
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 $$x$$.

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

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

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