+1) ! \\cdot 3^{n+1}}{\\sqrt{(2(n+1)) !}} / \\frac{n ! \\cdot 3^{n}}{\\sqrt{(2 n) !}}=\\frac{(n+1) !}{n !} \\cdot \\frac{3^{n+1}}{3^{n}} \\cdot \\frac{\\sqrt{(2 n) !}}{\\sqrt{(2(n+1)) !}}\n$$\n\nWe can simplify $\\frac{(n+1) !}{n !} \\cdot \\frac{3^{n+1}}{3^{n}}=3(n+1)$. Combining the square roots and working out the brackets carefully gives that\n\n$$\n\\frac{\\sqrt{(2 n) !}}{\\sqrt{(2(n+1)) !}}=\\sqrt{\\frac{(2 n) !}{(2 n+2) !}}=\\sqrt{\\frac{1 \\cdot 2 \\cdots 2 n}{1 \\cdot 2 \\cdots 2 n \\cdot(2 n+1) \\cdot(2 n+2)}}=\\frac{1}{\\sqrt{(2 n+1)(2 n+2)}}\n$$\n\nHence,\n\n$$\n\\left|\\frac{a_{n+1}}{a_{n}}\\right|=\\frac{3(n+1)}{\\sqrt{(2 n+1)(2 n+2)}}=\\frac{3(1+1 / n)}{\\sqrt{\\frac{1}{n^{2}}(2 n+1)(2 n+2)}}=\\frac{3(1+1 / n)}{\\sqrt{(2+1 / n)(2+2 / n)}} \\rightarrow \\frac{3}{\\sqrt{4}}=\\frac{3}{2}>1\n$$\n\nHence, the sequence does not converge by the ratio test.\n\nIf the sequence is a geometric sequence, we can also apply the ratio test, which will give the same answer as the geometric sequence test.\n\nExample 2.4.15. Let $x \\in \\mathbb{R}$ and let $a_{n}=x^{n}$. Then,\n\n$$\n\\left|\\frac{a_{n+1}}{a_{n}}\\right|=\\frac{\\left|x^{n+1}\\right|}{\\left|x^{n}\\right|}=\\frac{|x|^{n+1}}{|x|^{n}}=|x| \\rightarrow|x|\n$$\n\nHence, if $|x|<1$ the sequence converges by the ratio test, if $|x|>1$ the sequence does not converge by the ratio test, and if $|x|=1$ the ratio test is inconclusive. Of course, we would have gotten this result immediately by using the geometric sequence test.\n\n\n\n\n\n\n\n\n\n\n\n\n::: {.content-hidden}\n\n\nUse abbot, marcellini, and analisi sapienza note. For next year remove integral test and add cauchy condensation test.\n\n\n## Divergence to infinity\n\nDo this from abbott or marcellini. For the moment we have only done\ndivergent as non-convergent. Would need definition of Divergence to $\\pm \\infty$.\n\n\n## Some notable limits\n\nMarcellini Page 101. Also see what is the english name for limiti notevoli.\n\n\n## Example: Euler's Number\n\n\nMarcellini Page 106\n\n\n## Recurrence relations\n\nMarcellini Page 110\n\n\n\n## Example: Heron's Method\n\n\nThe first explicit algorithm for approximating \n$$\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$$\n\n\n:::\n\n\n\n## Convergent sequences\n\n\nWe have notice that the sequence $\\left(\\frac{1}{n}\\right)_{n \\in \\N}$ gets close to $0$ as $n$ gets large. We would like to say that $a_{n}$ converges to $0$ as $n$ tends to infinity.\n\nTo make this precise, we first have to say what it means for two numbers to be *close*. For this we use the notion of absolute value, and say that:\n\n- $x$ and $y$ are close if $|x-y|$ is small. \n- $|x-y|$ is called the distance between $x$ and $y$\n- For $x$ to be close to $0$, we need that $|x-0|=|x|$ is small.\n\nSaying that $|x|$ is *small* is not very precise. Let us now give the formal definition of convergent sequence.\n\n::: Definition \n### Convergent sequence {#definition-convergent-sequence}\n\nWe say that a sequence $\\left(a_{n}\\right)_{n \\in \\N}$ in $\\mathbb{R}$ **converges** to $a \\in \\mathbb{R}$ or equivalently has limit $a$, denoted by\n\n$$\n\\lim _{n \\rightarrow \\infty} a_{n}=a \\text {, }\n$$\n\nif for all $\\e \\in \\mathbb{R}, \\e>0$, there exists an $N \\in \\N$ such that for all $n \\in \\N, n \\geq N$ it holds that \n$$\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$$\n\nIf there exists $a \\in \\R$ such that $\\lim _{n \\rightarrow \\infty} a_{n}=a$, then we say that the sequence $\\left(a_{n}\\right)_{n \\in \\N}$ is **convergent**.\n\n:::\n\n\n::: Remark\n\n- Informally, Definition \\ref{definition-convergent-sequence} says that, no matter how small we choose $\\e$ (as long as it is strictly positive), we always have that $a_{n}$ has a distance to $a$ of less than or equal to $\\e$ from a certain point onwards (i.e., from $N$ onward). The sequence $\\left(a_{n}\\right)$ may fluctuate wildly in the beginning, but from $N$ onward it should stay within a distance of $\\e$ of $a$.\n\n- In general $N$ depends on $\\e$. If $\\e$ is chosen smaller, we might have to take $N$ larger: this means we need to wait longer before the sequence stays within a distance $\\e$ from $a$.\n\n:::\n\n\nWe now prove that the sequence\n$$\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$${#eq-ratio-test-proof}\n\n- $L<1$: Choose $r>0$ such that \n$$\nL1$: Choose $r>0$ such that \n$$\n1 0 \\,.\n$$\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\n\n\n\n\n\n::: {.content-hidden}\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 $\\pm \\infty$.\n\n\n## Some notable limits\n\nMarcellini Page 101. Also see what is the english name for limiti notevoli.\n\n\n## Example: Euler's Number\n\n\nMarcellini Page 106\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", "supporting": [ "chap_6_files/figure-html" ], diff --git a/_quarto.yml b/_quarto.yml index 5e05956..31f95d0 100644 --- a/_quarto.yml +++ b/_quarto.yml @@ -48,7 +48,8 @@ book: - sections/chap_4.qmd - sections/chap_5.qmd - sections/chap_6.qmd - #- sections/chap_7.qmd + - sections/chap_7.qmd + - sections/chap_8.qmd - sections/license.qmd - sections/references.qmd diff --git a/docs/Numbers,-Sequences-and-Series.pdf b/docs/Numbers,-Sequences-and-Series.pdf index 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  • - 6  Sequences + 6  Sequences in \(\mathbb{R}\) +
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    • +
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    + + 7  Sequences in \(\mathbb{C}\) +
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    • +
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    + + 8  Series
  • diff --git a/docs/search.json b/docs/search.json index 732a261..e06f8ef 100644 --- a/docs/search.json +++ b/docs/search.json @@ -266,60 +266,74 @@ "text": "5.9 Roots in \\(\\mathbb{C}\\)\n\nProblemLet \\(n \\in \\mathbb{N}\\) and \\(c \\in \\mathbb{C}\\). We want to find the \\(n\\)-th roots of \\(c\\). This means we want to find all complex solutions to \\[\nz^{n}=c \\,.\n\\]\n\n\nThe Fundamental Theorem of Algebra ensures that the above has \\(n\\) complex solutions. To find these solutions, we pass to the exponential form.\n\nTheorem 71Let \\(n \\in \\mathbb{N}\\), \\(c \\in \\mathbb{C}\\) and consider the equation \\[\nz^{n}=c \\,.\n\\tag{5.17}\\] All the \\(n\\) solutions to (5.17) are given by \\[\nz_k = \\sqrt[n]{|c|} \\, \\exp \\left(i \\, \\frac{\\theta + 2 \\pi k}{n}\\right) \\,, \\quad k = 0, \\ldots, n-1 \\,,\n\\] where \\(\\sqrt[n]{|c|}\\) is the \\(n\\)-th root of the real number \\(|c|\\), and \\(\\theta = \\arg(c)\\).\n\n\n\nProofWrite \\(c\\) in exponential form: \\[\nc=|c|e^{i \\theta} =|c| e^{i(\\theta + 2 \\pi k)} \\, , \\quad k \\in \\mathbb{Z}\\,,\n\\] where \\(\\theta = \\arg(c)\\). Therefore (5.17) is equivalent to \\[\nz^n = |c| e^{i(\\theta + 2 \\pi k)} \\,.\n\\] By the properties of the exponential, we see that the above is solved by \\[\nz_k= \\sqrt[n]{|c|} \\exp \\left(i \\, \\frac{\\theta + 2 \\pi k}{n}\\right), \\quad k \\in \\mathbb{Z}\\,.\n\\] By choosing \\(k = 0, \\ldots, n-1\\) we obtain \\(n\\) different solutions.\n\n\n\nExample 72We want to find all the \\(z \\in \\mathbb{C}\\) such that \\[\nz^{5}=-32 \\,.\n\\] Let \\(c = -32\\). We have \\[\n|c| = |-32|=32=2^{5}\\,, \\quad \\theta = \\arg (-32)=\\pi \\,.\n\\] Hence, the solutions are given by \\[\nz_k = \\left(2^{5}\\right)^{\\frac{1}{5}} \\exp \\left(i \\pi \\, \\frac{1+2 k}{5} \\right) \\,, \\quad k \\in \\mathbb{Z}\\,.\n\\]\nBy taking \\(k=0,1,2,3,4\\) we get the solutions \\[\\begin{align*}\nz_0 & = 2 e^{i \\frac{\\pi}{5}} \\, & \\quad & z_1 = 2 e^{i \\frac{3 \\pi}{5}} \\\\\nz_2 & = 2 e^{i \\pi}=-2 \\, & \\quad & z_3=2 e^{i \\frac{7 \\pi}{5}} \\\\\nz_4 &= 2 e^{i \\frac{9 \\pi}{5}} & \\quad &\n\\end{align*}\\]\n\n\n\nExample 73We want to find all the \\(z \\in \\mathbb{C}\\) such that \\[\nz^{4}=9\\left(\\cos \\left(\\frac{\\pi}{3}\\right)+i \\sin \\left(\\frac{\\pi}{3}\\right)\\right) \\,.\n\\] Set \\[\nc:=9\\left(\\cos \\left(\\frac{\\pi}{3}\\right)+i \\sin \\left(\\frac{\\pi}{3}\\right)\\right) \\,.\n\\] The complex number \\(c\\) is already in the trigonometric form, so that we can immediately obtain \\[\n|c| = 9 \\,, \\quad \\theta = \\arg(c) = \\frac{\\pi}{3} \\,.\n\\] Hence the solutions are given by \\[\\begin{align*}\nz_k & = \\sqrt[4]{9} \\, \\exp \\left( i \\, \\frac{ \\pi/3 + 2 \\pi k}{4} \\right) \\\\\n& = \\sqrt{3} \\exp \\left( i \\pi \\, \\frac{1+6 k}{12} \\right)\n\\end{align*}\\] for \\(k \\in \\mathbb{Z}\\). Choosing \\(k=0,1,2,3\\) gives the \\(4\\) solutions \\[\\begin{align*}\nz_0 & = \\sqrt{3} e^{i \\pi \\frac{1}{12}} & \\quad & z_1 = \\sqrt{3} e^{i \\pi \\frac{7}{12}} \\\\\nz_2 & = \\sqrt{3} e^{i \\pi \\frac{13}{12}} & \\quad & z_3 = \\sqrt{3} e^{i \\pi \\frac{19}{12}}\n\\end{align*}\\]" }, { - "objectID": "sections/chap_6.html#sequences-in-mathbbr", - "href": "sections/chap_6.html#sequences-in-mathbbr", - "title": "6  Sequences", - "section": "6.1 Sequences in \\(\\mathbb{R}\\)", - "text": "6.1 Sequences in \\(\\mathbb{R}\\)\nWe start with the definition of sequence of Real numbers.\n\nDefinition 2: Sequence of Real numbersA sequence \\(a\\) in \\(\\mathbb{R}\\) is a function \\[\na \\colon \\mathbb{N}\\to \\mathbb{R}\\,.\n\\] For \\(n \\in \\mathbb{N}\\), we denote the \\(n\\)-th element of the sequence \\(a\\) by \\[\na_{n}=a(n)\n\\] and write the sequence as \\[\n\\left(a_{n}\\right)_{n \\in \\mathbb{N}} \\,.\n\\]\n\n\n\nNotation 3We will sometimes omit the subscript \\(n \\in \\mathbb{N}\\) and simply write \\[\n\\left(a_{n}\\right) \\,.\n\\] In certain situations, we will also write \\[\n\\left(a_{n}\\right)_{n=1}^{\\infty} \\,.\n\\]\n\n\n\nExample 4\n\nIn general \\(\\left(a_{n}\\right)_{n \\in \\mathbb{N}}\\) is the sequence \\[\n\\left(a_{1}, a_{2}, a_{3}, \\ldots\\right)\\,.\n\\]\nConsider the function \\[\na \\colon \\mathbb{N}\\to \\mathbb{N}\\, , \\quad n \\mapsto 2 n \\,.\n\\] This is also a sequence of real numbers. It can be written as \\[\n(2 n)_{n \\in \\mathbb{N}}\n\\] and it represents the sequence of even numbers \\[\n(2,4,6,8,10, \\ldots) \\,.\n\\]\nLet \\[\na_{n}=(-1)^{n}\n\\] Then \\(\\left(a_{n}\\right)\\) is the sequence \\[\n(-1,1,-1,1,-1,1, \\ldots) \\,.\n\\]\n\\(\\left(\\frac{1}{n}\\right)_{n \\in \\mathbb{N}}\\) is the sequence \\[\n\\left(1, \\frac{1}{2}, \\frac{1}{4}, \\frac{1}{5}, \\ldots\\right) \\,.\n\\]" + "objectID": "sections/chap_6.html#definition-of-sequence", + "href": "sections/chap_6.html#definition-of-sequence", + "title": "6  Sequences in \\(\\mathbb{R}\\)", + "section": "6.1 Definition of sequence", + "text": "6.1 Definition of sequence\nWe start with the definition of sequence of Real numbers.\n\nDefinition 2: Sequence of Real numbersA sequence \\(a\\) in \\(\\mathbb{R}\\) is a function \\[\na \\colon \\mathbb{N}\\to \\mathbb{R}\\,.\n\\] For \\(n \\in \\mathbb{N}\\), we denote the \\(n\\)-th element of the sequence \\(a\\) by \\[\na_{n}=a(n)\n\\] and write the sequence as \\[\n\\left(a_{n}\\right)_{n \\in \\mathbb{N}} \\,.\n\\]\n\n\n\nNotation 3We will sometimes omit the subscript \\(n \\in \\mathbb{N}\\) and simply write \\[\n\\left(a_{n}\\right) \\,.\n\\] In certain situations, we will also write \\[\n\\left(a_{n}\\right)_{n=1}^{\\infty} \\,.\n\\]\n\n\n\nExample 4\n\nIn general \\(\\left(a_{n}\\right)_{n \\in \\mathbb{N}}\\) is the sequence \\[\n\\left(a_{1}, a_{2}, a_{3}, \\ldots\\right)\\,.\n\\]\nConsider the function \\[\na \\colon \\mathbb{N}\\to \\mathbb{N}\\, , \\quad n \\mapsto 2 n \\,.\n\\] This is also a sequence of real numbers. It can be written as \\[\n(2 n)_{n \\in \\mathbb{N}}\n\\] and it represents the sequence of even numbers \\[\n(2,4,6,8,10, \\ldots) \\,.\n\\]\nLet \\[\na_{n}=(-1)^{n}\n\\] Then \\(\\left(a_{n}\\right)\\) is the sequence \\[\n(-1,1,-1,1,-1,1, \\ldots) \\,.\n\\]\n\\(\\left(\\frac{1}{n}\\right)_{n \\in \\mathbb{N}}\\) is the sequence \\[\n\\left(1, \\frac{1}{2}, \\frac{1}{4}, \\frac{1}{5}, \\ldots\\right) \\,.\n\\]" }, { "objectID": "sections/chap_6.html#convergent-sequences", "href": "sections/chap_6.html#convergent-sequences", - "title": "6  Sequences", + "title": "6  Sequences in \\(\\mathbb{R}\\)", "section": "6.2 Convergent sequences", "text": "6.2 Convergent sequences\nWe have notice that the sequence \\(\\left(\\frac{1}{n}\\right)_{n \\in \\mathbb{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.\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\\)\nFor \\(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\nDefinition 5: Convergent sequenceWe say that a sequence \\(\\left(a_{n}\\right)_{n \\in \\mathbb{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 \\text {, }\n\\]\nif for all \\(\\varepsilon\\in \\mathbb{R}, \\varepsilon>0\\), there exists an \\(N \\in \\mathbb{N}\\) such that for all \\(n \\in \\mathbb{N}, n \\geq N\\) it holds that \\[\n\\left|a_{n}-a\\right| < \\varepsilon\\,.\n\\] Using quantifiers, we can write this as \\[\n\\forall \\, \\varepsilon>0 , \\, \\exists \\, N \\in \\mathbb{N}\\, \\text{ s.t. } \\, \\forall \\, n \\geq N \\,, \\,\\left|a_{n}-a\\right| < \\varepsilon\\,.\n\\]\nIf there exists \\(a \\in \\mathbb{R}\\) such that \\(\\lim _{n \\rightarrow \\infty} a_{n}=a\\), then we say that the sequence \\(\\left(a_{n}\\right)_{n \\in \\mathbb{N}}\\) is convergent.\n\n\n\nRemark 6\n\nInformally, Definition 5 says that, no matter how small we choose \\(\\varepsilon\\) (as long as it is strictly positive), we always have that \\(a_{n}\\) has a distance to \\(a\\) of less than or equal to \\(\\varepsilon\\) 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 \\(\\varepsilon\\) of \\(a\\).\nIn general \\(N\\) depends on \\(\\varepsilon\\). If \\(\\varepsilon\\) is chosen smaller, we might have to take \\(N\\) larger: this means we need to wait longer before the sequence stays within a distance \\(\\varepsilon\\) from \\(a\\).\n\n\n\nWe now prove that the sequence \\[\na_n = \\frac1n\n\\] converges to \\(0\\), according to Definition 5.\n\nTheorem 7The sequence \\(\\left(\\frac{1}{n}\\right)_{n \\in \\mathbb{N}}\\) converges to \\(0\\) , i.e.,\n\\[\n\\lim_{n \\rightarrow \\infty} \\frac{1}{n}=0 \\,.\n\\]\n\n\nWe give two proofs of the above theorem:\n\nLong proof, with all the details.\nShort proof, with less details, but still acceptable.\n\n\nProof: Proof of Theorem 7 (Long version)We have to show that\n\\[\n\\lim _{n \\rightarrow \\infty} \\frac{1}{n}=0,\n\\]\nwhich by definition is equivalent to showing that\n\\[\n\\forall \\, \\varepsilon>0 \\,, \\, \\exists \\, N \\in \\mathbb{N}\\, \\text{ s.t. } \\, \\forall \\, n \\geq N \\,, \\, \\left|\\frac{1}{n}-0\\right| < \\varepsilon\\,.\n\\tag{6.1}\\]\nLet \\(\\varepsilon\\in \\mathbb{R}\\) with \\(\\varepsilon>0\\). Choose \\(N \\in \\mathbb{N}\\) such that \\[\nN > \\frac{1}{\\varepsilon}\\,.\n\\] Such natural number \\(N\\) exists thanks to the Archimedean property. The above implies \\[\n\\frac{1}{N} < \\varepsilon\\,,\n\\tag{6.2}\\] Let \\(n \\in \\mathbb{N}\\) with \\(n \\geq N\\). By (6.2) we have \\[\n\\frac{1}{n} \\leq \\frac{1}{N} < \\varepsilon\\,.\n\\] From this we deduce \\[\n\\left|\\frac{1}{n}-0\\right|=\\frac{1}{n} \\leq \\frac{1}{N} < \\varepsilon\n\\]\nSince \\(n \\in \\mathbb{N}, n \\geq N\\) was arbitrary, we have proven that \\[\n\\left|\\frac{1}{n}-0\\right| < \\varepsilon\\,, \\quad \\forall \\, n \\geq N \\,.\n\\tag{6.3}\\] Condition (6.3) holds for all \\(\\varepsilon>0\\), for the choice of \\(N \\in \\mathbb{N}\\) such that \\[\nN > \\frac{1}{\\varepsilon} \\,.\n\\] We have hence shown (6.1), and the proof is concluded.\n\n\nAs the above proof is quite long and includes lots of details, it is acceptable to shorten it. For example:\n\nWe skip some intermediate steps.\nWe do not mention the Archimedean property.\nWe leave out the conclusion when it is obvious that the statement has been proven.\n\n\nProof: Proof of Theorem 7 (Short version)We have to show that\n\\[\n\\forall \\, \\varepsilon>0 \\,, \\, \\exists \\, N \\in \\mathbb{N}\\, \\text{ s.t. } \\, \\forall \\, n \\geq N \\,, \\, \\left|\\frac{1}{n}-0\\right| < \\varepsilon\\,.\n\\]\nLet \\(\\varepsilon>0\\). Choose \\(N \\in \\mathbb{N}\\) such that \\[\nN > \\frac{1}{\\varepsilon}\\,.\n\\] Let \\(n \\geq N\\). Then \\[\n\\left|\\frac{1}{n}-0\\right|=\\frac{1}{n} \\leq \\frac{1}{N} < \\varepsilon\\,.\n\\]\n\n\nIn Theorem 7 we showed that \\[\n\\lim_{n \\to \\infty} \\, \\frac{1}{n} = 0 \\,.\n\\] We can generalise this statement to prove that \\[\n\\lim_{n \\to \\infty} \\, \\frac{1}{n^{p}} = 0\n\\] for any \\(p>0\\) fixed.\n\nTheorem 8For all \\(p>0\\), the sequence \\(\\left(\\frac{1}{n^{p}}\\right)_{n \\in \\mathbb{N}}\\) converges to \\(0\\) , i.e.,\n\\[\n\\lim _{n \\rightarrow \\infty} \\frac{1}{n^{p}}=0\n\\]\n\n\n\nProofLet \\(p>0\\). We have to show that \\[\n\\forall \\varepsilon>0 \\,, \\, \\exists \\, N \\in \\mathbb{N}\\, \\text{ s.t. } \\, \\forall \\, n \\geq N \\,, \\, \\left|\\frac{1}{n^{p}}-0\\right| < \\varepsilon\\,.\n\\] Let \\(\\varepsilon>0\\). Choose \\(N \\in \\mathbb{N}\\) such that \\[\nN > \\frac{1}{\\varepsilon^{1 / p}} \\,.\n\\tag{6.4}\\] Let \\(n \\geq N\\). Since \\(p>0\\), we have \\(n^{p} \\geq N^{p}\\), which implies \\[\n\\frac{1}{n^p} \\leq \\frac{1}{N^p} \\,.\n\\] By (6.4) we deduce \\[\n\\frac{1}{N^p} < \\varepsilon\\,.\n\\] Then \\[\n\\left|\\frac{1}{n^{p}}-0\\right|=\\frac{1}{n^{p}} \\leq \\frac{1}{N^{p}} < \\varepsilon\\,.\n\\]\n\n\n\nQuestion 9Why did we choose \\(N \\in \\mathbb{N}\\) such that \\[\nN > \\frac{1}{\\varepsilon^p}\n\\] in 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 \\mathbb{N}\\) such that the inequality \\[\n|a_N - a | < \\varepsilon\n\\] is satisfied.\n\nImportantAny rough work required to prove convergence must be shown before the formal proof (in assignments).\n\n\n\nExample 10Prove that, as \\(n \\rightarrow \\infty\\),\n\\[\n\\frac{n}{2n+3} \\to \\frac{1}{2} \\,.\n\\]\nPart 1. Rough Work.\nLet \\(\\varepsilon>0\\). We want to find \\(N \\in \\mathbb{N}\\) such that \\[\n\\left|\\frac{n}{2n+3}-\\frac{1}{2}\\right| < \\varepsilon\\,, \\quad \\forall \\, n \\geq N \\,.\n\\] To this end, we compute: \\[\\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*}\\] Therefore \\[\\begin{align*}\n\\left|\\frac{n}{2n+3}-\\frac{1}{2}\\right| < \\varepsilon\n\\quad \\iff & \\quad\n\\frac{3}{4n + 6} < \\varepsilon\\\\\n\\quad \\iff & \\quad\n\\frac{4n + 6}{3} > \\frac{1}{\\varepsilon} \\\\\n\\quad \\iff & \\quad\n4n + 6 > \\frac{3}{\\varepsilon} \\\\\n\\quad \\iff & \\quad\n4n > \\frac{3}{\\varepsilon} - 6 \\\\\n\\quad \\iff & \\quad\nn > \\frac{3}{4\\varepsilon} - \\frac{6}{4} \\,.\n\\end{align*}\\] Looking at the above equivalences, it is clear that \\(N \\in \\mathbb{N}\\) has to be chosen so that \\[\nN > \\frac{3}{4\\varepsilon} - \\frac{6}{4} \\,.\n\\]\nPart 2. Formal Proof. We have to show that \\[\n\\forall \\varepsilon>0 \\,, \\, \\exists \\, N \\in \\mathbb{N}\\, \\text{ s.t. } \\, \\forall \\, n \\geq N \\,, \\, \\left|\\frac{n}{2n+3}-\\frac{1}{2}\\right| < \\varepsilon\\,.\n\\] Let \\(\\varepsilon>0\\). Choose \\(N \\in \\mathbb{N}\\) such that \\[\nN > \\frac{3}{4\\varepsilon} - \\frac{6}{4} \\,.\n\\tag{6.5}\\] By the rough work shown above, inequality (6.5) is equivalent to \\[\n\\frac{3}{4N + 6} < \\varepsilon\\,.\n\\] Let \\(n \\geq N\\). Then \\[\\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& < \\varepsilon\\,,\n\\end{align*}\\] where in the third line we used that \\(n \\geq N\\).\n\n\nWe conclude by showing that constant sequences always converge.\n\nTheorem 11Let \\(c \\in \\mathbb{R}\\) and define the constant sequence \\[\na_n := c \\,, \\quad \\forall \\, n \\in \\mathbb{N}\\,.\n\\] We have that \\[\n\\lim_{n \\to \\infty} \\, a_n = c \\,.\n\\]\n\n\n\nProofWe have to prove that \\[\n\\forall \\varepsilon>0 \\,, \\, \\exists \\, N \\in \\mathbb{N}\\, \\text{ s.t. } \\, \\forall \\, n \\geq N \\,, \\, \\left|a_n - c \\right| < \\varepsilon\\,.\n\\tag{6.6}\\] Let \\(\\varepsilon>0\\). We have \\[\n\\left|a_n - c \\right| = \\left|c - c \\right| = 0 < \\varepsilon\\,, \\quad \\forall \\, n \\in \\mathbb{N}\\,.\n\\] Therefore we can choose \\(N=1\\) and (6.6) is satisfied." }, { "objectID": "sections/chap_6.html#divergent-sequences", "href": "sections/chap_6.html#divergent-sequences", - "title": "6  Sequences", + "title": "6  Sequences in \\(\\mathbb{R}\\)", "section": "6.3 Divergent sequences", "text": "6.3 Divergent sequences\nThe opposite of convergent sequences are divergent sequences.\n\nDefinition 12: Divergent sequenceWe say that a sequence \\(\\left(a_{n}\\right)_{n \\in \\mathbb{N}}\\) in \\(\\mathbb{R}\\) is divergent if it is not convergent.\n\n\n\nRemark 13Proving 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 \\((a_n)\\) cannot converge to \\(a\\) for any \\(a \\in \\mathbb{R}\\).\nIn other words, we have to show that there does not exist an \\(a \\in \\mathbb{R}\\) such that \\[\n\\lim _{n \\rightarrow \\infty} \\, a_n = a \\,.\n\\] Using quantifiers, this means \\[\n\\nexists \\, a \\in \\mathbb{R}\\, \\text{ s.t. } \\, \\forall \\, \\varepsilon>0 \\,, \\,\n\\exists \\, N \\in \\mathbb{N}\\, \\text{ s.t. } \\, \\forall \\, n \\geq N \\,, \\,\n\\left|a_{n}-a\\right| < \\varepsilon\\,.\n\\] The above is equivalent to showing that \\[\n\\forall \\, a \\in \\mathbb{R}\\,, \\, \\exists \\, \\varepsilon> 0 \\, \\text{ s.t. } \\, \\forall \\, N \\in \\mathbb{N}\\,, \\, \\exists \\, n \\geq N \\, \\text{ s.t. } \\, \\left|a_{n}-a\\right| \\geq \\varepsilon\\,.\n\\]\n\n\n\nTheorem 14Let \\(\\left(a_{n}\\right)\\) be the sequence defined by \\[\na_{n}=(-1)^{n} \\,.\n\\] Then \\(\\left(a_{n}\\right)\\) does not converge.\n\n\n\nProof\nTo prove that \\(\\left(a_{n}\\right)\\) does not converge, we have to show that \\[\n\\forall \\, a \\in \\mathbb{R}\\,, \\, \\exists \\, \\varepsilon> 0 \\, \\text{ s.t. } \\, \\forall \\, N \\in \\mathbb{N}\\,, \\, \\exists \\, n \\geq N \\, \\text{ s.t. } \\, \\left|a_{n}-a\\right| \\geq \\varepsilon\\,.\n\\] Let \\(a \\in \\mathbb{R}\\). Choose \\[\n\\varepsilon=\\frac{1}{2} \\,.\n\\] Let \\(N \\in \\mathbb{N}\\). We distinguish two cases:\n\n\\(a \\geq 0\\): Choose \\(n=2 N+1\\). Note that \\(n \\geq N\\). Then \\[\\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} = \\varepsilon\\,,\n\\end{align*}\\] where we used that \\(a \\geq 0\\), and therefore \\[\n|-1-a|=1+a \\geq 1 \\,.\n\\]\n\\(a<0\\): Choose \\(n=2 N\\). Note that \\(n \\geq N\\). Then \\[\\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} = \\varepsilon\\,,\n\\end{align*}\\] where we used that \\(a < 0\\), and therefore \\[\n|1-a|=1-a > 1 \\,.\n\\]" }, { "objectID": "sections/chap_6.html#uniqueness-of-limit", "href": "sections/chap_6.html#uniqueness-of-limit", - "title": "6  Sequences", + "title": "6  Sequences in \\(\\mathbb{R}\\)", "section": "6.4 Uniqueness of limit", "text": "6.4 Uniqueness of limit\nIn Definition 5 of convergence, we used the notation \\[\n\\lim_{n \\rightarrow \\infty} a_{n}=a \\,.\n\\] The 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\\] for some \\[\na \\neq b \\,.\n\\] In the next theorem we will show that the limit is unique, if it exists.\n\nTheorem 15: Uniqueness of limitLet \\(\\left(a_{n}\\right)_{n \\in \\mathbb{N}}\\) be a sequence. Suppose that \\[\n\\lim_{n \\rightarrow \\infty} a_{n} = a \\,, \\quad\n\\lim_{n \\rightarrow \\infty} a_{n} = b \\,.\n\\] Then \\(a=b\\).\n\n\n\nProofAssume that, \\[\n\\lim_{n \\rightarrow \\infty} a_{n} = a \\,, \\quad\n\\lim_{n \\rightarrow \\infty} a_{n} = b \\,.\n\\] Suppose by contradiction that \\[\na \\neq b \\,.\n\\] Choose \\[\n\\varepsilon:= \\frac{1}{2} \\, |a-b| \\,.\n\\] Therefore \\(\\varepsilon>0\\), since \\(|a-b|>0\\). By the convergence \\(a_{n} \\rightarrow a\\), \\[\n\\exists \\, N_1 \\in \\mathbb{N}\\, \\text{ s.t. } \\, \\forall \\, n \\geq N_1 \\,, \\,\n\\left|a_{n}-a\\right| < \\varepsilon\\,.\n\\] By the convergence \\(a_{n} \\rightarrow b\\), \\[\n\\exists \\, N_2 \\in \\mathbb{N}\\, \\text{ s.t. } \\, \\forall \\, n \\geq N_2 \\,, \\,\n\\left|a_{n} - b \\right| < \\varepsilon\\,.\n\\] Define \\[\nN := \\max \\{ N_1, N_2 \\} \\,.\n\\] Choose an \\(n \\in \\mathbb{N}\\) such that \\(n \\geq N\\). In particular \\[\nn \\geq N_1 \\,, \\quad n \\geq N_2 \\,.\n\\] Then \\[\\begin{align*}\n2 \\varepsilon& = |a-b| \\\\\n& = \\left|a-a_{n}+a_{n}-b\\right| \\\\\n& \\leq\\left|a-a_{n}\\right|+\\left|a_{n}-b\\right| \\\\\n& < \\varepsilon+ \\varepsilon\\\\\n& = 2 \\varepsilon\\,,\n\\end{align*}\\] where we used the triangle inequality in the first inequality. Hence \\(2 \\varepsilon< 2 \\varepsilon\\), which gives a contradiction.\n\n\n\nExample 16Prove that\n\\[\n\\lim _{n \\rightarrow \\infty} \\frac{n^{2}-1}{2 n^{2}-3}=\\frac{1}{2}\n\\]\nAccording to Theorem 15, it suffices to show that the sequence \\[\n\\left(\\frac{n^{2}-1}{2 n^{2}-3}\\right)_{n \\in \\mathbb{N}}\n\\] converges to \\(\\frac{1}{2}\\), since then \\(\\frac{1}{2}\\) can be the only limit.\nPart 1. Rough Work.\nLet \\(\\varepsilon>0\\). We want to find \\(N \\in \\mathbb{N}\\) such that \\[\n\\left|\\frac{n^{2}-1}{2 n^{2}-3} - \\frac{1}{2}\\right| < \\varepsilon\\,, \\quad \\forall \\, n \\geq N \\,.\n\\] To this end, we compute: \\[\\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*}\\] which holds if \\(n \\geq 3\\), since in this case \\(n^2 - 6 \\geq 0\\). Therefore \\[\\begin{align*}\n\\left|\\frac{n^{2}-1}{2 n^{2}-3}-\\frac{1}{2}\\right| < \\varepsilon\n\\quad \\impliedby & \\quad\n\\frac{1}{3n^2} < \\varepsilon\\\\\n\\quad \\iff & \\quad\n3n^2 > \\frac{1}{\\varepsilon} \\\\\n\\quad \\iff & \\quad\nn^2 > \\frac{1}{3\\varepsilon} \\\\\n\\quad \\iff & \\quad\nn > \\frac{1}{\\sqrt{3\\varepsilon}} \\,.\n\\end{align*}\\] Looking at the above implications, it is clear that \\(N \\in \\mathbb{N}\\) has to be chosen so that \\[\nN > \\frac{1}{\\sqrt{3\\varepsilon}} \\,.\n\\] Moreover we need to recall that \\(N\\) has to satisfy \\[\nN \\geq 3\n\\] for the estimates to hold.\nPart 2. Formal Proof. We have to show that \\[\n\\forall \\varepsilon>0 \\,, \\, \\exists \\, N \\in \\mathbb{N}\\, \\text{ s.t. } \\, \\forall \\, n \\geq N \\,, \\, \\left|\\frac{n^{2}-1}{2 n^{2}-3}-\\frac{1}{2}\\right| < \\varepsilon\\,.\n\\] Let \\(\\varepsilon>0\\). Choose \\(N \\in \\mathbb{N}\\) such that \\[\nN > \\max \\left\\{ \\frac{1}{\\sqrt{3\\varepsilon}}, 3 \\right\\} \\,.\n\\] Let \\(n \\geq N\\). Then \\[\\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& < \\varepsilon\\,,\n\\end{align*}\\] where we used that \\[\nn \\geq N \\geq 3\n\\] which implies \\[\nn^2 - 6 \\geq 0\\,,\n\\] in the third line. The last inequality holds, since it is equivalent to \\[\nN > \\frac{1}{\\sqrt{3 \\varepsilon}} \\,.\n\\]" }, { "objectID": "sections/chap_6.html#bounded-sequences", "href": "sections/chap_6.html#bounded-sequences", - "title": "6  Sequences", + "title": "6  Sequences in \\(\\mathbb{R}\\)", "section": "6.5 Bounded sequences", "text": "6.5 Bounded sequences\nAn important property of sequences is boundedness.\n\nDefinition 17: Bounded sequenceA sequence \\(\\left(a_{n}\\right)_{n \\in \\mathbb{N}}\\) is called bounded if there exists a constant \\(M \\in \\mathbb{R}\\), with \\(M>0\\), such that \\[\n\\left|a_{n}\\right| \\leq M \\,, \\quad \\forall \\, n \\in \\mathbb{N}\\,.\n\\]\n\n\nDefinition 17 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\\left|a_{n}\\right| \\leq M \\,,\n\\] or equivalently, that \\[\n-M \\leq a_{n} \\leq M \\,.\n\\]\nWe now show that any sequence that converges is also bounded\n\nTheorem 18If a sequence \\(\\left(a_{n}\\right)_{n \\in \\mathbb{N}}\\) converges, then the sequence is bounded.\n\n\n\nProofSuppose the sequence \\(\\left(a_{n}\\right)_{n \\in \\mathbb{N}}\\) converges and let \\[\na: = \\lim_{n \\rightarrow \\infty} a_{n}\n\\] By definition of convergence we have that \\[\n\\forall \\, \\varepsilon>0 \\,, \\, \\exists N \\in \\mathbb{N}\\, \\text{ s.t. } \\, \\forall \\, n \\geq N \\,, \\, \\left|a_{n}-a \\right| < \\varepsilon\\,.\n\\] In particular, we can choose \\[\n\\varepsilon= 1\n\\] and let \\(N \\in \\mathbb{N}\\) be that value such that \\[\n\\left|a_{n}-a \\right| < 1 \\,, \\quad \\forall \\, n \\geq N \\,.\n\\] If \\(n \\geq N\\) we have, by the triangle inequality, \\[\\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*}\\] Set \\[\nM:=\\max \\left\\{\\left|a_{1}\\right|,\\left|a_{2}\\right|, \\ldots,\\left|a_{N-1}\\right|, 1+|a|\\right\\} \\,.\n\\] Note that such maximum exists, being the set finite. Then \\[\n\\left|a_{n}\\right| \\leq M \\,, \\quad \\forall \\, n \\in \\mathbb{N}\\,,\n\\] showing that \\((a_n)\\) is bounded.\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\nExample 19In Theorem 7 we have shown that \\[\n\\lim_{n \\to \\infty} \\, \\frac{1}{n} = 0\n\\] Hence, it follows from Theorem 18 that the sequence \\((1/n)\\) is bounded.\nIndeed, we have that \\[\n\\left|\\frac{1}{n}\\right|=\\frac{1}{n} \\leq 1 \\,, \\quad \\forall \\, n \\in \\mathbb{N}\\,,\n\\] since \\(n \\geq 1\\) for all \\(n \\in \\mathbb{N}\\).\n\n\n\nWarningThe converse of Theorem 18 does not hold: There exist sequences \\((a_n)\\) which are bounded, but not convergent.\n\n\n\nExample 20Define the sequence \\[\na_n = (-1)^n \\,.\n\\] We have proven in Theorem 14 that \\((a_n)\\) is not convergent. However \\((a_n)\\) is bounded, with \\(M=1\\), since \\[\n|a_n| = |(-1)^n| = 1 = M \\,, \\quad \\forall \\, n \\in \\mathbb{N}\\,.\n\\]\n\n\nTaking the contrapositive of the statement in Theorem 18 we get the following corollary:\n\nCorollary 21If a sequence \\(\\left(a_{n}\\right)_{n \\in \\mathbb{N}}\\) is not bounded, then the sequence does not converge.\n\n\n\nRemark 22For a sequence \\((a_n)\\) to be unbounded, it means that \\[\n\\forall \\, M > 0 \\,, \\,\\, \\exists \\, n \\in \\mathbb{N}\\, \\text{ s.t. } \\, \\left|a_{n}\\right| > M \\,.\n\\] The above is saying that no real number \\(M>0\\) can be a bound for \\(|a_n|\\), since there is always an index \\(n \\in \\mathbb{N}\\) such that \\[\n|a_n| > M \\,.\n\\]\n\n\nWe can use Corollary 21 to show that certain sequences do not converge.\n\nTheorem 23For all \\(p>0\\), the sequence \\(\\left(n^{p}\\right)_{n \\in \\mathbb{N}}\\) does not converge.\n\n\n\nProofLet \\(p>0\\). We prove that the sequence \\(\\left(n^{p}\\right)_{n \\in \\mathbb{N}}\\) is unbounded, that is, \\[\n\\forall \\, M > 0 \\,, \\,\\, \\exists \\, n \\in \\mathbb{N}\\, \\text{ s.t. } \\, \\left|a_{n}\\right| > M \\,.\n\\] To this end, let \\(M > 0\\). Choose \\(n \\in \\mathbb{N}\\) such that \\[\nn > M^{1/p} \\,.\n\\] Then \\[\na_n = n^{p}>\\left(M^{1/p}\\right)^{p}=M \\,.\n\\] This proves that the sequence \\((n^p)\\) is unbounded. Hence \\((n^p)\\) cannot converge, by Corollary 21.\n\n\n\nTheorem 24The sequence \\((\\log n)_{n \\in \\mathbb{N}}\\) does not converge.\n\n\n\nProofLet us show that \\((\\log n)_{n \\in \\mathbb{N}}\\) is unbounded, that is, \\[\n\\forall \\, M > 0 \\,, \\,\\, \\exists \\, n \\in \\mathbb{N}\\, \\text{ s.t. } \\, \\left|a_{n}\\right| > M \\,.\n\\] To this end let \\(M > 0\\). Choose \\(n \\in \\mathbb{N}\\) such that \\[\nn \\geq e^{M+1} \\,.\n\\] Then \\[\n|a_n| = |\\log n| \\geq \\left| \\log e^{M+1} \\right| = M + 1 > M \\,.\n\\] This proves that the sequence \\((\\log n)\\) is unbounded. Hence \\((\\log n)\\) cannot converge, by Corollary 21." }, { "objectID": "sections/chap_6.html#algebra-of-limits", "href": "sections/chap_6.html#algebra-of-limits", - "title": "6  Sequences", + "title": "6  Sequences in \\(\\mathbb{R}\\)", "section": "6.6 Algebra of limits", "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 25: 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> 0 \\,, \\, \\exists \\, N \\in \\mathbb{N}\\, \\text{ s.t. } \\, \\forall \\, n \\geq N \\,, \\, \\left|(a_{n} + b_n) - (a+b) \\right| < \\varepsilon\\,.\n\\] Let \\(\\varepsilon>0\\) and set \\[\n\\widetilde{\\varepsilon} := \\frac{\\varepsilon}{2} \\,.\n\\] Since \\(a_{n} \\rightarrow a\\) and \\(\\widetilde{\\varepsilon}>0\\), there exists \\(N_1 \\in \\mathbb{N}\\) such that \\[\n|a_n - a|< \\widetilde{\\varepsilon} \\,, \\quad \\forall \\, n \\geq N_1 \\,.\n\\] Since \\(b_{n} \\rightarrow b\\) and \\(\\widetilde{\\varepsilon}>0\\), there exists \\(N_2 \\in \\mathbb{N}\\) such that \\[\n|b_n - b|< \\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& < \\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> 0 \\,, \\, \\exists \\, N \\in \\mathbb{N}\\, \\text{ s.t. } \\, \\forall \\, n \\geq N \\,, \\, \\left|a_{n} b_{n} - a b \\right| < \\varepsilon\\,.\n\\] Let \\(\\varepsilon>0\\). The sequence \\(\\left(a_{n}\\right)\\) converges, and hence is bounded, by Theorem 18. This means there exists some \\(M>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} > 0\\), there exists \\(N_1 \\in \\mathbb{N}\\) such that \\[\n| a_n - a | < \\widetilde{\\varepsilon} \\,, \\quad \\forall \\, n \\geq N_1 \\,.\n\\] Since \\(b_{n} \\rightarrow b\\) and \\(\\widetilde{\\varepsilon} > 0\\), there exists \\(N_2 \\in \\mathbb{N}\\) such that \\[\n| b_n - b | < \\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& < 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> 0 \\,, \\, \\exists \\, N \\in \\mathbb{N}\\, \\text{ s.t. } \\, \\forall \\, n \\geq N \\,, \\, \\left|\\frac{a_{n}}{ b_{n}} - \\frac{a}{b} \\right| < \\varepsilon\\,.\n\\] We suppose in addition that \\(b>0\\). The proof is very similar for the case \\(b <0\\). Let \\(\\varepsilon> 0\\). Set \\[\n\\delta := \\frac{b}{2} \\,.\n\\] Since \\(b_n \\to b\\) and \\(\\delta>0\\), there exists \\(N_1 \\in \\mathbb{N}\\) such that \\[\n|b_n - b| < \\delta \\, \\quad \\forall \\, n \\geq N_1 \\,.\n\\] In particular we have \\[\nb_n > 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}>0\\) and \\(a_n \\to a\\), there exists \\(N_2 \\in \\mathbb{N}\\) such that \\[\n|a_n - a |< \\widetilde{\\varepsilon} \\,, \\quad \\forall \\, n \\geq N_2 \\,.\n\\] Since \\(\\widetilde{\\varepsilon}>0\\) and \\(b_n \\to b\\), there exists \\(N_3 \\in \\mathbb{N}\\) such that \\[\n|b_n - b |< \\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& < \\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 25 as the Algebra of Limits. We now show how to use Theorem 25 for computing certain limits.\n\nExample 26\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 11 we know that \\[\n3 \\rightarrow 3\\,, \\quad 4 \\to 4 \\,, \\quad 7 \\to 7 \\,.\n\\] From Theorem 7 we know that \\[\n\\frac{1}{n} \\rightarrow 0 \\,.\n\\] Hence, it follows from Theorem 25 Point 2 that \\[\n\\frac{4}{n} = 4 \\cdot \\frac1n \\rightarrow 4 \\cdot 0 = 0 \\,.\n\\] By Theorem 25 Point 1 we have \\[\n7 + \\frac{4}{n} \\rightarrow 7 + 0 = 7 \\,.\n\\] Finally we can use Theorem 25 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 26 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 27\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 8 we have \\[\n\\frac{1}{n^2} \\to 0 \\,.\n\\] We can then use the Algebra of Limits Theorem 25 Point 2 to infer \\[\n\\frac{3}{n^2} \\to 3 \\cdot 0 = 0\n\\] and Theorem 25 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 25 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 28\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 25 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 25 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 25 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 27, 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 29\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 25 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 25 Point 2 ensures that \\[\na_{n} \\to \\frac{8}{5} \\cdot \\frac{1}{3}=\\frac{8}{15} \\,.\n\\]" }, { "objectID": "sections/chap_6.html#fractional-powers", "href": "sections/chap_6.html#fractional-powers", - "title": "6  Sequences", + "title": "6  Sequences in \\(\\mathbb{R}\\)", "section": "6.7 Fractional powers", "text": "6.7 Fractional powers\nThe Algebra of Limits Theorem 25 can also be used when fractional powers of \\(n\\) are involved.\n\nExample 30\nProve that \\[\na_n = \\frac{n^{7 / 3}+2 \\sqrt{n}+7}{4 n^{3 / 2}+5 n}\n\\] does not converge.\n\nProof. The largest power of \\(n\\) in the denominator is \\(n^{3/2}\\). Hence we factor out \\(n^{3/2}\\) \\[\\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*}\\] where we set \\[\nb_n := n^{5/6}+2 n^{- 1} + 7 n^{-3 / 2} \\,, \\quad\nc_n := 4 + 5 n^{-3/2} \\,.\n\\] We 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\nWe now present a general result about the square root of a sequence.\n\nTheorem 31Let \\(\\left(a_{n}\\right)_{n \\in \\mathbb{N}}\\) be a sequence in \\(\\mathbb{R}\\) such that \\[\n\\lim_{n \\to \\infty} \\, a_n = a \\,,\n\\] for some \\(a \\in \\mathbb{R}\\). If \\(a_{n} \\geq 0\\) for all \\(n \\in \\mathbb{N}\\) and \\(a \\geq 0\\), then \\[\n\\lim _{n \\rightarrow \\infty} \\sqrt{a_{n}}=\\sqrt{a} \\,.\n\\]\n\n\n\nProof\nLet \\(\\varepsilon>0\\). We the two cases \\(a>0\\) and \\(a=0\\):\n\n\\(a>0\\): Define \\[\n\\delta := \\frac{a}{2} \\,.\n\\] Since \\(\\delta > 0\\) and \\(a_n \\to a\\), there exists \\(N_1 \\in \\mathbb{N}\\) such that \\[\n\\left|a_{n}-a\\right| < \\delta \\,, \\quad \\forall \\, n \\geq N_1 \\,.\n\\] In particular \\[\na_n > a - \\delta = a - \\frac{a}{2} = \\frac{a}{2} \\,, \\quad \\forall \\, n \\geq N_1 \\,,\n\\] from which we infer \\[\n\\sqrt{a_n} > \\sqrt{a/2} \\,, \\quad \\forall \\, n \\geq N_1 \\,,\n\\] Now set \\[\n\\widetilde{\\varepsilon} := \\left(\\sqrt{a/2} + \\sqrt{a} \\right) \\, \\varepsilon\\,.\n\\] Since \\(\\widetilde{\\varepsilon}>0\\) and \\(a_n \\to a\\), there exists \\(N_2 \\in \\mathbb{N}\\) such that \\[\n\\left|a_{n}-a\\right| < \\widetilde{\\varepsilon} \\,, \\quad \\forall \\, n \\geq N_2 \\,.\n\\] Let \\[\nN:= \\max\\{ N_1, N_2 \\} \\,.\n\\] For \\(n \\geq N\\) we have \\[\\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{\\varepsilon} }{\\sqrt{a/2} + \\sqrt{a}} \\\\\n& = \\varepsilon\\,.\n\\end{align*}\\]\n\\(a=0\\): In this case \\[\na_n \\to a = 0 \\,.\n\\] Since \\(\\varepsilon^2>0\\), there exists \\(N \\in \\mathbb{N}\\) such that \\[\n|a_n - 0 | = |a_n| < \\varepsilon^2 \\,, \\quad \\forall \\, n \\geq N \\,.\n\\] Therefore \\[\n|\\sqrt{a_n} - \\sqrt{0} | = | \\sqrt{a_n}| < \\sqrt{\\varepsilon^2} = \\varepsilon\\,, \\quad \\forall \\, n \\geq N \\,.\n\\]\n\n\n\nLet us show an application of Theorem 31.\n\nExample 32\nDefine the sequence \\[\na_n = \\sqrt{9 n^{2}+3 n+1}-3 n \\,.\n\\] Prove that \\[\n\\lim_{n \\to \\infty} \\, a_n = \\frac12 \\,.\n\\]\n\nProof. We first rewrite \\[\\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*}\\] The biggest power of \\(n\\) in the denominator is \\(n\\). Therefore we factor out \\(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*}\\] By the Algebra of Limits we have \\[\n9+ \\frac{3}{n} + \\frac{1}{n^{2}} \\to 9 + 0 + 0 = 9 \\,.\n\\] Therefore we can use Theorem 31 to infer \\[\n\\sqrt{ 9 + \\frac{3}{n} + \\frac{1}{n^{2}} } \\to \\sqrt{9} \\,.\n\\] By the Algebra of Limits we conclude: \\[\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\nExample 33\nProve that the sequence \\[\na_n = \\sqrt{9 n^{2}+3 n+1}-2 n\n\\] does not converge.\n\nProof. We rewrite \\(a_n\\) as \\[\\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*}\\] where we factored \\(n\\), being it the largest power of \\(n\\) in the denominator, and defined \\[\nb_n : = 5 n + 3 + \\dfrac{1}{n}\\,, \\quad\nc_n := \\sqrt{9+ \\dfrac{3}{n} + \\dfrac{1}{ n^2 } } + 2 \\,.\n\\] Note that \\[\n9+ \\dfrac{3}{n} + \\dfrac{1}{ n^2 } \\to 9\n\\] by the Algebra of Limits. Therefore \\[\n\\sqrt{9+ \\dfrac{3}{n} + \\dfrac{1}{ n^2 }} \\to \\sqrt{9} = 3\n\\] by Theorem 31. Hence \\[\nc_n = \\sqrt{9+ \\dfrac{3}{n} + \\dfrac{1}{ n^2 }} + 2 \\to 3 + 2 = 5 \\,.\n\\] The numerator \\[\nb_n = 5 n + 3 + \\dfrac{1}{n}\n\\] is instead unbounded. Therefore \\((a_n)\\) is not convergent, by the Algebra of Limits and the usual contradiction argument." }, { "objectID": "sections/chap_6.html#limit-tests", "href": "sections/chap_6.html#limit-tests", - "title": "6  Sequences", + "title": "6  Sequences in \\(\\mathbb{R}\\)", "section": "6.8 Limit tests", - "text": "6.8 Limit tests\nIn this section we discuss a number of tests to determine whether a sequence converges or not. These are known as limit tests.\n\n6.8.1 Squeeze Theorem\nWhen a sequence \\((a_n)\\) oscillates, it is difficult to compute the limit. Examples of terms which produce oscillations are \\[\n(-1)^n \\,, \\quad \\sin(n) \\,, \\quad \\cos(n) \\,.\n\\] In 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\nTheorem 34: Squeeze theoremLet \\(\\left(a_{n}\\right), \\left(b_{n}\\right)\\) and \\(\\left(c_{n}\\right)\\) be sequences in \\(\\mathbb{R}\\). Suppose that \\[\nb_{n} \\leq a_{n} \\leq c_{n} \\,, \\quad \\forall \\, n \\in \\mathbb{N}\\,,\n\\] and that \\[\n\\lim_{n \\rightarrow \\infty} b_{n} = \\lim_{n \\rightarrow \\infty} c_{n} = L \\, .\n\\] Then \\[\n\\lim_{n \\rightarrow \\infty} a_{n}= L \\, .\n\\]\n\n\n\nProofLet \\(\\varepsilon>0\\). Since \\(b_{n} \\to L\\) and \\(c_n \\to L\\) , there exist \\(N_1, N_2 \\in \\mathbb{N}\\) such that \\[\\begin{align*}\n-\\varepsilon< b_n - L < \\varepsilon\\,, \\,\\, \\forall \\, n \\geq N_1 \\,, \\\\\n- \\varepsilon< c_n - L < \\varepsilon\\,, \\,\\, \\forall \\, n \\geq N_2 \\,.\n\\end{align*}\\] Set \\[\nN := \\max\\{N_1,N_2\\} \\,.\n\\] Let \\(n \\geq N\\). Using the assumption that \\(b_n \\leq a_{n} \\leq c_{n}\\), we get \\[\nb_n - L \\leq a_{n} - L \\leq c_{n} - L \\,.\n\\] In particular \\[\n- \\varepsilon< b_n - L \\leq a_n - L \\leq b_n - L < \\varepsilon\\,.\n\\] The above implies \\[\n- \\varepsilon< a_n - L < \\varepsilon\\quad \\implies \\quad \\left|a_{n}-L\\right| < \\varepsilon\\,.\n\\]\n\n\n\nExample 35\nProve that \\[\n\\lim_{n \\to \\infty} \\, \\frac{(-1)^{n}}{n} = 0 \\,.\n\\]\n\nProof. For all \\(n \\in \\mathbb{N}\\) we can estimate \\[\n-1 \\leq(-1)^{n} \\leq 1 \\,.\n\\] Therefore \\[\n\\frac{-1}{n} \\leq \\frac{(-1)^{n}}{n} \\leq \\frac{1}{n} \\,, \\quad \\forall \\, n \\in \\mathbb{N}\\,.\n\\] Moreover \\[\n\\lim_{n \\to \\infty} \\frac{-1}{n}= -1 \\cdot 0=0 \\,, \\quad\n\\lim_{n \\to \\infty} \\frac{1}{n}=0 \\,.\n\\] By the Squeeze Theorem 34 we conclude \\[\n\\lim_{n \\to \\infty} \\frac{(-1)^{n}}{n}=0 \\,.\n\\]\n\n\n\n\nExample 36\nProve that \\[\n\\lim_{n \\to \\infty} \\frac{\\cos (3 n) + 9 n^{2}}{11 n^{2}+15 \\sin (17 n)} = \\frac{9}{11} \\,.\n\\]\n\nProof. We know that \\[\n-1 \\leq \\cos(x) \\leq 1 \\,, \\quad - 1 \\leq \\sin(x) \\leq 1 \\,, \\quad \\forall \\, x \\in \\mathbb{R}\\,.\n\\] Therefore, for all \\(n \\in \\mathbb{N}\\) \\[\n- 1 \\leq \\cos(3n) \\leq 1 \\,, \\quad -1 \\leq \\sin(17n) \\leq 1 \\,.\n\\] We can use the above to estimate the numerator in the given sequence: \\[\n-1 + 9 n^{2} \\leq \\cos (3 n)+9 n^{2} \\leq 1+ 9 n^{2} \\,.\n\\tag{6.7}\\] Concerning the denominator, we have \\[\n11 n^{2}-15 \\leq 11 n^{2}+15 \\sin (17 n) \\leq 11 n^{2} + 15\n\\] and therefore \\[\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\\tag{6.8}\\] Putting together (6.7)-(6.8) we obtain \\[\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\\] By the Algebra of Limits we infer \\[\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\\] and \\[\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\\] Applying the Squeeze Theorem 34 we conclude \\[\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\nWarningSuppose that the sequences \\((a_n), (b_n), (c_n)\\) satisfy \\[\nb_n \\leq a_n \\leq c_n \\,, \\quad \\forall n \\in \\mathbb{N}\\,,\n\\] and \\[\nb_n \\to L_1 \\,, \\quad c_n \\to L_2 \\,, \\quad L_1 \\neq L_2 \\,.\n\\] In general, we cannot conclude that \\(a_n\\) converges.\n\n\n\nExample 37\nConsider the sequence \\[\na_n = \\left( 1 + \\frac{1}{n} \\right) (-1)^n \\,.\n\\] For all \\(n \\in \\mathbb{N}\\) we can bound \\[\n- 1 - \\frac{1}{n} \\leq \\left( 1 + \\frac{1}{n} \\right) (-1)^n \\leq 1 + \\frac{1}{n} \\,.\n\\] However \\[\n- 1 - \\frac{1}{n} \\longrightarrow - 1 - 0 = -1\n\\] and \\[\n1 + \\frac{1}{n} \\longrightarrow 1 + 0 = 1 \\,.\n\\] Since \\[\n- 1 \\neq 1 \\,,\n\\] we cannot apply the Squeeze Theorem 34 to conclude convergence of \\((a_n)\\). Indeed, \\((a_n)\\) is a divergent sequence.\n\nProof. Suppose by contradiction that \\(a_n \\to a\\). We have \\[\na_n = (-1)^n + \\frac{(-1)^n}{n} = b_n + c_n\n\\] where \\[\nb_n := (-1)^n \\,, \\quad c_n := \\frac{(-1)^n}{n} \\,.\n\\] We have seen in Example 36 that \\(c_n \\to 0\\). Therefore, by the Algebra of Limits, we have \\[\nb_n = a_n - c_n \\longrightarrow a - 0 = a \\,.\n\\] However, Theorem 14 says that the sequence \\(b_n = (-1)^n\\) diverges. Contradiction. Hence \\((a_n)\\) diverges.\n\n\n\n\n\n6.8.2 Geometric sequences\n\nDefinition 38A sequence \\(\\left(a_{n}\\right)\\) is called a geometric sequence if \\[\na_{n}=x^{n} \\,,\n\\] for some \\(x \\in \\mathbb{R}\\).\n\n\nThe value of \\(|x|\\) determines whether or not a geometric sequence converges, as shown in the following theorem.\n\nTheorem 39: Geometric sequence test\nLet \\(x \\in \\mathbb{R}\\) and let \\(\\left(a_{n}\\right)\\) be the sequence defined by \\[\na_{n}:=x^{n} \\,.\n\\] We have:\n\nIf \\(|x|<1\\), then \\[\n\\lim_{n \\to \\infty} a_{n} = 0 \\,.\n\\]\nIf \\(|x|>1\\), then sequence \\(\\left(a_{n}\\right)\\) is unbounded, and hence divergent.\n\n\n\n\nWarning\nThe Geometric sequence test in Theorem 39 does not address the case \\[\n|x|=1 \\,.\n\\] This is because, in this case, the sequence \\[\na_n = x^n\n\\] might converge or diverge, depending on the value of \\(x\\). Indeed, \\[\n|x| = 1 \\quad \\implies \\quad x = \\pm 1 \\,.\n\\] We therefore have two cases:\n\n\\(x = 1\\): Then \\[\na_n = 1^n = 1\n\\] so that \\(a_n \\to 1\\) and \\((a_n)\\) is convergent.\n\\(x=-1\\): Then \\[\na_n = x^n = (-1)^n\n\\] which is divergent by Theorem 14.\n\n\n\nTo prove Theorem 39 we need the following inequality, known as Bernoulli’s inequality.\n\nLemma 40: Bernoulli’s inequalityLet \\(x \\in \\mathbb{R}\\) with \\(x>-1\\). Then \\[\n(1+x)^{n} \\geq 1+n x \\,, \\quad \\forall \\, n \\in \\mathbb{N}\\,.\n\\tag{6.9}\\]\n\n\n\nProofLet \\(x \\in \\mathbb{R}, x>-1\\). We prove the statement by induction:\n\nBase case: (6.9) holds with equality when \\(n=1\\).\nInduction hypothesis: Let \\(k \\in \\mathbb{N}\\) and suppose that (6.9) holds for \\(n=k\\), i.e., \\[\n(1+x)^{k} \\geq 1+k x \\,.\n\\] Then \\[\\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*}\\] where we used that \\(kx^2 \\geq 0\\). Then (6.9) holds for \\(n=k+1\\).\n\nBy induction we conclude (6.9).\n\n\nWe are ready to prove Theorem 39.\n\nProof: Proof of Theorem 39Part 1. The case \\(|x|<1\\).\nIf \\(x=0\\), then \\[\na_n = x^n = 0\n\\] so that \\(a_n \\to 0\\). Hence assume \\(x \\neq 0\\). We need to prove that \\[\n\\forall \\, \\varepsilon> 0 \\,, \\, \\exists \\, N \\in \\mathbb{N}\\, \\text{ s.t. } \\,\n\\forall \\, n \\geq N \\,, \\,\\, |x^n - 0| < \\varepsilon\\,.\n\\] Let \\(\\varepsilon>0\\). We have \\[\n|x| < 1 \\quad \\implies \\quad \\frac{1}{|x|} > 1 \\,.\n\\] Therefore \\[\n|x|= \\frac{1}{1+u} \\,, \\quad u:=\\frac{1}{|x|} - 1 > 0 \\,.\n\\] Let \\(N \\in \\mathbb{N}\\) be such that \\[\nN > \\frac{1}{\\varepsilon u} \\,,\n\\] so that \\[\n\\frac{1}{N u} < \\varepsilon\\,.\n\\] Let \\(n \\geq N\\). Then \\[\\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 & < \\varepsilon\\,,\n\\end{align*}\\] where we used Bernoulli’s inequality (6.9) in the first inequality.\nPart 2. The case \\(|x|>1\\).\nTo prove that (a_n) does not converge, we prove that it is unbounded. This means showing that \\[\n\\forall \\, M > 0 \\,, \\, \\exists n \\in \\mathbb{N}\\, \\text{ s.t. } \\, \\left| a_{n}\\right| >M \\,.\n\\] Let \\(M > 0\\). We have two cases:\n\n\\(0 < M \\leq 1\\): Choose \\(n=1\\). Then \\[\n\\left|a_{1}\\right|=|x|>1 \\geq M \\,.\n\\]\n\\(M>1\\): Choose \\(n \\in \\mathbb{N}\\) such that \\[\nn> \\frac{\\log M}{\\log |x|} \\,.\n\\] Note that \\(\\log |x|>0\\) since \\(|x|>1\\). Therefore \\[\\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*}\\] Then \\[\n\\left|a_{n}\\right|=\\left|x^{n}\\right|=|x|^{n} > M \\,.\n\\]\n\nHence \\((a_n)\\) is unbounded. By Corollary 21 we conclude that \\((a_n)\\) is divergent.\n\n\n\nExample 41\nWe can apply Theorem 39 to prove convergence or divergence for the following sequences.\n\nWe have \\[\n\\left(\\frac{1}{2}\\right)^{n} \\longrightarrow 0\n\\] since \\[\n\\left|\\frac{1}{2}\\right|=\\frac{1}{2}<1 \\,.\n\\]\nWe have \\[\n\\left(\\frac{-1}{2}\\right)^{n} \\longrightarrow 0\n\\] since \\[\n\\left|\\frac{-1}{2}\\right|=\\frac{1}{2}<1 \\,.\n\\]\nThe sequence \\[\na_n = \\left(\\frac{-3}{2}\\right)^{n}\n\\] does not converge, since \\[\n\\left|\\frac{-3}{2}\\right|=\\frac{3}{2}>1 \\,.\n\\]\nAs \\(n \\rightarrow \\infty\\), \\[\n\\frac{3^{n}}{(-5)^{n}}=\\left(-\\frac{3}{5}\\right)^{n} \\longrightarrow 0\n\\] since \\[\n\\left|-\\frac{3}{5}\\right|=\\frac{3}{5}<1 \\,.\n\\]\nThe sequence \\[\na_{n}=\\frac{(-7)^{n}}{2^{2 n}}\n\\] does not converge, since \\[\n\\frac{(-7)^{n}}{2^{2 n}}=\\frac{(-7)^{n}}{\\left(2^{2}\\right)^{n}}=\\left(-\\frac{7}{4}\\right)^{n}\n\\] and \\[\n\\left|-\\frac{7}{4}\\right|=\\frac{7}{4}>1 \\,.\n\\]\n\n\n\n\n\n6.8.3 Ratio test\n\nTheorem 42: Ratio test\nLet \\(\\left(a_{n}\\right)\\) be a sequence in \\(\\mathbb{R}\\) such that \\[\na_{n} \\neq 0 \\,, \\quad \\forall \\, n \\in \\mathbb{N}\\,.\n\\]\n\nSuppose that the following limit exists: \\[\nL:=\\lim_{n \\to \\infty}\\left|\\frac{a_{n+1}}{a_{n}}\\right| \\,.\n\\] Then,\n\nIf \\(L<1\\) we have \\[\n\\lim_{n \\to\\infty} a_{n}=0 \\,.\n\\]\nIf \\(L>1\\), the sequence \\(\\left(a_{n}\\right)\\) is unbounded, and hence does not converge.\n\nSuppose that there exists \\(N \\in \\mathbb{N}\\) and \\(M>1\\) such that \\[\n\\left|\\frac{a_{n+1}}{a_{n}}\\right| \\geq M \\,, \\quad \\forall \\, n \\geq N \\,. \\,.\n\\] Then the sequence \\(\\left(a_{n}\\right)\\) is unbounded, and hence does not converge.\n\n\n\n\nProofDefine the sequence \\(b_{n}=\\left|a_{n}\\right|\\). Then,\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\\]\nPart 1. Suppose that there exists the limit \\[\nL:=\\lim_{n \\to \\infty}\\left|\\frac{a_{n+1}}{a_{n}}\\right| \\,.\n\\] Therefore \\[\n\\lim_{n \\to \\infty} \\frac{b_{n+1}}{b_{n}} = L \\,.\n\\tag{6.10}\\]\n\n\\(L<1\\): Choose \\(r>0\\) such that \\[\nL<r<1 \\,.\n\\] Set \\[\n\\varepsilon:= r-L\n\\] By the convergence at (6.10) there exists \\(N \\in \\mathbb{N}\\) such that \\[\n\\left|\\frac{b_{n+1}}{b_{n}}-L\\right| < \\varepsilon= r - L \\,, \\quad \\forall \\, n \\geq N \\,.\n\\] In particular \\[\n\\frac{b_{n+1}}{b_{n}}-L < r-L \\,, \\quad \\forall \\, n \\geq N \\,,\n\\] which implies \\[\nb_{n+1} < r \\,{b_{n}} \\,, \\quad \\forall \\, n \\geq N \\,.\n\\tag{6.11}\\] Let \\(n \\geq N\\), we can use (6.11) recursively and obtain \\[\n0 \\leq b_{n} < \\, r b_{n-1} < \\ldots < r^{n-N} \\, b_{N} = r^{n} \\, \\frac{b_{N}}{r^{N}} \\,.\n\\] In particular, we have proven that \\[\n0 \\leq b_{n} < r^{n} \\, \\frac{b_{N}}{r^{N}} \\,, \\quad \\forall \\, n \\in \\mathbb{N}\\,.\n\\tag{6.12}\\] Since \\(|r|<1\\), by the Geometric sequence test Theorem 39 we infer \\[\nr^{n} \\rightarrow 0 \\,.\n\\] The Algebra of Limits the yields \\[\nr^{n} \\, \\frac{b_{N}}{r^{N}} \\rightarrow 0 \\,.\n\\] By the Squeeze Theorem 34 applied to (6.12), it follows that \\[ \nb_{n} = |a_n| \\to 0 \\,.\n\\] Since \\[\n-\\left|a_{n}\\right| \\leq a_{n} \\leq\\left|a_{n}\\right| \\,,\n\\] and \\[\n-|a_n| \\to 0 \\,, \\quad |a_n| \\to 0 \\,,\n\\] we can again apply the Squeeze Theorem 34 to infer \\[\na_{n} \\rightarrow 0 \\,.\n\\]\n\\(L>1\\): Choose \\(r>0\\) such that \\(1<r<L\\). Let \\(\\varepsilon=L-r>0\\). If \\(\\frac{b_{n+1}}{b_{n}} \\rightarrow L\\), there exists \\(n_{0} \\in \\mathbb{N}\\) such that for all \\(n \\geq n_{0}\\)\n\n\\[\n\\left|\\frac{b_{n+1}}{b_{n}}-L\\right| \\leq \\varepsilon=L-r\n\\]\nIn particular,\n\\[\n-(L-r) \\leq \\frac{b_{n+1}}{b_{n}}-L \\Longleftrightarrow \\frac{b_{n+1}}{b_{n}} \\geq r \\Longleftrightarrow b_{n+1} \\geq r b_{n}\n\\]\n(In the other case we assumed this is the case, so this statement also holds there.) Hence, for all \\(n \\geq n_{0}\\)\n\\[\nb_{n} \\geq r^{n-n_{0}} b_{n_{0}}=r^{n} \\frac{b_{n_{0}}}{r^{n_{0}}}\n\\]\nSince \\(|r|>1\\), it follows from the geometric sequence test that the right hand side is unbounded and hence also \\(b_{n}\\) is unbounded.\n\n\n\nExample 43Let \\(a_{n}=\\frac{3^{n}}{n !}\\). Then,\n\\[\n\\left|\\frac{a_{n+1}}{a_{n}}\\right|=\\frac{3^{n+1}}{(n+1) !} / \\frac{3^{n}}{n !}=\\frac{3^{n+1}}{3^{n}} \\frac{n !}{(n+1) !}=3 \\frac{1 \\cdot 2 \\cdots n}{1 \\cdot 2 \\cdots n \\cdot(n+1)}=\\frac{3}{n+1} \\longrightarrow 0,\n\\]\nas \\(n \\rightarrow \\infty\\). Hence, \\(L=0<1\\) so \\(\\lim _{n \\rightarrow \\infty} a_{n}=0\\) by the ratio test.\n\nLet \\(b_{n}=\\frac{n !}{100^{n}}\\). Then\n\n\\[\n\\left|\\frac{b_{n+1}}{b_{n}}\\right|=\\frac{(n+1) !}{100^{n+1}} / \\frac{n !}{100^{n}}=\\frac{100^{n}}{100^{n+1}} \\frac{(n+1) !}{(n !)}=\\frac{n+1}{100} .\n\\]\nChoose \\(n_{0}=101\\). Then for all \\(n \\geq n_{0}\\),\n\\[\n\\left|\\frac{b_{n+1}}{b_{n}}\\right| \\geq \\frac{101}{100}>1\n\\]\nHence, \\(b_{n}\\) is an unbounded sequence and does not converge.\n\n\n\nWarningFor part (c) of the theorem we give some examples. If you try to apply the ratio test for a fraction of two polynomials, the limit will always be equal to 1 , so the ratio test is inconclusive for this type of sequence.\n\n\nExample 2.4.13. \\(\\quad\\) Let \\(a_{n}=\\frac{1}{n}\\). Then,\n\\[\n\\left|\\frac{a_{n+1}}{a_{n}}\\right|=\\frac{1}{n+1} / \\frac{1}{n}=\\frac{n}{n+1} \\rightarrow 1\n\\]\nso the ratio test is inconclusive. But we know that \\(\\frac{1}{n} \\rightarrow 0\\).\n\nLet \\(b_{n}=\\frac{n^{2}}{n+1}\\). Then,\n\n\\[\n\\left|\\frac{b_{n+1}}{b_{n}}\\right|=\\frac{(n+1)^{2}}{(n+1)+1} / \\frac{n^{2}}{n+1}=\\frac{(n+1)^{3}}{(n+2) n^{2}}=\\frac{(1+1 / n)^{3}}{(1+2 / n) 1} \\rightarrow 1\n\\]\nso the ratio test is inconclusive. But we can compute \\(\\frac{n^{2}}{n+1}=\\frac{n}{1+1 / n}\\), which is unbounded so \\(b_{n}\\) does not converge.\n\nLet \\(c_{n}=\\frac{n}{2 n+1}\\). Then,\n\n\\[\n\\left|\\frac{c_{n+1}}{c_{n}}\\right|=\\frac{n+1}{2(n+1)+1} / \\frac{n}{2 n+1}=\\frac{(n+1)(2 n+1)}{(2 n+3) n}=\\frac{(1+1 / n)(2+1 / n)}{(2+3 / n) 1} \\rightarrow 1,\n\\]\nso the ratio test is inconclusive. But we can compute \\(\\frac{n}{2 n+1}=\\frac{1}{2+1 / n} \\rightarrow \\frac{1}{2}\\).\nAs mentioned, the ratio test is especially useful if factorials and geometric terms are involved. It is important to make sure you use brackets correctly, it is very easy to make mistakes.\nExample 2.4.14. Let\n\\[\na_{n}=\\frac{n ! \\cdot 3^{n}}{\\sqrt{(2 n) !}}\n\\]\nThen,\n\\[\n\\left|\\frac{a_{n+1}}{a_{n}}\\right|=\\frac{(n+1) ! \\cdot 3^{n+1}}{\\sqrt{(2(n+1)) !}} / \\frac{n ! \\cdot 3^{n}}{\\sqrt{(2 n) !}}=\\frac{(n+1) !}{n !} \\cdot \\frac{3^{n+1}}{3^{n}} \\cdot \\frac{\\sqrt{(2 n) !}}{\\sqrt{(2(n+1)) !}}\n\\]\nWe can simplify \\(\\frac{(n+1) !}{n !} \\cdot \\frac{3^{n+1}}{3^{n}}=3(n+1)\\). Combining the square roots and working out the brackets carefully gives that\n\\[\n\\frac{\\sqrt{(2 n) !}}{\\sqrt{(2(n+1)) !}}=\\sqrt{\\frac{(2 n) !}{(2 n+2) !}}=\\sqrt{\\frac{1 \\cdot 2 \\cdots 2 n}{1 \\cdot 2 \\cdots 2 n \\cdot(2 n+1) \\cdot(2 n+2)}}=\\frac{1}{\\sqrt{(2 n+1)(2 n+2)}}\n\\]\nHence,\n\\[\n\\left|\\frac{a_{n+1}}{a_{n}}\\right|=\\frac{3(n+1)}{\\sqrt{(2 n+1)(2 n+2)}}=\\frac{3(1+1 / n)}{\\sqrt{\\frac{1}{n^{2}}(2 n+1)(2 n+2)}}=\\frac{3(1+1 / n)}{\\sqrt{(2+1 / n)(2+2 / n)}} \\rightarrow \\frac{3}{\\sqrt{4}}=\\frac{3}{2}>1\n\\]\nHence, the sequence does not converge by the ratio test.\nIf the sequence is a geometric sequence, we can also apply the ratio test, which will give the same answer as the geometric sequence test.\nExample 2.4.15. Let \\(x \\in \\mathbb{R}\\) and let \\(a_{n}=x^{n}\\). Then,\n\\[\n\\left|\\frac{a_{n+1}}{a_{n}}\\right|=\\frac{\\left|x^{n+1}\\right|}{\\left|x^{n}\\right|}=\\frac{|x|^{n+1}}{|x|^{n}}=|x| \\rightarrow|x|\n\\]\nHence, if \\(|x|<1\\) the sequence converges by the ratio test, if \\(|x|>1\\) the sequence does not converge by the ratio test, and if \\(|x|=1\\) the ratio test is inconclusive. Of course, we would have gotten this result immediately by using the geometric sequence test." + "text": "6.8 Limit tests\nIn this section we discuss a number of tests to determine whether a sequence converges or not. These are known as limit tests.\n\n6.8.1 Squeeze Theorem\nWhen a sequence \\((a_n)\\) oscillates, it is difficult to compute the limit. Examples of terms which produce oscillations are \\[\n(-1)^n \\,, \\quad \\sin(n) \\,, \\quad \\cos(n) \\,.\n\\] In 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\nTheorem 34: Squeeze theoremLet \\(\\left(a_{n}\\right), \\left(b_{n}\\right)\\) and \\(\\left(c_{n}\\right)\\) be sequences in \\(\\mathbb{R}\\). Suppose that \\[\nb_{n} \\leq a_{n} \\leq c_{n} \\,, \\quad \\forall \\, n \\in \\mathbb{N}\\,,\n\\] and that \\[\n\\lim_{n \\rightarrow \\infty} b_{n} = \\lim_{n \\rightarrow \\infty} c_{n} = L \\, .\n\\] Then \\[\n\\lim_{n \\rightarrow \\infty} a_{n}= L \\, .\n\\]\n\n\n\nProofLet \\(\\varepsilon>0\\). Since \\(b_{n} \\to L\\) and \\(c_n \\to L\\) , there exist \\(N_1, N_2 \\in \\mathbb{N}\\) such that \\[\\begin{align*}\n-\\varepsilon< b_n - L < \\varepsilon\\,, \\,\\, \\forall \\, n \\geq N_1 \\,, \\\\\n- \\varepsilon< c_n - L < \\varepsilon\\,, \\,\\, \\forall \\, n \\geq N_2 \\,.\n\\end{align*}\\] Set \\[\nN := \\max\\{N_1,N_2\\} \\,.\n\\] Let \\(n \\geq N\\). Using the assumption that \\(b_n \\leq a_{n} \\leq c_{n}\\), we get \\[\nb_n - L \\leq a_{n} - L \\leq c_{n} - L \\,.\n\\] In particular \\[\n- \\varepsilon< b_n - L \\leq a_n - L \\leq b_n - L < \\varepsilon\\,.\n\\] The above implies \\[\n- \\varepsilon< a_n - L < \\varepsilon\\quad \\implies \\quad \\left|a_{n}-L\\right| < \\varepsilon\\,.\n\\]\n\n\n\nExample 35\nProve that \\[\n\\lim_{n \\to \\infty} \\, \\frac{(-1)^{n}}{n} = 0 \\,.\n\\]\n\nProof. For all \\(n \\in \\mathbb{N}\\) we can estimate \\[\n-1 \\leq(-1)^{n} \\leq 1 \\,.\n\\] Therefore \\[\n\\frac{-1}{n} \\leq \\frac{(-1)^{n}}{n} \\leq \\frac{1}{n} \\,, \\quad \\forall \\, n \\in \\mathbb{N}\\,.\n\\] Moreover \\[\n\\lim_{n \\to \\infty} \\frac{-1}{n}= -1 \\cdot 0=0 \\,, \\quad\n\\lim_{n \\to \\infty} \\frac{1}{n}=0 \\,.\n\\] By the Squeeze Theorem 34 we conclude \\[\n\\lim_{n \\to \\infty} \\frac{(-1)^{n}}{n}=0 \\,.\n\\]\n\n\n\n\nExample 36\nProve that \\[\n\\lim_{n \\to \\infty} \\frac{\\cos (3 n) + 9 n^{2}}{11 n^{2}+15 \\sin (17 n)} = \\frac{9}{11} \\,.\n\\]\n\nProof. We know that \\[\n-1 \\leq \\cos(x) \\leq 1 \\,, \\quad - 1 \\leq \\sin(x) \\leq 1 \\,, \\quad \\forall \\, x \\in \\mathbb{R}\\,.\n\\] Therefore, for all \\(n \\in \\mathbb{N}\\) \\[\n- 1 \\leq \\cos(3n) \\leq 1 \\,, \\quad -1 \\leq \\sin(17n) \\leq 1 \\,.\n\\] We can use the above to estimate the numerator in the given sequence: \\[\n-1 + 9 n^{2} \\leq \\cos (3 n)+9 n^{2} \\leq 1+ 9 n^{2} \\,.\n\\tag{6.7}\\] Concerning the denominator, we have \\[\n11 n^{2}-15 \\leq 11 n^{2}+15 \\sin (17 n) \\leq 11 n^{2} + 15\n\\] and therefore \\[\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\\tag{6.8}\\] Putting together (6.7)-(6.8) we obtain \\[\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\\] By the Algebra of Limits we infer \\[\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\\] and \\[\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\\] Applying the Squeeze Theorem 34 we conclude \\[\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\nWarningSuppose that the sequences \\((a_n), (b_n), (c_n)\\) satisfy \\[\nb_n \\leq a_n \\leq c_n \\,, \\quad \\forall n \\in \\mathbb{N}\\,,\n\\] and \\[\nb_n \\to L_1 \\,, \\quad c_n \\to L_2 \\,, \\quad L_1 \\neq L_2 \\,.\n\\] In general, we cannot conclude that \\(a_n\\) converges.\n\n\n\nExample 37\nConsider the sequence \\[\na_n = \\left( 1 + \\frac{1}{n} \\right) (-1)^n \\,.\n\\] For all \\(n \\in \\mathbb{N}\\) we can bound \\[\n- 1 - \\frac{1}{n} \\leq \\left( 1 + \\frac{1}{n} \\right) (-1)^n \\leq 1 + \\frac{1}{n} \\,.\n\\] However \\[\n- 1 - \\frac{1}{n} \\longrightarrow - 1 - 0 = -1\n\\] and \\[\n1 + \\frac{1}{n} \\longrightarrow 1 + 0 = 1 \\,.\n\\] Since \\[\n- 1 \\neq 1 \\,,\n\\] we cannot apply the Squeeze Theorem 34 to conclude convergence of \\((a_n)\\). Indeed, \\((a_n)\\) is a divergent sequence.\n\nProof. Suppose by contradiction that \\(a_n \\to a\\). We have \\[\na_n = (-1)^n + \\frac{(-1)^n}{n} = b_n + c_n\n\\] where \\[\nb_n := (-1)^n \\,, \\quad c_n := \\frac{(-1)^n}{n} \\,.\n\\] We have seen in Example 36 that \\(c_n \\to 0\\). Therefore, by the Algebra of Limits, we have \\[\nb_n = a_n - c_n \\longrightarrow a - 0 = a \\,.\n\\] However, Theorem 14 says that the sequence \\(b_n = (-1)^n\\) diverges. Contradiction. Hence \\((a_n)\\) diverges.\n\n\n\n\n\n6.8.2 Geometric sequences\n\nDefinition 38A sequence \\(\\left(a_{n}\\right)\\) is called a geometric sequence if \\[\na_{n}=x^{n} \\,,\n\\] for some \\(x \\in \\mathbb{R}\\).\n\n\nThe value of \\(|x|\\) determines whether or not a geometric sequence converges, as shown in the following theorem.\n\nTheorem 39: Geometric sequence test\nLet \\(x \\in \\mathbb{R}\\) and let \\(\\left(a_{n}\\right)\\) be the sequence defined by \\[\na_{n}:=x^{n} \\,.\n\\] We have:\n\nIf \\(|x|<1\\), then \\[\n\\lim_{n \\to \\infty} a_{n} = 0 \\,.\n\\]\nIf \\(|x|>1\\), then sequence \\(\\left(a_{n}\\right)\\) is unbounded, and hence divergent.\n\n\n\n\nWarning\nThe Geometric sequence test in Theorem 39 does not address the case \\[\n|x|=1 \\,.\n\\] This is because, in this case, the sequence \\[\na_n = x^n\n\\] might converge or diverge, depending on the value of \\(x\\). Indeed, \\[\n|x| = 1 \\quad \\implies \\quad x = \\pm 1 \\,.\n\\] We therefore have two cases:\n\n\\(x = 1\\): Then \\[\na_n = 1^n = 1\n\\] so that \\(a_n \\to 1\\) and \\((a_n)\\) is convergent.\n\\(x=-1\\): Then \\[\na_n = x^n = (-1)^n\n\\] which is divergent by Theorem 14.\n\n\n\nTo prove Theorem 39 we need the following inequality, known as Bernoulli’s inequality.\n\nLemma 40: Bernoulli’s inequalityLet \\(x \\in \\mathbb{R}\\) with \\(x>-1\\). Then \\[\n(1+x)^{n} \\geq 1+n x \\,, \\quad \\forall \\, n \\in \\mathbb{N}\\,.\n\\tag{6.9}\\]\n\n\n\nProofLet \\(x \\in \\mathbb{R}, x>-1\\). We prove the statement by induction:\n\nBase case: (6.9) holds with equality when \\(n=1\\).\nInduction hypothesis: Let \\(k \\in \\mathbb{N}\\) and suppose that (6.9) holds for \\(n=k\\), i.e., \\[\n(1+x)^{k} \\geq 1+k x \\,.\n\\] Then \\[\\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*}\\] where we used that \\(kx^2 \\geq 0\\). Then (6.9) holds for \\(n=k+1\\).\n\nBy induction we conclude (6.9).\n\n\nWe are ready to prove Theorem 39.\n\nProof: Proof of Theorem 39Part 1. The case \\(|x|<1\\).\nIf \\(x=0\\), then \\[\na_n = x^n = 0\n\\] so that \\(a_n \\to 0\\). Hence assume \\(x \\neq 0\\). We need to prove that \\[\n\\forall \\, \\varepsilon> 0 \\,, \\, \\exists \\, N \\in \\mathbb{N}\\, \\text{ s.t. } \\,\n\\forall \\, n \\geq N \\,, \\,\\, |x^n - 0| < \\varepsilon\\,.\n\\] Let \\(\\varepsilon>0\\). We have \\[\n|x| < 1 \\quad \\implies \\quad \\frac{1}{|x|} > 1 \\,.\n\\] Therefore \\[\n|x|= \\frac{1}{1+u} \\,, \\quad u:=\\frac{1}{|x|} - 1 > 0 \\,.\n\\] Let \\(N \\in \\mathbb{N}\\) be such that \\[\nN > \\frac{1}{\\varepsilon u} \\,,\n\\] so that \\[\n\\frac{1}{N u} < \\varepsilon\\,.\n\\] Let \\(n \\geq N\\). Then \\[\\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 & < \\varepsilon\\,,\n\\end{align*}\\] where we used Bernoulli’s inequality (6.9) in the first inequality.\nPart 2. The case \\(|x|>1\\).\nTo prove that (a_n) does not converge, we prove that it is unbounded. This means showing that \\[\n\\forall \\, M > 0 \\,, \\, \\exists n \\in \\mathbb{N}\\, \\text{ s.t. } \\, \\left| a_{n}\\right| >M \\,.\n\\] Let \\(M > 0\\). We have two cases:\n\n\\(0 < M \\leq 1\\): Choose \\(n=1\\). Then \\[\n\\left|a_{1}\\right|=|x|>1 \\geq M \\,.\n\\]\n\\(M>1\\): Choose \\(n \\in \\mathbb{N}\\) such that \\[\nn> \\frac{\\log M}{\\log |x|} \\,.\n\\] Note that \\(\\log |x|>0\\) since \\(|x|>1\\). Therefore \\[\\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*}\\] Then \\[\n\\left|a_{n}\\right|=\\left|x^{n}\\right|=|x|^{n} > M \\,.\n\\]\n\nHence \\((a_n)\\) is unbounded. By Corollary 21 we conclude that \\((a_n)\\) is divergent.\n\n\n\nExample 41\nWe can apply Theorem 39 to prove convergence or divergence for the following sequences.\n\nWe have \\[\n\\left(\\frac{1}{2}\\right)^{n} \\longrightarrow 0\n\\] since \\[\n\\left|\\frac{1}{2}\\right|=\\frac{1}{2}<1 \\,.\n\\]\nWe have \\[\n\\left(\\frac{-1}{2}\\right)^{n} \\longrightarrow 0\n\\] since \\[\n\\left|\\frac{-1}{2}\\right|=\\frac{1}{2}<1 \\,.\n\\]\nThe sequence \\[\na_n = \\left(\\frac{-3}{2}\\right)^{n}\n\\] does not converge, since \\[\n\\left|\\frac{-3}{2}\\right|=\\frac{3}{2}>1 \\,.\n\\]\nAs \\(n \\rightarrow \\infty\\), \\[\n\\frac{3^{n}}{(-5)^{n}}=\\left(-\\frac{3}{5}\\right)^{n} \\longrightarrow 0\n\\] since \\[\n\\left|-\\frac{3}{5}\\right|=\\frac{3}{5}<1 \\,.\n\\]\nThe sequence \\[\na_{n}=\\frac{(-7)^{n}}{2^{2 n}}\n\\] does not converge, since \\[\n\\frac{(-7)^{n}}{2^{2 n}}=\\frac{(-7)^{n}}{\\left(2^{2}\\right)^{n}}=\\left(-\\frac{7}{4}\\right)^{n}\n\\] and \\[\n\\left|-\\frac{7}{4}\\right|=\\frac{7}{4}>1 \\,.\n\\]\n\n\n\n\n\n6.8.3 Ratio test\n\nTheorem 42: Ratio test\nLet \\(\\left(a_{n}\\right)\\) be a sequence in \\(\\mathbb{R}\\) such that \\[\na_{n} \\neq 0 \\,, \\quad \\forall \\, n \\in \\mathbb{N}\\,.\n\\]\n\nSuppose that the following limit exists: \\[\nL:=\\lim_{n \\to \\infty}\\left|\\frac{a_{n+1}}{a_{n}}\\right| \\,.\n\\] Then,\n\nIf \\(L<1\\) we have \\[\n\\lim_{n \\to\\infty} a_{n}=0 \\,.\n\\]\nIf \\(L>1\\), the sequence \\(\\left(a_{n}\\right)\\) is unbounded, and hence does not converge.\n\nSuppose that there exists \\(N \\in \\mathbb{N}\\) and \\(L>1\\) such that \\[\n\\left|\\frac{a_{n+1}}{a_{n}}\\right| \\geq L \\,, \\quad \\forall \\, n \\geq N \\,.\n\\] Then the sequence \\(\\left(a_{n}\\right)\\) is unbounded, and hence does not converge.\n\n\n\n\nProofDefine the sequence \\(b_{n}=\\left|a_{n}\\right|\\). Then,\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\\]\nPart 1. Suppose that there exists the limit \\[\nL:=\\lim_{n \\to \\infty}\\left|\\frac{a_{n+1}}{a_{n}}\\right| \\,.\n\\] Therefore \\[\n\\lim_{n \\to \\infty} \\frac{b_{n+1}}{b_{n}} = L \\,.\n\\tag{6.10}\\]\n\n\\(L<1\\): Choose \\(r>0\\) such that \\[\nL<r<1 \\,.\n\\] Set \\[\n\\varepsilon:= r-L\n\\] By the convergence at (6.10) there exists \\(N \\in \\mathbb{N}\\) such that \\[\n\\left|\\frac{b_{n+1}}{b_{n}}-L\\right| < \\varepsilon= r - L \\,, \\quad \\forall \\, n \\geq N \\,.\n\\] In particular \\[\n\\frac{b_{n+1}}{b_{n}}-L < r-L \\,, \\quad \\forall \\, n \\geq N \\,,\n\\] which implies \\[\nb_{n+1} < r \\,{b_{n}} \\,, \\quad \\forall \\, n \\geq N \\,.\n\\tag{6.11}\\] Let \\(n \\geq N\\), we can use (6.11) recursively and obtain \\[\n0 \\leq b_{n} < \\, r b_{n-1} < \\ldots < r^{n-N} \\, b_{N} = r^{n} \\, \\frac{b_{N}}{r^{N}} \\,.\n\\] In particular, we have proven that \\[\n0 \\leq b_{n} < r^{n} \\, \\frac{b_{N}}{r^{N}} \\,, \\quad \\forall \\, n \\in \\mathbb{N}\\,.\n\\tag{6.12}\\] Since \\(|r|<1\\), by the Geometric sequence test Theorem 39 we infer \\[\nr^{n} \\rightarrow 0 \\,.\n\\] The Algebra of Limits the yields \\[\nr^{n} \\, \\frac{b_{N}}{r^{N}} \\rightarrow 0 \\,.\n\\] By the Squeeze Theorem 34 applied to (6.12), it follows that \\[ \nb_{n} = |a_n| \\to 0 \\,.\n\\] Since \\[\n-\\left|a_{n}\\right| \\leq a_{n} \\leq\\left|a_{n}\\right| \\,,\n\\] and \\[\n-|a_n| \\to 0 \\,, \\quad |a_n| \\to 0 \\,,\n\\] we can again apply the Squeeze Theorem 34 to infer \\[\na_{n} \\rightarrow 0 \\,.\n\\]\n\\(L>1\\): Choose \\(r>0\\) such that \\[\n1<r<L \\,.\n\\] Define \\[\n\\varepsilon:= L - r > 0 \\,.\n\\] By the convergence (6.10), there exists \\(N \\in \\mathbb{N}\\) such that \\[\n\\left|\\frac{b_{n+1}}{b_{n}}-L\\right| < \\varepsilon= L - r \\,, \\quad \\forall \\, n \\geq N \\,.\n\\] In particular, \\[\n-(L-r) < \\frac{b_{n+1}}{b_{n}}-L \\,, \\quad \\forall \\, n \\geq N \\,,\n\\] which implies \\[\nb_{n+1} > r \\, b_{n} \\,, \\quad \\forall \\, n \\geq N \\,.\n\\tag{6.13}\\] Let \\(n \\geq N\\). Applying (6.13) recursively we get \\[\nb_{n} > r^{n-N} \\, b_{N} = r^{n} \\, \\frac{b_{N}}{r^{N}} \\,, \\quad \\forall \\, n \\geq N \\,.\n\\tag{6.14}\\] Since \\(|r|>1\\), by the Geometric sequence test we have that the sequence \\[\n(r^n)\n\\] is unbounded. Therefore also the right hand side of (6.14) is unbounded, proving that \\((b_n)\\) is unbounded. Since \\[\nb_n = |a_n|\\,,\n\\] we conclude that \\((a_n)\\) is unbounded. By Corollary 21 we conclude that \\((a_n)\\) does not converge.\n\nPart 2. Suppose that there exists \\(N \\in \\mathbb{N}\\) and \\(M>1\\) such that \\[\n\\left|\\frac{a_{n+1}}{a_{n}}\\right| \\geq M \\,, \\quad \\forall \\, n \\geq N \\,.\n\\] Since \\(b_n = |a_n|\\), we infer \\[\nb_{n+1} \\geq L \\, b_n \\,, \\quad \\forall \\, n \\geq N \\,.\n\\] Arguing as above, we obtain \\[\nb_n \\geq L^n \\, \\frac{b_N}{L^N} \\,, \\quad \\forall \\, n \\geq N \\,.\n\\] Since \\(L>1\\), we have that the sequence \\[ \nL^n \\, \\frac{b_N}{L^N}\n\\] is unbounded, by the Geometric sequence test. Hence also \\((b_n)\\) is unbounded, from which we conclude that \\((a_n)\\) is unbounded. By Corollary 21 we conclude that \\((a_n)\\) does not converge.\n\n\nLet us apply the Ratio test to some concrete examples.\n\nExample 43\nLet \\[\na_{n}=\\frac{3^{n}}{n !} \\,,\n\\] where we recall that \\(n!\\) (pronounced \\(n\\) factorial) is defined by \\[\nn! := n \\cdot (n-1) \\cdot (n-2) \\cdot \\ldots \\cdot 3 \\cdot 2 \\cdot 1 \\,.\n\\] Prove that \\[\n\\lim_{n \\to \\infty} a_n = 0 \\,.\n\\]\n\nProof. We have \\[\\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*}\\] Hence, \\(L=0<1\\) so \\(a_{n} \\to 0\\) by the Ratio test in Theorem 42.\n\n\n\n\nExample 44\nConsider the sequence \\[\na_{n}=\\frac{n ! \\cdot 3^{n}}{\\sqrt{(2 n) !}} \\,.\n\\] Prove that \\((a_n)\\) is divergent.\n\nProof. We have \\[\\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*}\\] For the first two fractions we have \\[\n\\frac{(n+1) !}{n !} \\cdot \\frac{3^{n+1}}{3^{n}} = 3(n+1) \\,,\n\\] while for the third fraction \\[\\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*}\\] Therefore, using the Algebra of Limits, \\[\\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*}\\] By the Ratio test we conclude that \\((a_n)\\) is divergent.\n\n\n\n\nExample 45\nLet \\[\na_{n}=\\frac{n !}{100^{n}} \\,.\n\\] Prove that \\((a_n)\\) is divergent.\n\nProof. \\[\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\\] Choose \\(N=101\\). Then for all \\(n \\geq N\\), \\[\n\\left|\\frac{a_{n+1}}{a_{n}}\\right| \\geq \\frac{101}{100}>1 \\,.\n\\] Hence \\(a_{n}\\) is divergent by the Ratio test.\n\n\n\n\nWarning\nThe Ratio test in Theorem 42 does not address the case \\[\nL=1 \\,.\n\\] This is because, in this case, the sequence \\((a_n)\\) might converge or diverge.\nFor example:\n\nDefine the sequence \\[\na_n = \\frac1n \\,.\n\\] We have \\[\n\\left|\\frac{a_{n+1}}{a_{n}}\\right| = \\frac{n}{n+1} \\rightarrow L = 1 \\,.\n\\] Hence we cannot apply the Ratio test. However we know that \\[\n\\lim_{n \\to \\infty} \\, \\frac1n = 0 \\,.\n\\]\nConsider the sequence \\[\na_n = n \\,.\n\\] We have \\[\n\\frac{|a_{n+1}|}{|a_n|} = \\frac{n+1}{n} \\rightarrow L = 1 \\,.\n\\] Hence we cannot apply the Ratio test. However we know that \\((a_n)\\) is unbounded, and thus divergent.\n\n\n\nIf the sequence \\((a_n)\\) is geometric, the Ratio test of Theorem 42 will give the same answer as the Geometric sequence test of Theorem 39. This is the content of the following remark.\n\nRemark 46Let \\(x \\in \\mathbb{R}\\) and define the geometric sequence \\[\na_{n}=x^{n} \\,.\n\\] Then \\[\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\\] Hence:\n\nIf \\(|x|<1\\), the sequence \\((a_n)\\) converges by the Ratio test\nIf \\(|x|>1\\), the sequence \\((a_n)\\) diverges by the Ratio test.\nIf \\(|x|=1\\), the sequence \\((a_n)\\) might be convergent or divergent.\n\nThese results are in agreement with the Geometric sequence test." + }, + { + "objectID": "sections/chap_7.html", + "href": "sections/chap_7.html", + "title": "7  Sequences in \\(\\mathbb{C}\\)", + "section": "", + "text": "Coming soon" + }, + { + "objectID": "sections/chap_8.html", + "href": "sections/chap_8.html", + "title": "8  Series", + "section": "", + "text": "Coming soon" }, { "objectID": "sections/license.html#reuse", diff --git a/docs/sections/chap_1.html b/docs/sections/chap_1.html index bb9532c..898e5b9 100644 --- a/docs/sections/chap_1.html +++ b/docs/sections/chap_1.html @@ -66,63 +66,55 @@ } } @@ -215,7 +215,19 @@
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    • 6.1 Sequences in \(\mathbb{R}\)
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    • 6.3 Divergent sequences
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  • Suppose that there exists \(N \in \mathbb{N}\) and \(M>1\) such that \[ -\left|\frac{a_{n+1}}{a_{n}}\right| \geq M \,, \quad \forall \, n \geq N \,. \,. +

  • Suppose that there exists \(N \in \mathbb{N}\) and \(L>1\) such that \[ +\left|\frac{a_{n+1}}{a_{n}}\right| \geq L \,, \quad \forall \, n \geq N \,. \] Then the sequence \(\left(a_{n}\right)\) is unbounded, and hence does not converge.

    • @@ -1479,88 +1491,133 @@

    we can again apply the Squeeze Theorem 34 to infer \[ a_{n} \rightarrow 0 \,. \]

    • -

  • \(L>1\): Choose \(r>0\) such that \(1<r<L\). Let \(\varepsilon=L-r>0\). If \(\frac{b_{n+1}}{b_{n}} \rightarrow L\), there exists \(n_{0} \in \mathbb{N}\) such that for all \(n \geq n_{0}\)

    • +

  • \(L>1\): Choose \(r>0\) such that \[ +1<r<L \,. +\] Define \[ +\varepsilon:= L - r > 0 \,. +\] By the convergence (6.10), there exists \(N \in \mathbb{N}\) such that \[ +\left|\frac{b_{n+1}}{b_{n}}-L\right| < \varepsilon= L - r \,, \quad \forall \, n \geq N \,. +\] In particular, \[ +-(L-r) < \frac{b_{n+1}}{b_{n}}-L \,, \quad \forall \, n \geq N \,, +\] which implies \[ +b_{n+1} > r \, b_{n} \,, \quad \forall \, n \geq N \,. +\tag{6.13}\] Let \(n \geq N\). Applying (6.13) recursively we get \[ +b_{n} > r^{n-N} \, b_{N} = r^{n} \, \frac{b_{N}}{r^{N}} \,, \quad \forall \, n \geq N \,. +\tag{6.14}\] Since \(|r|>1\), by the Geometric sequence test we have that the sequence \[ +(r^n) +\] is unbounded. Therefore also the right hand side of (6.14) is unbounded, proving that \((b_n)\) is unbounded. Since \[ +b_n = |a_n|\,, +\] we conclude that \((a_n)\) is unbounded. By Corollary 21 we conclude that \((a_n)\) does not converge.

    • -

    \[ -\left|\frac{b_{n+1}}{b_{n}}-L\right| \leq \varepsilon=L-r -\]

    -

    In particular,

    -

    \[ --(L-r) \leq \frac{b_{n+1}}{b_{n}}-L \Longleftrightarrow \frac{b_{n+1}}{b_{n}} \geq r \Longleftrightarrow b_{n+1} \geq r b_{n} -\]

    -

    (In the other case we assumed this is the case, so this statement also holds there.) Hence, for all \(n \geq n_{0}\)

    -

    \[ -b_{n} \geq r^{n-n_{0}} b_{n_{0}}=r^{n} \frac{b_{n_{0}}}{r^{n_{0}}} +

    Part 2. Suppose that there exists \(N \in \mathbb{N}\) and \(M>1\) such that \[ +\left|\frac{a_{n+1}}{a_{n}}\right| \geq M \,, \quad \forall \, n \geq N \,. +\] Since \(b_n = |a_n|\), we infer \[ +b_{n+1} \geq L \, b_n \,, \quad \forall \, n \geq N \,. +\] Arguing as above, we obtain \[ +b_n \geq L^n \, \frac{b_N}{L^N} \,, \quad \forall \, n \geq N \,. +\] Since \(L>1\), we have that the sequence \[ +L^n \, \frac{b_N}{L^N} +\] is unbounded, by the Geometric sequence test. Hence also \((b_n)\) is unbounded, from which we conclude that \((a_n)\) is unbounded. By Corollary 21 we conclude that \((a_n)\) does not converge.

    + + +

    Let us apply the Ratio test to some concrete examples.

    +
    +
    Example 43
    +

    Let \[ +a_{n}=\frac{3^{n}}{n !} \,, +\] where we recall that \(n!\) (pronounced \(n\) factorial) is defined by \[ +n! := n \cdot (n-1) \cdot (n-2) \cdot \ldots \cdot 3 \cdot 2 \cdot 1 \,. +\] Prove that \[ +\lim_{n \to \infty} a_n = 0 \,. \]

    -

    Since \(|r|>1\), it follows from the geometric sequence test that the right hand side is unbounded and hence also \(b_{n}\) is unbounded.

    +
    +

    Proof. We have \[\begin{align*} +\left|\frac{a_{n+1}}{a_{n}}\right| & = \dfrac{\left( \dfrac{3^{n+1}}{(n+1) !} \right) }{ \left( \dfrac{3^{n}}{n !} \right) } \\ +& = \frac{3^{n+1}}{3^{n}} \, \frac{n !}{(n+1) !} \\ +& = \frac{3 \cdot 3^n}{3^n} \, \frac{n!}{(n+1) n!} \\ +& =\frac{3}{n+1} \longrightarrow L = 0 \,. +\end{align*}\] Hence, \(L=0<1\) so \(a_{n} \to 0\) by the Ratio test in Theorem 42.

    +
    -
    -

    Example 43
    Let \(a_{n}=\frac{3^{n}}{n !}\). Then,

    -

    \[ -\left|\frac{a_{n+1}}{a_{n}}\right|=\frac{3^{n+1}}{(n+1) !} / \frac{3^{n}}{n !}=\frac{3^{n+1}}{3^{n}} \frac{n !}{(n+1) !}=3 \frac{1 \cdot 2 \cdots n}{1 \cdot 2 \cdots n \cdot(n+1)}=\frac{3}{n+1} \longrightarrow 0, -\]

    -

    as \(n \rightarrow \infty\). Hence, \(L=0<1\) so \(\lim _{n \rightarrow \infty} a_{n}=0\) by the ratio test.

    -
      -
    • Let \(b_{n}=\frac{n !}{100^{n}}\). Then
      • -
    -

    \[ -\left|\frac{b_{n+1}}{b_{n}}\right|=\frac{(n+1) !}{100^{n+1}} / \frac{n !}{100^{n}}=\frac{100^{n}}{100^{n+1}} \frac{(n+1) !}{(n !)}=\frac{n+1}{100} . -\]

    -

    Choose \(n_{0}=101\). Then for all \(n \geq n_{0}\),

    -

    \[ -\left|\frac{b_{n+1}}{b_{n}}\right| \geq \frac{101}{100}>1 -\]

    -

    Hence, \(b_{n}\) is an unbounded sequence and does not converge.

    +
    +
    Example 44
    +

    Consider the sequence \[ +a_{n}=\frac{n ! \cdot 3^{n}}{\sqrt{(2 n) !}} \,. +\] Prove that \((a_n)\) is divergent.

    +
    +

    Proof. We have \[\begin{align*} +\left|\frac{a_{n+1}}{a_{n}}\right| & =\frac{(n+1) ! \cdot 3^{n+1}}{\sqrt{(2(n+1)) !}} \frac{\sqrt{(2 n) !}}{n ! \cdot 3^{n}} \\ +& =\frac{(n+1) !}{n !} \cdot \frac{3^{n+1}}{3^{n}} \cdot \frac{\sqrt{(2 n) !}}{\sqrt{(2(n+1)) !}} +\end{align*}\] For the first two fractions we have \[ +\frac{(n+1) !}{n !} \cdot \frac{3^{n+1}}{3^{n}} = 3(n+1) \,, +\] while for the third fraction \[\begin{align*} +\frac{\sqrt{(2 n) !}}{\sqrt{(2(n+1)) !}} & =\sqrt{\frac{(2 n) !}{(2 n+2) !}} \\ +& = \sqrt{\frac{ (2n)! }{ (2n+2) \cdot (2n+1) \cdot (2n)! }} \\ +& = \frac{1}{\sqrt{(2 n+1)(2 n+2)}} \,. +\end{align*}\] Therefore, using the Algebra of Limits, \[\begin{align*} +\left|\frac{a_{n+1}}{a_{n}}\right| & = \frac{3(n+1)}{\sqrt{(2 n+1)(2 n+2)}}\\ +& = \frac{3n \left(1+ \dfrac{1}{n} \right)}{\sqrt{n^2 \left( 2 + \dfrac{1}{n}\right) \left(2 + \dfrac{2}{n} \right)}} \\ +& = \frac{3 \left(1+ \dfrac{1}{n} \right)}{\sqrt{ \left( 2 + \dfrac{1}{n}\right) \left(2 + \dfrac{2}{n} \right)}} \longrightarrow \frac{3}{\sqrt{4}} = \frac{3}{2} > 1 \,. +\end{align*}\] By the Ratio test we conclude that \((a_n)\) is divergent.

    +
    -
    -

    Warning
    For part (c) of the theorem we give some examples. If you try to apply the ratio test for a fraction of two polynomials, the limit will always be equal to 1 , so the ratio test is inconclusive for this type of sequence.

    +
    +
    Example 45
    +

    Let \[ +a_{n}=\frac{n !}{100^{n}} \,. +\] Prove that \((a_n)\) is divergent.

    +
    +

    Proof. \[ +\left|\frac{a_{n+1}}{a_{n}}\right| += \frac{100^{n}}{100^{n+1}} \frac{(n+1) !}{n !} +=\frac{n+1}{100} \,. +\] Choose \(N=101\). Then for all \(n \geq N\), \[ +\left|\frac{a_{n+1}}{a_{n}}\right| \geq \frac{101}{100}>1 \,. +\] Hence \(a_{n}\) is divergent by the Ratio test.

    +
    -

    Example 2.4.13. \(\quad\) Let \(a_{n}=\frac{1}{n}\). Then,

    -

    \[ -\left|\frac{a_{n+1}}{a_{n}}\right|=\frac{1}{n+1} / \frac{1}{n}=\frac{n}{n+1} \rightarrow 1 -\]

    -

    so the ratio test is inconclusive. But we know that \(\frac{1}{n} \rightarrow 0\).

    +
    +
    Warning
    +

    The Ratio test in Theorem 42 does not address the case \[ +L=1 \,. +\] This is because, in this case, the sequence \((a_n)\) might converge or diverge.

    +

    For example:

      -
    • Let \(b_{n}=\frac{n^{2}}{n+1}\). Then,
      • +

    • Define the sequence \[ +a_n = \frac1n \,. +\] We have \[ +\left|\frac{a_{n+1}}{a_{n}}\right| = \frac{n}{n+1} \rightarrow L = 1 \,. +\] Hence we cannot apply the Ratio test. However we know that \[ +\lim_{n \to \infty} \, \frac1n = 0 \,. +\]

      • +

    • Consider the sequence \[ +a_n = n \,. +\] We have \[ +\frac{|a_{n+1}|}{|a_n|} = \frac{n+1}{n} \rightarrow L = 1 \,. +\] Hence we cannot apply the Ratio test. However we know that \((a_n)\) is unbounded, and thus divergent.

    -

    \[ -\left|\frac{b_{n+1}}{b_{n}}\right|=\frac{(n+1)^{2}}{(n+1)+1} / \frac{n^{2}}{n+1}=\frac{(n+1)^{3}}{(n+2) n^{2}}=\frac{(1+1 / n)^{3}}{(1+2 / n) 1} \rightarrow 1 -\]

    -

    so the ratio test is inconclusive. But we can compute \(\frac{n^{2}}{n+1}=\frac{n}{1+1 / n}\), which is unbounded so \(b_{n}\) does not converge.

    +
    +
    +

    If the sequence \((a_n)\) is geometric, the Ratio test of Theorem 42 will give the same answer as the Geometric sequence test of Theorem 39. This is the content of the following remark.

    +
    +

    Remark 46
    Let \(x \in \mathbb{R}\) and define the geometric sequence \[ +a_{n}=x^{n} \,. +\] Then \[ +\left|\frac{a_{n+1}}{a_{n}}\right| = \frac{\left|x^{n+1}\right|}{\left|x^{n}\right|} += \frac{|x|^{n+1}}{|x|^{n}} +=|x| \rightarrow |x| \,. +\] Hence:

      -
    • Let \(c_{n}=\frac{n}{2 n+1}\). Then,
      • +
    • If \(|x|<1\), the sequence \((a_n)\) converges by the Ratio test
      • +
    • If \(|x|>1\), the sequence \((a_n)\) diverges by the Ratio test.
      • +
    • If \(|x|=1\), the sequence \((a_n)\) might be convergent or divergent.
    -

    \[ -\left|\frac{c_{n+1}}{c_{n}}\right|=\frac{n+1}{2(n+1)+1} / \frac{n}{2 n+1}=\frac{(n+1)(2 n+1)}{(2 n+3) n}=\frac{(1+1 / n)(2+1 / n)}{(2+3 / n) 1} \rightarrow 1, -\]

    -

    so the ratio test is inconclusive. But we can compute \(\frac{n}{2 n+1}=\frac{1}{2+1 / n} \rightarrow \frac{1}{2}\).

    -

    As mentioned, the ratio test is especially useful if factorials and geometric terms are involved. It is important to make sure you use brackets correctly, it is very easy to make mistakes.

    -

    Example 2.4.14. Let

    -

    \[ -a_{n}=\frac{n ! \cdot 3^{n}}{\sqrt{(2 n) !}} -\]

    -

    Then,

    -

    \[ -\left|\frac{a_{n+1}}{a_{n}}\right|=\frac{(n+1) ! \cdot 3^{n+1}}{\sqrt{(2(n+1)) !}} / \frac{n ! \cdot 3^{n}}{\sqrt{(2 n) !}}=\frac{(n+1) !}{n !} \cdot \frac{3^{n+1}}{3^{n}} \cdot \frac{\sqrt{(2 n) !}}{\sqrt{(2(n+1)) !}} -\]

    -

    We can simplify \(\frac{(n+1) !}{n !} \cdot \frac{3^{n+1}}{3^{n}}=3(n+1)\). Combining the square roots and working out the brackets carefully gives that

    -

    \[ -\frac{\sqrt{(2 n) !}}{\sqrt{(2(n+1)) !}}=\sqrt{\frac{(2 n) !}{(2 n+2) !}}=\sqrt{\frac{1 \cdot 2 \cdots 2 n}{1 \cdot 2 \cdots 2 n \cdot(2 n+1) \cdot(2 n+2)}}=\frac{1}{\sqrt{(2 n+1)(2 n+2)}} -\]

    -

    Hence,

    -

    \[ -\left|\frac{a_{n+1}}{a_{n}}\right|=\frac{3(n+1)}{\sqrt{(2 n+1)(2 n+2)}}=\frac{3(1+1 / n)}{\sqrt{\frac{1}{n^{2}}(2 n+1)(2 n+2)}}=\frac{3(1+1 / n)}{\sqrt{(2+1 / n)(2+2 / n)}} \rightarrow \frac{3}{\sqrt{4}}=\frac{3}{2}>1 -\]

    -

    Hence, the sequence does not converge by the ratio test.

    -

    If the sequence is a geometric sequence, we can also apply the ratio test, which will give the same answer as the geometric sequence test.

    -

    Example 2.4.15. Let \(x \in \mathbb{R}\) and let \(a_{n}=x^{n}\). Then,

    -

    \[ -\left|\frac{a_{n+1}}{a_{n}}\right|=\frac{\left|x^{n+1}\right|}{\left|x^{n}\right|}=\frac{|x|^{n+1}}{|x|^{n}}=|x| \rightarrow|x| -\]

    -

    Hence, if \(|x|<1\) the sequence converges by the ratio test, if \(|x|>1\) the sequence does not converge by the ratio test, and if \(|x|=1\) the ratio test is inconclusive. Of course, we would have gotten this result immediately by using the geometric sequence test.

    +

    These results are in agreement with the Geometric sequence test.

    +
    +
    @@ -1936,8 +1993,8 @@

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