MATH 409 Lecture 8

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Lecture Slides

New Challenges

Challenge 7

Build a sequence $\left\{x_{n}\right\}$ of real numbers such that every real number is a limit point of $\left\{x_{n}\right\}$ (a limit point f a sequence is, by definition, the limit of a convergent subsequence).

Challenge 8

Let $\left\{F_{n}\right\}$ be the sequence of Fibbonacci numbers. Prove that

$\lim _{n\to \infty }{\frac {F_{n+1}}{F_{n}}}={\frac {{\sqrt {5}}+1}{2}}$

Review

• MATH 409 Lecture 7#Monotone Sequences

Examples

Theorem. If $0<a<1$, then $a^{n}\to 0$ as $n\to \infty$.

Proof. Since $a<1$ and $a>0$, it follows that $a^{n+1}<a^{n}$ and $a^{n}>0$ for all $n\in \mathbb {N}$. Hence the sequence $\left\{a^{n}\right\}$ is strictly decreasing and bounded. Therefore it converges to some $x\in \mathbb {R}$.

Since $a^{n+1}=a^{n}\,a$ for all $n$, it follows that $a^{n+1}\to x\,a$ as $n\to \infty$. However, $\left\{a^{n+1}\right\}$ is a subsequence of $\{a^{n}\}$, hence it converges to the same limit as $\left\{a^{n}\right\}$. Thus $x\,a=x$, which implies that $x=0$.

quod erat demonstrandum

Theorem. If $a>1$, then $a^{n}\to +\infty$ as $n\to \infty$.

Proof. Since $a>1$, it follows that $a^{n+1}>a^{n}>1$ for all $n\in \mathbb {N}$. Hence $\left\{a^{n}\right\}$ is strictly increasing. Then $\left\{a^{n}\right\}$ either diverges to $+\infty$ or converges to a limit $x$. In the latter case, we argue as above to obtain that $x=0$. However, this contradicts with $a^{n}>1$. Thus $\left\{a^{n}\right\}$ diverges to $+\infty$.

quod erat demonstrandum

Theorem. If $a>0$, then ${\sqrt[{n}]{a}}\to 1$ as $n\to \infty$.

Observe that by definition, ${\sqrt[{n}]{a}}$ is a unique positive number $r$ such that $r^{n}=a$.

Proof. If $a\geq 1$, then $a^{n+1}\geq a^{n}\geq 1$ for all $n\in \mathbb {N}$, which implies that ${\sqrt[{n(n+1)}]{a^{n+1}}}\geq {\sqrt[{n(n+1)}]{a^{n}}}\geq 1$. Notice that ${\sqrt[{n(n+1)}]{a^{n+1}}}={\sqrt[{n}]{a}}$ and ${\sqrt[{n(n+1)}]{a^{n}}}={\sqrt[{n+1}]{a}}$. Hence ${\sqrt[{n}]{a}}\geq {\sqrt[{n+1}]{a}}\geq 1$ for all Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} .

Similarly, in the case $0<a<1$, we obtain that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sqrt[n]{a} < \sqrt[n+1]{a} < 1} for all Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} . In either case, the sequence Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left\{ \sqrt[n]{a} \right\}} is monotone and bounded. Therefore it converges to a limit Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} . Then the sequence Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left\{ \sqrt[2n]{a} \right\}} also converges to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} since it is a subsequence of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left\{ \sqrt[n]{a} \right\}} . At the same time, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left( \sqrt[2n]{a} \right)^2 = \sqrt[n]{a}} , which implies that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x^2 = x} . Hence Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \in \{ 0, 1 \}} . However, the limit cannot be 0 since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sqrt[n]{a} \ge \min{(a,1)} > 0} (the first element sets the lower bound of an increasing function). Thus the limit is 1.

quod erat demonstrandum

Theorem. The sequence Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_n = \left( 1 + \frac{1}{n} \right)^n} for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n \in \mathbb{N}} is increasing and bounded, hence it is convergent.

Note: The limit is the number Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathrm{e} = 2.71828\ldots}

Proof. First let us show that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left\{ x_n \right\}} is increasing. For any Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n \in \mathbb{N}} ,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} x_n &= \left( 1 + \frac{1}{n} \right)^n \\ &= \left( \frac{n+1}{n} \right)^n \\ &= \frac{(n+1)^n}{n^n} \end{align}}

If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n \ge 2} , then similarly

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_{n-1} = \frac{n^{n-1}}{(n-1)^{n-1}}}

Hence

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \frac{x_n}{x_{n-1}} &= \frac{(n+1)^n}{n^n} \cdot \frac{(n-1)^{n-1}}{n^{n-1}} \\ &= \left( \frac{(n+1)(n-1)}{n^2} \right)^{n-1} \cdot \frac{n+1}{n} \\ &= \left( \frac{n^2-1}{n^2} \right)^{n-1} \cdot \frac{n+1}{n} \\ &= \left( 1 - \frac{1}{n^2} \right)^{n-1} \, \left( 1 + \frac{1}{n} \right) \end{align}}

To continue, we need the following lemma:

Lemma. If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0 < x < 1} , then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (1-x)^k \ge 1 - k\,x} for all Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k \in \mathbb{N}} .

Proof by induction. Basis. For Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k = 1} , we have Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (1-x)^1 = 1 - 1 \cdot x} .

Induction. Assume Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left( 1-x \right)^k \ge 1-k\,x} for some Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k \in \mathbb{N}} and all Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \in (0,1)} . Then

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} (1-x)^{k+1} &= \left( 1+x \right)^k \, (1-x) \\ &\ge (1-k\,x)\,(1-x) \\ &= 1 - k\,x - x + k\,x^2 > 1 - (k+1)\,x \end{align}}

Thus the lemma holds by induction on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k} .

Remark. According to the binomial formula,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (1-x)^k = 1 - k\,x + \frac{k(k-1)}{2} \, x^2 - \ldots}

Using this lemma, we obtain that

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \frac{x_n}{x_{n-1}} &= \left( 1 - \frac{1}{n^2} \right)^{n-1} \, \left( 1 + \frac{1}{n} \right) \\ &\ge \left( 1 - \frac{n-1}{n^2} \right) \, \left( 1 + \frac{1}{n} \right) \\ &= 1 - \frac{n-1}{n^2} + \frac{1}{n} - \frac{n-1}{n^3} \\ &= 1 + \frac{1}{n^2} - \frac{n-1}{n^3} \\ &= 1 + \frac{1}{n^3} > 1 \end{align}}

Thus the sequence Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left\{ x_n \right\}} is strictly increasing.

---

Now let us show that the sequence Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left\{ x_n \right\}} is bounded. Since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left\{ x_n \right\}} is increasing, it is enough to show that it is bounded above. By the binomial formula,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} x_n &= \left( 1 + \frac{1}{n} \right)^n \\ &= \sum_{k=0}^n \binom{n}{k} \, \left( \frac{1}{n} \right)^k \\ &= \sum_{k=0}^n \frac{n!}{k!(n-k)!} \, \left( \frac{1}{n} \right)^k \\ \end{align}}

Observe that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{n!}{k!(n-k)!} \, \left( \frac{1}{n} \right)^k \le 1} for all Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0 \le k \le n} because

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{n!}{k!(n-k)!} \, \left( \frac{1}{n} \right)^k = \frac{n(n-1)\dots(n-k+1)}{n\cdot n \dots n}}

Note there are Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k} terms in both the numerator and the denominator, and the denominator is obviously larger.

It follows that $x_{n}\leq \sum _{k=0}^{n}{\frac {1}{k!}}=1+{\frac {1}{1!}}+\dots +{\frac {1}{n!}}$

Further observe that $k!\geq 2^{k-1}$ for all $k\geq 0$ because $1\cdot 2\cdot 3\dots k\geq 2\cdot 2\dots 2$. There are $k-1$ factors greater than 2 on the LHS, and $k-1$ factors equal to 2 on the RHS.

Therefore we obtain

${\begin{aligned}x_{n}&\leq 1+1+{\frac {1}{2}}+{\frac {1}{2^{2}}}+\dots +{\frac {1}{2^{n-1}}}\\&=3-{\frac {1}{2^{n-1}}}\\&\leq 3\end{aligned}}$

So the sequence is bounded.

quod erat demonstrandum

Cauchy Sequences

A sequence $\left\{x_{n}\right\}$ of real numbers is called a Cauchy sequence if, for any $\epsilon >0$, there exists $N\in \mathbb {N}$ such that $\left|x_{n}-x_{m}\right|<\epsilon$ whenever $n,m\geq N$.

Note that the definition of cauchy sequences is redundant with equivalent to the theorem regarding convergent of sequences, only cauchy sequences do not need a limit to prove they converge.

Theorem. Any convergent sequence is Cauchy.

Proof. Let $\left\{x_{n}\right\}$ be a convergent sequence and $a$ be its limit. Then for any $\epsilon >0$ there exists $N\in \mathbb {N}$ such that $\left|x_{n}-a\right|<{\frac {\epsilon }{2}}$ whenever $n\geq N$. Now for any natural numbers $n,m\geq N$ we have

$\left|x_{n}-x_{m}\right|=\left|x_{n}-a+a-x_{m}\right|\leq \left|x_{n}-a\right|+\left|x_{m}-a\right|<{\frac {\epsilon }{2}}+{\frac {\epsilon }{2}}=\epsilon$

Thus $\left\{x_{n}\right\}$ is a Cauchy sequence.

quod erat demonstrandum

Theorem. Any Cauchy sequence is convergent.

Proof. (proved by Cauchy, but he called them fundamental sequences) Suppose $\left\{x_{n}\right\}$ is a Cauchy sequence. First let us show that the sequence is bounded. Since $\left\{x_{n}\right\}$ is Cauchy, there exists $N\in \mathbb {N}$ such that $\left|x_{n}-x_{m}\right|<1$ whenever Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n,m \ge N} . In particular, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left| x_n - x_N \right| < 1} for all Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n \ge N} . Then

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |x_n| = \left| \left( x_n - x_N \right) + x_N \right| \le |x_n - x_N| + |x_N| < |x_N| + 1}

It follows that for any natural number Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} , we have Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |x_n| \le M} , where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle M = \max{(|x_1|, |x_2|, \ldots, |x_{N-1}|, |x_N|+1)}} .

Now the Bolzano-Weierstrass theorem implies that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left\{ x_n \right\}} has a subsequence Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left\{ x_{n_k} \right\}_{k \in \mathbb{N}}} converging to some Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a \in \mathbb{R}} . Given Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon > 0} , there exists Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K_\epsilon \in \mathbb{N}} such that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left| x_{n_k} - a \right| < \frac{\epsilon}{2}} for all Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k \ge K_\epsilon} . Also, there exists Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle N_\epsilon \in \mathbb{N}} such that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left| x_n - x_m \right| < \frac{\epsilon}{2}} whenever Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n,m \ge N_\epsilon} . Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k = \max{(K_\epsilon, N_\epsilon)}} . Then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k \ge K_\epsilon} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n_k \ge k \ge N_\epsilon} . Therefore for any Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n \ge N_\epsilon} , we obtain

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left| x_n - a \right| = \left| \left( x_n - x_{n_k} \right) + \left( x_{n_k} - a \right) \right| \le \left| x_n - x_{n_k} \right| + \left| x_{n_k} - a \right| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon}

Thus the entire sequence Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left\{ x_n \right\}} converges to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} .

quod erat demonstrandum

Limit Points

A limit point of a sequence Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left\{ x_n \right\}} is the limit of any convergent subsequence of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left\{ x_n \right\}} .

• A convergent sequence has only one limit point, its limit.
• Any bounded sequence has at least one limit point (by Bolzano-Wierstrass theorem)
• If a bounded sequence is not convergent, then it has at least two limit points.
• The sequence Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left\{ \left( -1 \right)^n \right\}} has 2 limit points, namely 1 and −1
• If all elements of a sequence belong to a closed interval Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left[ a,b \right]} , then all its limits belong to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left[ a,b \right]} as well (by comparison theorem)
• The set of limit points of the sequence Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left\{ \sin{n} \right\}} is the entire interval Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left[ -1,1 \right]} (key: π is irrational)
• If a sequence diverges to infinity, then it has no limit points
• If a sequence does not diverge to infinity, then it has a bounded subsequence and hence it has a limit point.