# Sequences of real numbers

## Theorem 2.1.3 (Triangular inequality)

For all $x,y\in \mathbb{R}, \lvert x+y\rvert\leq \vert x\rvert+\vert y\rvert$

<Proof>

\begin{align*} 0\leq (x+y)^2 &=x^2+2xy + y^2 \\ &\leq \lvert x\rvert^2 + 2\lvert x\rvert\cdot \lvert y\rvert + \lvert y\rvert^2 =(\lvert x\rvert+\lvert y\rvert)^2 \end{align*}

Thus,

\begin{align*} \lvert x+y\rvert = \sqrt{(x+y)^2} \leq \sqrt{(\lvert x\rvert+\vert y\rvert)^2} = \lvert x\rvert +\lvert y\rvert \end{align*} $\tag*{\square}$

## Corollary 2.1.4

For all $x,y,z\in \mathbb{R}$, we have

\begin{align} \lvert x-y\rvert&\leq \lvert x-z\rvert +\lvert z-y\rvert \\ \lvert \lvert x\rvert-\lvert y\rvert \rvert &\leq \lvert x-y\rvert \end{align}

<Proof>

(1) By triangular inequality,

\begin{align*} \lvert x-y\rvert = \lvert (x-z) + (z-y) \rvert \leq \lvert x-z\rvert+\lvert z-y\rvert \end{align*}

(2) By triangular inequality, $\lvert x\rvert = \lvert x-y+y\rvert \leq \lvert x-y\rvert+\lvert y\rvert$ Thus, $\lvert x\rvert-\lvert y\rvert \leq \lvert x-y\rvert$. Similarly,

\begin{align*} \lvert y \rvert= \lvert y-x+x\rvert \leq \lvert x-y\rvert + \lvert x\rvert \end{align*}

So, $\lvert y\rvert-\lvert x\rvert \leq \lvert x-y\rvert$.

$\therefore \lvert\lvert x\rvert-\lvert y\rvert\rvert\leq \lvert x-y\rvert$

$\tag*{\square}$

## Definition 2.1.5

$d: \mathbb{R}\times\mathbb{R}\to\mathbb{R}$ is metric

(1) $d(x,y) \geq 0, d(x,y) = 0 \iff x=y$

(2) $d(x,y) = d(y,x)$

(3) $d(x,y) \leq d(x,z) + d(y,z)$

## Definition 2.1.6

Let $p \in \mathbb{R}$ and let $\epsilon >0$. The set

\begin{align*} N_\epsilon (p) &:=\{ x\in \mathbb{R}: \lvert x-p\rvert<\epsilon \} \\ &= \{x\in\mathbb{R}: p-\epsilon <x<p+\epsilon\} \\ &=(p-\epsilon,p+\epsilon) \end{align*}

is called an $\epsilon$-neighborhood of the point $p$.

## Definition 2.1.7

A sequence $(p_n)_{n=1}^\infty$ in $\mathbb{R}$ is said to converge ,

If $\exists p\in\mathbb{R}$ such that $\forall \epsilon >0, \exists N\in\mathbb{N}$ such that $n\geq N \Rightarrow \lvert p_n-p\rvert<\epsilon: (\iff p_n \in N_\epsilon (p))$. We write

\begin{align} \lim_{n\to\infty} p_n = p \end{align}

$(p_n )_{n=1}^\infty$ diverges if it does not converge.

## Theorem

Let $(p_n)_{n=1}^\infty$ be a sequence in metric space $(X,d)$. $p_n$ converges to $p\in X$ if and only if for every $r>0, N_r(p)$ contains almost all of the $(p_n)_{n=1}^\infty$, i.e., the neighborhood contains all but a finite number of $(p_n)_{n=1}^\infty$.

<proof>

$\Rightarrow$ Suppose that $p_n\to p$ as $n\to\infty$. Let $\epsilon>0$ be given. There exists $N\in\mathbb{N}$ such that $n\geq N \Rightarrow p_n\in N_\epsilon (p).$ Thus, there are $N-1$ points that lie outside of $N_\epsilon (p)$.

$\Leftarrow$ Suppose that any $N_\epsilon (p)$ contains almost all of the $(p_n)_{n=1}^\infty$ and only a finite number of $(p_n)_{n=1}^\infty$ lie outside of $N_\epsilon (p)$. Let $\epsilon >0$ be given and let $N$ be the greatest index of those who lie outside of $N_\epsilon (p)$. Then if $n\geq N+1$, then $p_n\in N_\epsilon (p)$.

$\therefore p_n\to p$

$\tag*{\square}$

## Examples 2.1.8

(a) $\lim_{n\to\infty}\frac{1}{n} = 0$

<proof>

Let $\epsilon >0$ be given. By Archimedean property, we can find $N\in\mathbb{N}$ s.t. $\frac{1}{N}< \epsilon$. If $n\geq N$, then $\lvert \frac{1}{n}\rvert=\frac{1}{n} \leq \frac{1}{N} <\epsilon$.

$\therefore \lim_{n\to\infty}\frac{1}{n} = 0$ $\tag*{\square}$

(b) $\lim_{n\to\infty}\frac{2n+1}{3n+2}=\frac{2}{3}$

<proof>

Let $\epsilon >0$ be given. By Archimedean property, $\exists N\in\mathbb{N}$ s.t. $\frac{1}{9N}<\epsilon$. If $n\geq N$, then

\begin{align*} \left\lvert\frac{2n+1}{3n+2} -\frac{2}{3}\right\rvert = \frac{1}{3(3n+2)} < \frac{1}{9n} \leq \frac{1}{9N} < \epsilon \end{align*}

$\therefore \frac{2n+1}{3n+2}=\frac{2}{3}$ $\tag*{\square}$

## Definition 2.1.9

A sequence $(p_n)_{n=1}^\infty$ in $\mathbb{R}$ is said to be bounded if there exists a constant $M>0$ such that $\lvert p_n\rvert\leq M$ for all $n\in\mathbb{N}$.

## Lemma 1

Let $a\in \mathbb{R}$ be a real number. If $\lvert a\rvert < \epsilon, \forall \epsilon >0$, then $a=0$.

<proof>

Assume that $a\neq 0$, i.e. $\lvert a\rvert >0$. Take $\epsilon_0 := \lvert a\rvert$. Then $\lvert a\rvert < \epsilon_0 = \lvert a\rvert$, which is a contradiction.

$\therefore a= 0$ $\tag*{\square}$

## Theorem 2.1.10

(a) If a sequence $(p_n)_{n=1}^\infty$ converges, then its limit is unique

(b) Every convergent sequence in $\mathbb{R}$ is bounded.

<proof>

(a) Suppose that $p_n\to L_1, p_n \to L_2$ as $n\to\infty$. Let $\epsilon >0$ be given.

Then there exists $N_1, N_2\in\mathbb{N}$ such that

\begin{align*} n\geq N_1 &\Rightarrow \lvert p_n-L_1\rvert <\epsilon/2 \\ n\geq N_2 &\Rightarrow \lvert p_n-L_2\rvert <\epsilon/2 \\ \end{align*}

Take \begin{align*} N:=\max\{N_1, N_2\} \end{align*}

If $n\geq N$, then

\begin{align*} \lvert L_1-L_2\rvert \leq \vert p_n-L_1\rvert + \vert p_n-L_2\rvert < \frac{\epsilon}{2} +\frac{\epsilon}{2} = \epsilon \end{align*}

Since $\epsilon >0$ was arbitrary, $L_1 = L_2$ by lemma 1.

$\tag*{\square}$

(b) Let $p_n \to L$. Then there is a $N\in\mathbb{N}$ s.t. $n> N \Rightarrow \lvert p_n -L\rvert <1$. Now,

\begin{align*} \lvert p_n\rvert -\lvert L\rvert \leq \lvert p_n-L\rvert < 1 \end{align*}

So, $\lvert p_n\rvert <\lvert L\rvert+1$. Define \begin{align*} M:= \max\{ \lvert p_1\rvert, \ldots, \lvert p_N\rvert, \lvert L\rvert+1\} > 0. \end{align*}

Then $\lvert p_n\rvert \leq M$ for all $n\in M$.

$\therefore (p_n)_{n=1}^\infty$ is bounded

$\tag*{\square}$

## Theorem 2.2.1

$\lim_{n\to\infty}a_n=a, \lim_{n\to\infty}b_n=b$

(a) $\lim_{n\to\infty}(a_n+b_n)= a + b$

(b) $\lim_{n\to\infty}a_n\cdot b_n = ab$

(c) $a\neq 0, a_n \neq 0, \forall n\in\mathbb{N} \Rightarrow \lim_{n\to\infty}\frac{b_n}{a_n}= \frac{b}{a}$

<proof>

(a) Let $\epsilon >0$ be given. There exists $N\in\mathbb{N}$ s.t. $n\geq N \Rightarrow \lvert a_n-a\rvert < \frac{\epsilon}{2}, \lvert b_n-b\rvert<\frac{\epsilon}{2}$. If $n\geq N$, then

\begin{align*} \lvert (a_n+b_n) -(a+b)\rvert &= \lvert a_n-a+b_n-b\rvert \\ &\leq \lvert a_n-a\rvert + \lvert b_n-b\rvert \\ &<\frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align*}

$\therefore \lim_{n\to\infty}(a_n+b_n) = a+b$

$\tag*{\square}$

(b) Let $\epsilon >0$ be given. Since $(a_n)_{n=1}^\infty$ converges, $(a_n)_{n=1}^\infty$ is a bounded sequence so that we can find $M>0$ s.t. $\lvert a_n\rvert \leq M$ for all $n\in \mathbb{N}$.

By our assumption, there exists $N\in\mathbb{N}$ s.t. $n\geq N\Rightarrow \lvert a_n-a\rvert<\frac{\epsilon}{\lvert b\rvert}$ and $\lvert b_n-b\rvert < \frac{\epsilon}{2M}$. If $n\geq N$, then

\begin{align*} \lvert a_nb_n -ab\rvert &= \lvert a_n b_n- a_nb + a_nb -ab\rvert \\ &\leq \lvert a_n\rvert\lvert b_n-b\rvert + \lvert b\rvert\lvert a_n -a\rvert \\ &\leq M\lvert b_n-b\rvert + \lvert b\rvert \lvert a_n-a\rvert \\ &< M \cdot \frac{\epsilon}{2M} + \lvert b\rvert \cdot \frac{\epsilon}{2\lvert b\rvert}\\ &=\frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align*}

$\therefore \lim_{n\to\infty}a_n b_n = b$ $\tag*{\square}$

(c) By (b), if suffices to show that $\lim_{n\to\infty}\frac{1}{a_n}=\frac{1}{a}$.

Let $\epsilon >0$ be given. Then, there exists $N_1 \in\mathbb{N}$ such that

\begin{align*} n\geq N_1 \Rightarrow \lvert a_n-a\rvert <\frac{\lvert a\rvert^2}{2}\epsilon \end{align*}

Also, there exists $N_2\in \mathbb{N}$such that

\begin{align*} n\geq N_2 \Rightarrow \lvert a_n-a\rvert < \frac{\lvert a\rvert}{2} \end{align*} Since $\lvert a\rvert-\lvert a_n\rvert \leq \lvert a_n-a\rvert, \lvert a_n\rvert>\frac{\lvert a\rvert}{2}$ for $n\geq N_2$. Put

\begin{align*} N:=\max\{ N_1, N_2\} \in \mathbb{N}. \end{align*}

If $n\geq N$, then

\begin{align} \lvert \frac{1}{a_n}=\frac{1}{a}\rvert = \frac{\lvert a_n-a\rvert}{\lvert a_n\rvert\lvert a\lvert} < \frac{2}{\lvert a\rvert^2}\cdot\frac{\lvert a\rvert^2}{2}\epsilon = \epsilon \end{align}

$\therefore \lim_{n\to\infty}\frac{1}{a_n}=\frac{1}{a}$ $\tag*{\square}$

## Corollary 2.2.

$\lim_{n\to\infty}a_n = a, c\in \mathbb{R}$

(a) $\lim_{n\to\infty}(a_n +c) = a+c$

(b) $\lim_{n\to\infty}ca_n = ca$

## Theorem 2.2.3

Let $\lim_{n\to\infty}a_n=0$ and let $(b_n)_{n=1}^\infty$ be a bounded sequence. Then $\lim_{n\to\infty}a_nb_n=0$

<proof> Let $\epsilon >0$ be given. Then there exists $N\in\mathbb{N}$ s.t. $n\geq N \Rightarrow \lvert a_n\rvert<\frac{\epsilon}{M}$. If $n\geq N$,

\begin{align*} \lvert a_n b_n\rvert &= \lvert a_n\rvert\lvert b_n\rvert\\ &\leq \lvert a_n\rvert M \\ &<\frac{\epsilon}{M}\cdot M = \epsilon \end{align*}

$\therefore \lim_{n\to\infty}a_nb_n=0$

$\tag*{\square}$

## Theorem 2.2.4 (Squeeze theorem)

Suppose the sequence $(a_n)_{n=1}^\infty, (b_n)_{n=1}^\infty,\text{and } (c_n)_{n=1}^\infty$ are sequences of real numbers for which there exists $n_0\in \mathbb{N}$ such that

\begin{align*} a_n \leq b_n \leq c_n \text{ for all } n\in \mathbb{N}, n\geq n_0 \end{align*}

and that $\lim_{n\to\infty}a_n = \lim_{n\to\infty}c_n = L.$ Then the sequence $(b_n)_{n=1}^\infty$ converges and $\lim_{n\to\infty}b_n=L$.

<proof>

Let $\epsilon >0$ be given. Then there exists $N_1,N_2\in\mathbb{N}$ such that

\begin{align*} n\geq N_1 \Rightarrow \lvert a_n-L\rvert <\epsilon \\ n\geq N_2 \Rightarrow \lvert b_n-L\rvert <\epsilon \end{align*}

Put

\begin{align*} N:=\max\{N_1, N_2, n_0\} \in \mathbb{N}. \end{align*}

If $n\geq N$, then \begin{align*} L-\epsilon < a_n\leq b_n \leq c_n < L+\epsilon \end{align*}

So, $\lvert b_n -L\rvert<\epsilon$ for $n\geq N$.

$\therefore \lim_{n\to\infty}b_n=L$. $\tag*{\square}$

## Theorem 2.2.6

(a) $\lim_{n\to\infty}\frac{1}{n^p} =0$, for $p> 0$

(b) $\lim_{n\to\infty}\sqrt[n]{p} =1$ for $p>0$

(c) $\lim_{n\to\infty}\sqrt[n]{n} = 1$

(d) $p>1, \alpha \in\mathbb{R}\Rightarrow \lim_{n\to\infty}\frac{n^\alpha}{p^n}=0$

(e) $\lvert p\rvert <1 \Rightarrow \lim_{n\to\infty}p^n =0$

(f) $p\in\mathbb{R} \Rightarrow \lim_{n\to\infty}\frac{p^n}{n!} = 0$

<proof>

(a) Let $\epsilon >0$ be given. By Archimedean property, there exists $N\in\mathbb{N}$ s.t. $\frac{1}{N}< \epsilon^{1/p}$.

For $n\geq N$,

\begin{align*} \frac{1}{n}\leq \frac{1}{N} < \epsilon^{1/p} \end{align*} Thus, $\left\lvert\frac{1}{n^p}\right\rvert < \frac{1}{N^p} <\epsilon$.

$\therefore \lim_{n\to\infty}\frac{1}{n^p}=0$ $\tag*{\square}$

(b) Assume $p>1$. Let $\epsilon >0$ be given and let $p:=1+\delta$ for some $\delta>0$. We want to show that

\begin{align*} 1\leq p^{\frac{1}{n}} \leq 1 + \frac{\delta}{n} \text{ for } n\geq 1 \end{align*}

Suppose $p^\frac{1}{n} > 1 +\frac{\delta}{n}$. Then $p=(p^\frac{1}{n})^n > (1+\frac{\delta}{n})^n$. By Bernoulli inequality,

\begin{align*} (1+\frac{\delta}{n})^n \geq 1 + n\cdot\frac{\delta}{n} = 1+\delta \end{align*} It implies that $p> 1+\delta$, which contradicts to the assumption.

$\therefore p^\frac{1}{n} \leq 1+\frac{\delta}{n}$

$\therefore \lim_{n\to\infty}p^\frac{1}{n}=1$

Now, assume $0<p<1$. Let $a:=\frac{1}{p}$. Then $a>1$.

$\therefore \lim_{n\to\infty}a^\frac{1}{n}=1$

By the limit theorem,

\begin{align*} \lim_{n\to\infty}p^\frac{1}{n} = \lim_{n\to\infty} \frac{1}{a^{1/n}} = \frac{1}{1} \end{align*}

$\therefore \lim_{n\to\infty}p^{\frac{1}{n}}=1$

$\tag*{\square}$

(c) Let $x_n:=\sqrt[n]{n}-1$ for $n\geq 2$. Then

\begin{align} n=(x_n+1)^n \geq \frac{n(n-1)}{2}x^2_n \end{align} Thus, $0<x^2_n \leq \frac{2}{n-1}$, i.e. $0<x_n \leq \sqrt{\frac{2}{n-1}}$.

$\therefore \lim_{n\to\infty}x_n=0$

$\therefore \lim_{n\to\infty}\sqrt[n]{n}=1$ $\tag*{\square}$

(d) Let $p:=1+q$ for some $q>0$ and let $k\in\mathbb{N}$ with $k>\alpha$. For $k< n/2$,

\begin{align} \begin{split} p^n = (1+q)^n &> {n \choose k}q^k \\ &=\frac{n(n-1)\cdots(n-k+1)}{k!}q^k \\ &> \frac{n^k}{2^kk!} q^k \:(\because n-k+1 > \frac{n}{2}+1 > \frac{n}{2})\\ \end{split} \end{align}

Thus,

\begin{align} 0< \frac{n^\alpha}{p^n} <\frac{2^k k!}{q^kn^{k-\alpha}} \end{align}

Since $k-\alpha>0, \lim_{n\to\infty}1/n^{k-\alpha}=0.$

$\therefore \lim_{n\to\infty}\frac{n^\alpha}{p^n} =0$ $\tag*{\square}$

(e) Let $p=\pm \frac{1}{q}$ for some $q>1$.

\begin{align} \lvert p^n\rvert = \vert p\rvert^n = \frac{1}{q^n} \end{align}

$\therefore \lim_{n\to\infty}p^n=0$ by (d) where $\alpha=0$.

(f) Let $k\in\mathbb{N}$ with $k>\lvert p\rvert$. For $n>k$

\begin{align*} \left\lvert\frac{p^n}{n!}\right\rvert &= \frac{\left\lvert p\right\rvert^n}{n!} = \frac{\left\lvert p\right\rvert^n}{(k-1)!k\cdots (n-1)n}\\ &< \frac{\left\lvert p\right\rvert^n}{(k-1)!k^{n-k+1}}\\ &=\frac{k^{k-1}}{(k-1)!}\frac{\left\lvert p\right\rvert^n}{k^n} \end{align*}

Since $0<\frac{\lvert p\rvert}{k}<1, \lim_{n\to\infty}\frac{\lvert p\rvert^n}{k}=0$

$\therefore \lim_{n\to\infty}\frac{p^n}{n!} = 0$ $\tag*{\square}$

## Reference

• Manfred Stoll, Introduction to Real Analysis, Pearson

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