2.1: Sequences

Definition

Let $\left\{a_{i}\right\}_{i \in I}$ be a sequence of real numbers. We say $\left\{a_{i}\right\}_{i \in I}$ converges, and has limit $L,$ if for every real number $\epsilon>0$ there exists an integer $N$ such that

$$\left|a_{i}-L\right|<\epsilon \text { whenever } i>N.$$

We say a sequence $\left\{a_{i}\right\}_{i \in I}$ which does not converge diverges.

Definitions: nondecreasing and nonincreasing sequences

• We say a sequence $\left\{a_{i}\right\}_{i \in I}$ is nondecreasing if $a_{i+1} \geq a_{i}$ for every $i \in I$ and increasing if $a_{i+1}>a_{i}$ for every $i \in I$.
• We say a sequence $\left\{a_{i}\right\}_{i \in I}$ is nonincreasing if $a_{i+1} \leq a_{i}$ for every $i \in I$ and decreasing if $a_{i+1}<a_{i}$ for every $i \in I$.

Definition: bounded sequences

We say a set $A \subset \mathbb{R}$ is bounded if there exists a real number $M$ such that $|a| \leq M$ for every $a \in A .$ We say a sequence $\left\{a_{i}\right\}_{i \in I}$ of real numbers is bounded if there exists a real number $M$ such that $\left|a_{i}\right| \leq M$ for all $i \in I .$

Theorem $\PageIndex{1}$

If $\left\{a_{i}\right\}_{i \in I}$ is a nondecreasing, bounded sequence of real numbers, then $\left\{a_{i}\right\}_{i \in I}$ converges.

Proof

Since $\left\{a_{i}\right\}_{i \in I}$ is bounded, the set of $A=\left\{a_{i}: i \in I\right\}$ has a supremum. Let $L=\sup A .$ For any $\epsilon>0,$ there must exist $N \in I$ such that $a_{N}>L-\epsilon$ (or else $L-\epsilon$ would be an upper bound for $A$ which is smaller than $L$ ). But then

$$L-\epsilon<a_{N} \leq a_{i} \leq L<L+\epsilon$$

for all $i \geq N,$ that is,

$$\left|a_{i}-L\right|<\epsilon$$

for all $i \geq N .$ Thus $\left\{a_{i}\right\}_{i \in I}$ converges and

$$L=\lim _{i \rightarrow \infty} a_{i}.$$

Q.E.D.

We conclude from the previous theorem that every nondecreasing sequence of real numbers either has a limit or is not bounded, that is, is unbounded.

Exercise $\PageIndex{1}$

Show that a nonincreasing, bounded sequence of real numbers must converge.

Definition

Let $\left\{a_{i}\right\}_{i \in I}$ be a sequence of real numbers. If for every real number $M$ there exists an integer $N$ such that $a_{i}>M$ whenever $i>N,$ then we say the sequence $\left\{a_{i}\right\}_{i \in I}$ diverges to positive infinity, denoted by

$$\lim _{i \rightarrow \infty} a_{i}=+\infty.$$

Similarly, if for every real number $M$ there exists an integer $N$ such that $a_{i}<M$ whenever $i>N,$ then we say the sequence $\left\{a_{i}\right\}_{i \in I}$ diverges to negative infinity, denoted by

$$\lim _{i \rightarrow \infty} a_{i}=-\infty.$$

Exercise $\PageIndex{2}$

Show that a nondecreasing sequence of real numbers either converges or diverges to positive infinity.

Exercise $\PageIndex{3}$

Show that a nonincreasing sequence of real numbers either converges or diverges to negative infinity.

2.1.1 Extended Real Numbers

It is convenient to add the symbols $+\infty$ and $-\infty$ to the real numbers $\mathbb{R} .$ Although neither $+\infty$ nor $-\infty$ is a real number, we agree to the following operational conventions:

1. Given any real number $r,-\infty<r<\infty$.

2. For any real number $r$,

$$r+(+\infty)=r-(-\infty)=r+\infty=+\infty,$$

$$r+(-\infty)=r-(+\infty)=r-\infty=-\infty,$$

and

$$\frac{r}{+\infty}=\frac{r}{-\infty}=0.$$

3. For any real number $r>0, r \cdot(+\infty)=+\infty$ and $r \cdot(-\infty)=-\infty$.

4. For any real number $r<0, r \cdot(+\infty)=-\infty$ and $r \cdot(-\infty)=+\infty$.

5. If $a_{i}=-\infty, i=1,2,3, \ldots,$ then $\lim _{i \rightarrow \infty} a_{i}=-\infty$.

6. If $a_{i}=+\infty, i=1,2,3, \ldots,$ then $\lim _{i \rightarrow \infty} a_{i}=+\infty$.

Note that with the order relation defined in this manner, $+\infty$ is an upper

bound and $-\infty$ is a lower bound for any set $A \subset \mathbb{R}$. Thus if $A \subset \mathbb{R}$ does not have a finite upper bound, then $\sup A=+\infty ;$ similarly, if $A \subset \mathbb{R}$ does not have a finite lower bound, then inf $A=-\infty$.

When working with extended real numbers, we refer to the elements of $\mathbb{R}$ as finite real numbers and the elements $+\infty$ and $-\infty$ as infinite real numbers.

Exercise $\PageIndex{4}$

Do the extended real numbers form a field?

2.1.2 Limit Superior and Inferior

Definition

Let $\left\{a_{i}\right\}_{i \in I}$ be a sequence of real numbers and, for each $i \in I,$ let $u_{i}=\sup \left\{a_{j}: j \geq i\right\} .$ If $u_{i}=+\infty$ for every $i \in I,$ we let

$$\limsup _{i \rightarrow \infty} a_{i}=+\infty;$$

otherwise, we let

$$\limsup _{i \rightarrow \infty} a_{i}=\inf \left\{u_{i}: i \in I\right\}.$$

In either case, we call $\liminf _{n \rightarrow \infty} a_{n}$ the limit inferior of the sequence $\left\{a_{i}\right\}_{i \in I}$.

Exercise $\PageIndex{5}$

Given a sequence $\left\{a_{i}\right\}_{i \in I},$ define $\left\{u_{i}\right\}_{i \in I}$ and $\left\{l_{i}\right\}_{i \in I}$ as in the previous two definitions. Show that

$$\limsup _{i \rightarrow \infty} a_{i}=\lim _{i \rightarrow \infty} u_{i}$$

and

$$\liminf _{i \rightarrow \infty} a_{i}=\lim _{i \rightarrow \infty} l_{i}.$$

Exercise $\PageIndex{6}$

Find $\limsup _{i \rightarrow \infty} a_{i}$ and $\liminf _{i \rightarrow \infty} a_{i}$ for the sequences $\left\{a_{i}\right\}_{i=1}^{\infty}$ as de fined below.

a. $a_{i}=(-1)^{i}$

b. $a_{i}=i$

c. $a_{i}=2^{-i}$

d. $a_{i}=\frac{1}{i}$

The following proposition is often called the squeeze theorem.

Proposition $\PageIndex{2}$

Suppose $\left\{a_{i}\right\}_{i \in I},\left\{b_{j}\right\}_{j \in J},$ and $\left\{c_{k}\right\}_{k \in K}$ are sequences of real numbers for which there exists an integer $N$ such that $a_{i} \leq c_{i} \leq b_{i}$ whenever $i>N .$ If

$$\lim _{i \rightarrow \infty} a_{i}=\lim _{i \rightarrow \infty} b_{i},$$

then

$$\lim _{i \rightarrow \infty} c_{i}=\lim _{i \rightarrow \infty} a_{i}=\lim _{i \rightarrow \infty} b_{i}.$$

Proof

Let $L=\lim _{i \rightarrow \infty} a_{i}=\lim _{i \rightarrow \infty} b_{i} .$ Suppose $L$ is finite. Given $\epsilon>0,$ there exists an integer $M$ such that

$$\left|a_{i}-L\right|<\frac{\epsilon}{3}$$

and

$$\left|b_{i}-L\right|<\frac{\epsilon}{3}$$

whenever $i>M .$ Then

$$\left|a_{i}-b_{i}\right| \leq\left|a_{i}-L\right|+\left|L-b_{i}\right|<\frac{\epsilon}{3}+\frac{\epsilon}{3}=\frac{2 \epsilon}{3}$$

whenever $i>M .$ Let $K$ be the larger of $N$ and $M .$ Then

$$\left|c_{i}-L\right| \leq\left|c_{i}-b_{i}\right|+\left|b_{i}-L\right| \leq\left|a_{i}-b_{i}\right|+\left|b_{i}-L\right|<\frac{2 \epsilon}{3}+\frac{\epsilon}{3}=\epsilon$$

whenever $i>K .$ Thus lim $c_{i}=L .$ The result when $L$ is infinite is a consequence of the next two exercises. $\quad$ Q.E.D.

Exercise $\PageIndex{7}$

Suppose $\left\{a_{i}\right\}_{i \in I}$ and $\left\{c_{k}\right\}_{k \in K}$ are sequences for which there exists an integer $N$ such that $a_{i} \leq c_{i}$ whenever $i>N .$ Show that if $\lim _{i \rightarrow \infty} a_{i}=+\infty,$ then $\lim _{i \rightarrow \infty} c_{i}=+\infty .$

Exercise $\PageIndex{8}$

Suppose $\left\{b_{j}\right\}_{j \in J}$ and $\left\{c_{k}\right\}_{k \in K}$ are sequences for which there exists an integer $N$ such that $c_{i} \leq b_{i}$ whenever $i>N .$ Show that if $\lim _{i \rightarrow \infty} b_{i}=-\infty,$ then $\lim _{i \rightarrow \infty} c_{i}=-\infty .$

Exercise $\PageIndex{9}$

Suppose $\left\{a_{i}\right\}_{i \in I}$ and $\left\{b_{j}\right\}_{j \in J}$ are sequences of real numbers with $a_{i} \leq b_{i}$ for all $i$ larger than some integer $N .$ Assuming both sequences converge, show that

$$\lim _{\mathfrak{i} \rightarrow \infty} a_{i} \leq \lim _{i \rightarrow \infty} b_{i}.$$

Exercise $\PageIndex{10}$

Show that for any sequence $\left\{a_{i}\right\}_{i \in I}$,

$$\liminf _{i \rightarrow \infty} a_{i} \leq \limsup _{i \rightarrow \infty} n_{i}.$$

Proposition $\PageIndex{3}$

Suppose $\left\{a_{i}\right\}_{i \in I}$ is a sequence for which

$$\limsup _{i \rightarrow \infty} a_{i}=\liminf _{i \rightarrow \infty} a_{i}.$$

Then

$$\lim _{i \rightarrow \infty} a_{i}=\limsup _{i \rightarrow \infty} a_{i}=\liminf _{i \rightarrow \infty} a_{i}.$$

Proof

Let $u_{i}=\sup \left\{a_{k}: k \geq i\right\}$ and $l_{i}=\inf \left\{a_{k}: k \geq i\right\} .$ Then $l_{i} \leq a_{i} \leq u_{i}$ for all $i \in I .$ Now

$$\lim _{i \rightarrow \infty} l_{i}=\liminf _{i \rightarrow \infty} a_{i}=\limsup _{i \rightarrow \infty} a_{i}=\lim _{i \rightarrow \infty} u_{i},$$

so the result follows from the squeeze theorem. $\quad$ Q.E.D.

Exercise $\PageIndex{11}$

Suppose $u$ is a real number such that $u \geq 0$ and $u<\epsilon$ for any real number $\epsilon>0 .$ Show that $u=0$.

2.1.3 Completeness

Definition

Suppose $\left\{a_{i}\right\}_{i \in I}$ is a sequence in $\mathbb{R} .$ We call $\left\{a_{i}\right\}_{i \in I}$ a Cauchy sequence if for every $\epsilon>0$ there exists an integer $N$ such that

$$\left|a_{i}-a_{j}\right|<\epsilon$$

whenever both $i>N$ and $j>N$.

Theorem $\PageIndex{4}$

Suppose $\left\{a_{i}\right\}_{i \in I}$ is a Cauchy sequence in $\mathbb{R} .$ Then $\left\{a_{i}\right\}_{i \in I}$ converges to a limit $L \in \mathbb{R}$.

Proof

Let $u_{i}=\sup \left\{a_{k}: k \geq i\right\}$ and $l_{i}=\inf \left\{a_{k}: k \geq i\right\} .$ Given any $\epsilon>0$, there exists $N \in \mathbb{Z}$ such that $\left|a_{i}-a_{j}\right|<\epsilon$ for all $i, j>N .$ Thus, for all $i, j>N$, $a_{i}<a_{j}+\epsilon_{1}$ and so

$$a_{i} \leq \inf \left\{a_{j}+\epsilon: j \geq i\right\}=l_{i}+\epsilon$$

for all $i>N .$ Since $\left\{l_{i}\right\}_{i \in I}$ is a nondecreasing sequence,

$$a_{i} \leq \sup \left\{l_{i}+\epsilon: i \in I\right\}=\liminf _{i \rightarrow \infty} a_{i}+\epsilon$$

for all $i>N .$ Hence

$$u_{i}=\sup \left\{a_{k}: k \geq i\right\} \leq \liminf _{i \rightarrow \infty} a_{i}+\epsilon$$

for all $i>N .$ Thus

$$\limsup _{i \rightarrow \infty} a_{i}=\inf \left\{u_{i}: i \in I\right\} \leq \liminf _{i \rightarrow \infty} a_{i}+\epsilon.$$

Since $\liminf _{i \rightarrow \infty} a_{i} \leq \limsup _{i \rightarrow \infty} a_{i},$ it follows that

$$\left|\limsup _{i \rightarrow \infty} a_{i}-\liminf _{i \rightarrow \infty} a_{i}\right| \leq \epsilon.$$

since this is true for every $\epsilon>0,$ we have $\lim \sup _{i \rightarrow \infty} a_{i}=\lim _{i \rightarrow \infty} \inf a_{i},$ and so $\left\{a_{i}\right\}_{i \in I}$ converges by Proposition $2.1 .3 .$ $\quad$ Q.E.D.

As a consequence of the previous theorem, we say that $\mathbb{R}$ is a complete metric space.

Exercise $\PageIndex{12}$

Suppose $A \subset \mathbb{R}, A \neq \emptyset,$ and $s=\sup A .$ Show that there exists a sequence $\left\{a_{i}\right\}_{i=1}^{\infty}$ with $a_{i} \in A$ such that $\lim _{i \rightarrow \infty} a_{i}=s$.

Exercise $\PageIndex{13}$

Given a real number $x \geq 0,$ show that there exists a real number $s \geq 0$ such that $s^{2}=x$.

We let $\sqrt{x}$ denote the number $s$ in the previous exercise, the square root of $x$.

2.1.4 Some Basic Results About Sequences

Proposition $\PageIndex{5}$

Suppose $\left\{x_{i}\right\}_{i \in I}$ is a convergent sequence in $\mathbb{R}, \alpha$ is a real number, and $L=\lim _{i \rightarrow \infty} x_{i} .$ Then the sequence $\left\{\alpha x_{i}\right\}_{i \in I}$ oonverges and

$$\lim _{i \rightarrow \infty} \alpha x_{i}=\alpha L.$$

Proof

If $\alpha=0,$ then $\left\{\alpha x_{i}\right\}_{i \in I}$ clearly converges to $0 .$ So assume $\alpha \neq 0 .$ Given $\epsilon>0,$ choose an integer $N$ such that

$$\left|x_{i}-L\right|<\frac{\epsilon}{|\alpha|}$$

whenever $i>N .$ Then for any $i>N$ we have

$$\left|\alpha x_{i}-\alpha L\right|=|\alpha|\left|x_{i}-L\right|<|\alpha| \frac{\epsilon}{|\alpha|}=\epsilon.$$

Thus $\lim _{i \rightarrow \infty} \alpha x_{i}=\alpha L$. $\quad$ Q.E.D.

Proposition $\PageIndex{6}$

Suppose $\left\{x_{i}\right\}_{i \in I}$ and $\left\{y_{i}\right\}_{i \in I}$ are convergent sequences in $\mathbb{R}$ with $L=\lim _{i \rightarrow \infty} x_{i}$ and $M=\lim _{i \rightarrow \infty} y_{i} .$ Then the sequence $\left\{x_{i}+y_{i}\right\}_{i \in I}$ converges and

$$\lim _{i \rightarrow \infty}\left(x_{i}+y_{i}\right)=L+M.$$

Exercise $\PageIndex{14}$

Prove the previous proposition.

Proposition $\PageIndex{7}$

Suppose $\left\{x_{i}\right\}_{i \in I}$ and $\left\{y_{i}\right\}_{i \in I}$ are convergent sequences in $\mathbb{R}$ with $L=\lim _{i \rightarrow \infty} x_{i}$ and $M=\lim _{i \rightarrow \infty} y_{i} .$ Then the sequence $\left\{x_{i} y_{i}\right\}_{i \in I}$ converges and

$$\lim _{i \rightarrow \infty} x_{i} y_{i}=L M.$$

Exercise $\PageIndex{15}$

Prove the previous proposition.

Proposition $\PageIndex{8}$

Suppose $\left\{x_{i}\right\}_{i \in I}$ and $\left\{y_{i}\right\}_{i \in I}$ are convergent sequences in $\mathbb{R}$ with $L=\lim _{i \rightarrow \infty} x_{i}, M=\lim _{i \rightarrow \infty} y_{i},$ and $y_{i} \neq 0$ for all $i \in I .$ If $M \neq 0,$ then the sequence $\left\{\frac{x_{i}}{y_{i}}\right\}_{i \in I}$ converges and

$$\lim _{i \rightarrow \infty} \frac{x_{i}}{y_{i}}=\frac{L}{M}.$$

Proof

Since $M \neq 0$ and $M=\lim _{i \rightarrow \infty} y_{i},$ we may choose an integer $N$ such that

$$\left|y_{i}\right|>\frac{|M|}{2}$$

whenever $i>N .$ Let $B$ be an upper bound for $\left\{\left|x_{i}\right|: i \in I\right\} \cup\left\{\left|y_{i}\right|: i \in I\right\} .$ Given any $\epsilon>0,$ we may choose an integer $P$ such that

$$\left|x_{i}-L\right|<\frac{M^{2} \epsilon}{4 B}$$

and

$$\left|y_{i}-M\right|<\frac{M^{2} \epsilon}{4 B}$$

whenever $i>P .$ Let $K$ be the larger of $N$ and $P .$ Then, for any $i>K,$ we have

$$\begin{aligned}\left|\frac{x_{i}}{y_{i}}-\frac{L}{M}\right| &=\frac{\left|x_{i} M-y_{i} L\right|}{\left|y_{i} M\right|} \\ &=\frac{\left|x_{i} M-x_{i} y_{i}+x_{i} y_{i}-y_{i} L\right|}{\left|y_{i} M\right|} \\ & \leq \frac{\left|x_{i}\right|\left|M-y_{i}\right|+\left|y_{i}\right|\left|x_{i}-L\right|}{\left|y_{i} M\right|} \\ &<\frac{B \frac{M^{2} \epsilon}{4 B}+B \frac{M^{2} \epsilon}{4 B}}{\frac{M^{2}}{2}} \\ &=\epsilon . \end{aligned}$$

Thus

$$\lim _{i \rightarrow \infty} \frac{x_{i}}{y_{i}}=\frac{L}{M}.$$

Q.E.D.

Exercise $\PageIndex{16}$

a. Show that $\lim _{n \rightarrow \infty} \frac{1}{n}=0$.

b. Show that $\lim _{n \rightarrow \infty} \frac{1}{n^{2}}=0$ by (i) using the definition of limit directly and then (ii) using previous results.

Exercise $\PageIndex{17}$

Show that for any positive integer $k$,

$$\lim _{n \rightarrow \infty} \frac{1}{n^{k}}=0.$$

Example $\PageIndex{1}$

We may combine the properties of this section to compute

$$\begin{aligned} \lim _{n \rightarrow \infty} \frac{5 n^{3}+3 n-6}{2 n^{3}+2 n^{2}-7} &=\lim _{n \rightarrow \infty} \frac{5+\frac{3}{n^{2}}-\frac{6}{n^{3}}}{2+\frac{2}{n}-\frac{7}{n^{3}}} \\ &=\frac{\lim _{n \rightarrow \infty} 5+3 \lim _{n \rightarrow \infty} \frac{1}{n^{2}}-6 \lim _{n \rightarrow \infty} \frac{1}{n^{3}}}{\lim _{n \rightarrow \infty} 2+2 \lim _{n \rightarrow \infty} \frac{1}{n}-7 \lim _{n \rightarrow \infty} \frac{1}{n^{3}}} \\ &=\frac{5+0+0}{2+0+0} \\ &=\frac{5}{2}. \end{aligned}$$

Exercise $\PageIndex{18}$

Evaluate

$$\lim _{n \rightarrow \infty} \frac{3 n^{5}+8 n^{3}-6 n}{8 n^{5}+2 n^{4}-31},$$

carefully showing each step.

Proposition $\PageIndex{9}$

Suppose $\left\{x_{i}\right\}_{i \in I}$ is a convergent sequence of nonnegative real numbers with $L=\lim _{i \rightarrow \infty} x_{i} .$ Then the sequence $\{\sqrt{x_{i}}\}_{i \in I}$ converges and

$$\lim _{i \rightarrow \infty} \sqrt{x_{i}}=\sqrt{L}.$$

Proof

Let $\epsilon>0$ be given. Suppose $L>0$ and note that

$$\left|x_{i}-L\right|=|\sqrt{x_{i}}-\sqrt{L}||\sqrt{x_{i}}+\sqrt{L}|$$

implies that

$$|\sqrt{x_{i}}-\sqrt{L}|=\frac{\left|x_{i}-L\right|}{|\sqrt{x_{i}}+\sqrt{L}|}$$

for any $i \in I .$ Choose an integer $N$ such that

$$\left|x_{i}-L\right|<\sqrt{L} \epsilon$$

whenever $i>N .$ Then, for any $i>N$,

$$|\sqrt{x_{i}}-\sqrt{L}|=\frac{\left|x_{i}-L\right|}{|\sqrt{x_{i}}+\sqrt{L}|}<\frac{\sqrt{L} \epsilon}{\sqrt{L}}=\epsilon.$$

Hence $\lim _{i \rightarrow \infty} \sqrt{x_{i}}=\sqrt{L}$.

If $L=0, \lim _{i \rightarrow \infty} x_{i}=0,$ so we may choose an integer $N$ such that $\left|x_{i}\right|<\epsilon^{2}$ for all $i>N .$ Then

$$|\sqrt{x_{i}}|<\epsilon$$

whenever $i>N,$ so $\lim _{i \rightarrow \infty} \sqrt{x_{i}}=0$. $\quad$ Q.E.D.

Exercise $\PageIndex{19}$

Evaluate

$$\lim _{n \rightarrow \infty} \frac{\sqrt{3 n^{2}+1}}{5 n+6},$$

carefully showing each step.

Exercise $\PageIndex{20}$

Given real numbers $r>0$ and $\alpha,$ show that $(\mathrm{a}) \alpha r<r$ if $0<\alpha<1$ and $(\mathrm{b}) r<\alpha r$ if $\alpha>1$.

Proposition $\PageIndex{10}$

If $x \in \mathbb{R}$ and $|x|<1,$ then

$$\lim _{n \rightarrow \infty} x^{n}=0.$$

Proof

We first assume $x \geq 0$. Then the sequence $\left\{x^{n}\right\}_{n=1}^{\infty}$ is nonincreasing and bounded below by 0. Hence the sequence converges. Let $L=\lim _{n \rightarrow \infty} x^{n} .$ Then

$$L=\lim _{n \rightarrow \infty} x^{n}=x \lim _{n \rightarrow \infty} x^{n-1}=x L,$$

from which it follows that $L(1-x)=0 .$ since $1-x>0,$ we must have $L=0$. The result for $x<0$ follows from the next exercise. $\quad$ Q.E.D.

Exercise $\PageIndex{21}$

Show that $\lim _{n \rightarrow \infty}\left|a_{n}\right|=0$ if and only if $\lim _{n \rightarrow \infty} a_{n}=0$.

2.1.5 Subsequences

Definition

Given a sequence $\left\{x_{i}\right\}_{i=m}^{\infty},$ suppose $\left\{n_{k}\right\}_{k=1}^{\infty}$ is an increasing sequence of integers with

$$m \leq n_{1}<n_{2}<n_{3}<\cdots .$$

Then we call the sequence $\left\{x_{n_{k}}\right\}_{k=1}^{\infty}$ a subsequence of $\left\{x_{i}\right\}_{i=m}^{\infty} .$

Example $\PageIndex{2}$

The sequence $\left\{x_{2 k}\right\}_{k=1}^{\infty}$ is a subsequence of the sequence $\left\{x_{i}\right\}_{i=1}^{\infty} .$ For example, $\left\{\frac{1}{2 i}\right\}_{i=1}^{\infty}$ is a subsequence of $\left\{\frac{1}{i}\right\}_{i=1}^{\infty}$.

Exercise $\PageIndex{22}$

Suppose $\left\{x_{i}\right\}_{i=m}^{\infty}$ converges with $\lim _{i \rightarrow \infty} x_{i}=L .$ Show that every subsequence $\left\{x_{n_{k}}\right\}_{k=1}^{\infty}$ of $\left\{x_{i}\right\}_{i=m}^{\infty}$ also converges and $\lim _{k \rightarrow \infty} x_{n_{k}}=L$.

Exercise $\PageIndex{23}$

Suppose $\left\{x_{i}\right\}_{i=m}^{\infty}$ diverges to $+\infty .$ Show that every subsequence $\left\{x_{n_{k}}\right\}_{k=1}^{\infty}$ of $\left\{x_{i}\right\}_{i=m}^{\infty}$ also diverges to $+\infty$.

Exercise $\PageIndex{24}$

Suppose $\left\{x_{i}\right\}_{i=m}^{\infty}$ diverges to $-\infty .$ Show that every subsequence $\left\{x_{n_{k}}\right\}_{k=1}^{\infty}$ of $\left\{x_{i}\right\}_{i=m}^{\infty}$ also diverges to $-\infty$.

Definition

Given a sequence $\left\{x_{i}\right\}_{i=m}^{\infty},$ we call any extended real number $\lambda$ which is the limit of a subsequence of $\left\{x_{i}\right\}_{i=m}^{\infty}$ a subsequential limit of $\left\{x_{i}\right\}_{i=m}^{\infty}$.

Example $\PageIndex{3}$

$-1$ and 1 are both subsequential limits of $\left\{(-1)^{i}\right\}_{i=0}^{\infty}$.

Exercise $\PageIndex{25}$

Suppose the sequence $\left\{x_{i}\right\}_{i=m}^{\infty}$ is not bounded. Show that either $-\infty$ or $+\infty$ is a subsequential limit of $\left\{x_{i}\right\}_{i=m}^{\infty} .$.

Proposition $\PageIndex{11}$

Suppose $\Lambda$ is the set of all subsequential limits of the sequence $\left\{x_{i}\right\}_{i=m}^{\infty} .$ Then $\Lambda \neq \emptyset$.

Proof

By the previous exercise, the proposition is true if $\left\{x_{i}\right\}_{i=m}^{\infty}$ is not bounded. So suppose $\left\{x_{i}\right\}_{i=m}^{\infty}$ is bounded and choose real numbers $a$ and $b$ such that $a \leq x_{i} \leq b$ for all $i \geq m$. Construct sequences $\left\{a_{i}\right\}_{i=1}^{\infty}$ and $\left\{b_{i}\right\}_{i=1}^{\infty}$ as follows: Let $a_{1}=a$ and $b_{1}=b$. For $i \geq 1,$ let

$$c=\frac{a_{i-1}+b_{i-1}}{2}.$$

If there exists an integer $N$ such that $a_{i-1} \leq x_{j} \leq c$ for all $j>N,$ let $a_{i}=a_{i-1}$ and $b_{i}=c ;$ otherwise, let $a_{i}=c$ and $b_{i}=b_{i-1} .$ Let $n_{1}=m$ and, for $k=2,3,4, \ldots,$ let $n_{k}$ be the smallest integer for which $n_{k}>n_{k-1}$ and $a_{k} \leq x_{n_{k}} \leq b_{k}$ Then $\left\{x_{n_{k}}\right\}_{k=1}^{\infty}$ is a Cauchy sequence which is a subsequence of $\left\{x_{i}\right\}_{i=m}^{\infty} .$ Thus $\left\{x_{n_{k}}\right\}_{k=1}^{\infty}$ converges and $\Lambda \neq 0 .$ $\quad$ Q.E.D.

Exercise $\PageIndex{26}$

Suppose $A \subset \mathbb{R}$ and $B \subset \mathbb{R}$ with $a \leq b$ for every $a \in A$ and $b \in B .$ Show that $\sup A \leq \inf B$.

Proposition $\PageIndex{12}$

Let $\Lambda$ be the set of subsequential limits of a sequence $\left\{x_{i}\right\}_{i=m}^{\infty} .$ Then

$$\limsup _{i \rightarrow \infty} x_{i}=\sup \Lambda .$$

Proof

Let $s=\sup \Lambda$ and, for $i \geq m, u_{i}=\sup \left\{x_{j}: j \geq i\right\} .$ Now since $x_{j} \leq u_{i}$ for all $j \geq i,$ it follows that $\lambda \leq u_{i}$ for every $\lambda \in \Lambda$ and $i \geq m .$ Hence, from the previous exercise, $s \leq \inf \left\{u_{i}: i \geq m\right\}=\limsup _{i \rightarrow \infty} x_{i}$.

Now suppose $s<\limsup _{i \rightarrow \infty} x_{i}$. Then there exists a real number $t$ such that $s<t<\lim \sup _{i \rightarrow \infty} x_{i} .$ In particular, $t<u_{i}$ for every $i \geq m .$ Let $n_{1}$ be the smallest integer for which $n_{1} \geq m$ and $x_{n_{1}}>t .$ For $k=2,3,4, \ldots,$ let $n_{k}$ be the smallest integer for which $n_{k}>n_{k-1}$ and $x_{n_{k}}>t .$ Then $\left\{x_{n_{k}}\right\}_{k=1}^{\infty}$ is a subsequence of $\left\{x_{i}\right\}_{i=m}^{\infty}$ which has a subsequential limit $\lambda \geq t$. Since $\lambda$ is also then a subsequential limit of $\left\{x_{i}\right\}_{i=m}^{\infty},$ we have $\lambda \in \Lambda$ and $\lambda \geq t>s,$ contradicting $s=\sup \Lambda .$ Hence we must have $\lim \sup _{i \rightarrow \infty} x_{i}=\sup \Lambda .$ $\quad$ Q.E.D.

Exercise $\PageIndex{27}$

Let $\Lambda$ be the set of subsequential limits of a sequence $\left\{x_{i}\right\}_{i=m}^{\infty} .$ Show that

$$\liminf _{i \rightarrow \infty} x_{i}=\inf \Lambda .$$