2.1: Convergence

Definition $\PageIndex{1}$

Let $\left\{a_{n}\right\}$ be a sequence of real numbers. We say that the sequence $\left\{a_{n}\right\}$ converges to $a \in \mathbb{R}$ if, for any $\varepsilon>0$, there exists a positive integer $N$ such that for any $n \in \mathbb{N}$ with $n \geq N$, one has

$$\left|a_{n}-a\right|<\varepsilon\left(\text { or equivalently }, a-\varepsilon<a_{n}<a+\varepsilon\right).$$

In this case, we call $a$ the limit of the sequence (see Theorem 2.1.3 below) and write $\lim _{n \rightarrow \infty} a_{n}=a$. If the sequence $\left\{a_{n}\right\}$ does not converge, we call the sequence divergent.

Remark $\PageIndex{1}$

It follows directly from the definition, using the Archimedean property, that a sequence $\left\{a_{n}\right\}$ converges to $a$ if and only if for any $\varepsilon>0$, there exists a real number $N$ such that for any $n \in \mathbb{N}$ with $n>N$, one has

$$\left|a_{n}-a\right|<\varepsilon. \nonumber$$

Example $\PageIndex{1}$

Let $a_{n}=\frac{1}{n}$ for $n \in \mathbb{N}$. We claim that $\lim _{n \rightarrow \infty} a_{n}=0$. We verify it using the definition.

Solution

Let $\varepsilon>0$. Choose an integer $N>1 / \varepsilon$. (Note that such an integer $N$ exists due to the Archimidean Property.) Then, if $n \geq N$, we get

$$\left|a_{n}-0\right|=\left|\frac{1}{n}\right|=\frac{1}{n} \leq \frac{1}{N}<\frac{1}{1 / \varepsilon}=\varepsilon. \nonumber$$

Example $\PageIndex{2}$

We now generalize the previous example as follows. Let $\alpha>0$ and consider the sequence given by

$$a_{n}=\frac{1}{n^{\alpha}} \text { for } n \in \mathbb{N}. \nonumber$$

Solution

Let $\varepsilon>0$. Choose an integer $N$ such that $N>\left(\frac{1}{\varepsilon}\right)^{1 / \alpha}$. For every $n \geq N$, one has $n>\left(\frac{1}{\varepsilon}\right)^{1 / \alpha}$ and, hence, $n^{\alpha}>\frac{1}{\varepsilon}$. This implies

$$\left|\frac{1}{n^{\alpha}}-0\right|=\frac{1}{n^{\alpha}}<\frac{1}{1 / \varepsilon}=\varepsilon.$$

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

Example $\PageIndex{3}$

Consider the sequence $\left\{a_{n}\right\}$ where

$$a_{n}=\frac{3 n^{2}+4}{2 n^{2}+n+5} \nonumber$$

We will prove directly from the definition that this sequence converges to $a= \frac{3}{2}$.

Solution

Let $\varepsilon>0$. We first search for a suitable $N$. To that end, we simplify and estimate the expression $\left|a_{n}-a\right|$. Notice that

$$\begin{aligned} \left|a_{n}-\frac{3}{2}\right| &=\left|\frac{3 n^{2}+4}{2 n^{2}+n+5}-\frac{3}{2}\right|=\left|\frac{2\left(3 n^{2}+4\right)-3\left(2 n^{2}+n+5\right)}{2\left(2 n^{2}+n+5\right)}\right|=\left|\frac{-7-3 n}{2\left(2 n^{2}+n+5\right)}\right| \\ &=\frac{3 n+7}{2\left(2 n^{2}+n+5\right)}<\frac{10 n}{4 n^{2}}=\frac{10}{4 n} \end{aligned}$$

To guarantee that the last expression is less than $\varepsilon$, it will suffice to choose $N>\frac{10}{4 \varepsilon}$. Indeed, if $n \geq N$, we get

$$\left|a_{n}-a\right| \leq \frac{10}{4 n} \leq \frac{10}{4 N}<\frac{10}{4 \frac{10}{4 \varepsilon}}=\varepsilon.$$

Example $\PageIndex{4}$

Let $\left\{a_{n}\right\}$ be given by

$$a_{n}=\frac{4 n^{2}-1}{3 n^{2}-n}. \nonumber$$

We claim $\lim _{n \rightarrow \infty} a_{n}=\frac{4}{3}$.

Solution

Let $\varepsilon>0$. We search for a suitable $N$. First notice that

$$\left|\frac{4 n^{2}-1}{3 n^{2}-n}-\frac{4}{3}\right|=\left|\frac{12 n^{2}-3-12 n^{2}+4 n}{3\left(3 n^{2}-n\right)}\right|=\left|\frac{4 n-3}{3\left(3 n^{2}-n\right)}\right|$$

Since $n \geq 1$, we have $n^{2} \geq n$ and $4n>3$. Therefore we have

$$\left|\frac{4 n^{2}-1}{3 n^{2}-n}-\frac{4}{3}\right|=\frac{4 n-3}{3\left(3 n^{2}-n\right)} \leq \frac{4 n-3}{3\left(3 n^{2}-n^{2}\right)}<\frac{4 n}{6 n^{2}}=\frac{4}{6 n}.$$

Thus, if $N>\frac{4}{6 \varepsilon}$, we have, for $n \geq N$

$$\left|\frac{4 n^{2}-1}{3 n^{2}-n}-\frac{4}{3}\right| \leq \frac{4}{6 n} \leq \frac{4}{6 N}<\varepsilon.$$

Example $\PageIndex{5}$

Consider the sequence given by

$$a_{n}=\frac{n^{2}+5}{4 n^{2}+n}.$$

We prove directly from the definition that $\left\{a_{n}\right\}$ converges to $\frac{1}{4}$.

Solution

Let $\varepsilon>0$. Now,

$$\left|\frac{n^{2}+5}{4 n^{2}+n}-\frac{1}{4}\right|=\left|\frac{4 n^{2}+20-4 n^{2}-n}{4\left(4 n^{2}+n\right)}\right|=\frac{|20-n|}{4\left(4 n^{2}+n\right)}.$$

If $n \geq 20$, then $|20-n|=n-20$. Therfore, for such $n$ we have

$$\left|\frac{n^{2}+5}{4 n^{2}+n}-\frac{1}{4}\right|=\frac{n-20}{4\left(4 n^{2}+n\right)} \leq \frac{n}{16 n^{2}}=\frac{1}{16 n}.$$

Choose $N>\max \left\{\frac{1}{16 \varepsilon}, 20\right\}$. Then, for $n \geq N$ we get

$$\left|\frac{n^{2}+5}{4 n^{2}+n}-\frac{1}{4}\right| \leq \frac{1}{16 n} \leq \frac{1}{16 N}<\varepsilon.$$

The following result is quite useful in proving certain inequalities between numbers.

Lemma $\PageIndex{2}$

Let $\ell \geq 0$, If $\ell<\varepsilon$ for all $\varepsilon>0$, then $\ell =0$.

Proof

This is easily proved by contraposition. If $\ell >0$, then there is a positive number, for example $\varepsilon=\ell / 2$, such that $\varepsilon<\ell$. $\square$

Theorem $\PageIndex{3}$

A convergent sequence $\left\{a_{n}\right\}$ has at most one limit

Proof

Suppose $\left\{a_{n}\right\}$ converges to $a$ and $b$. Then given $\varepsilon>0$, there exist positive integers $N_{1}$ and $N_{2}$ such that

$$\left|a_{n}-a\right|<\varepsilon / 2 \text { for all } n \geq N_{1}$$

and

$$\left|a_{n}-b\right|<\varepsilon / 2 \text { for all } n \geq N_{2}.$$

Let $N=\max \left\{N_{1}, N_{2}\right\}$. Then

$$|a-b| \leq\left|a-a_{N}\right|+\left|a_{N}-b\right|<\varepsilon / 2+\varepsilon / 2=\varepsilon.$$

Since $\varepsilon>0$ is arbitrary, by Lemma 2.1.2, $|a-b|=0$ and, hence, $a=b$. $\square$

The following lemma is simple generalization of (2.1.2).

Lemma $\PageIndex{4}$

Given real numbers $a,b$, then $a \leq b$ if and only if $a<b+\varepsilon$ for all $\varepsilon>0$.

Proof

Suppose $a<b+\varepsilon$ for all $\varepsilon>0$. And suppose, by way of contradiction, that $a>b$. then set $\varepsilon_{0}=a-b$. Then $\varepsilon_{0}>0$. By assumption, we should have $a<b+\varepsilon_{0}=b+a-b=a$, which is a contradiction. It follows that $a \leq b$.

The other direction follows immediately from the order axioms. $\square$

The following comparison theorem shows that (non-strict) inequalities are preserved "in the limit".

Theorem $\PageIndex{5}$ - Comparison Theorem.

Suppose $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ converge to $a$ and $b$, respectively, and $a_{n} \leq b_{n}$ for all $n \in \mathbb{N}$. Then $a \leq b$.

Proof

For any $\varepsilon>0$, there exist $N_{1}, N_{2} \in \mathbb{N}$ such that

$$a-\frac{\varepsilon}{2}<a_{n}<a+\frac{\varepsilon}{2}, \quad \text { for } n \geq N_{1},$$

$$b-\frac{\varepsilon}{2}<b_{n}<b+\frac{\varepsilon}{2}, \quad \text { for } n \geq N_{2}.$$

Choose $N>\max \left\{N_{1}, N_{2}\right\}$. Then

$$a-\frac{\varepsilon}{2}<a_{N} \leq b_{N}<b+\frac{\varepsilon}{2}.$$

Thus, $a<b+\varepsilon$ for any $\varepsilon>0$. Using Lemma 2.1.4 we conclude $a \leq b$. $\square$

Theorem $\PageIndex{6}$ - The Squeeze Theorem.

Suppose the sequences $\left\{a_{n}\right\}$, $\left\{b_{n}\right\}$, and $\left\{c_{n}\right\}$ satisfy

$$a_{n} \leq b_{n} \leq c_{n} \text { for all } n \in \mathbb{N},$$

and $\lim _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty} c_{n}=\ell$. Then $\lim _{n \rightarrow \infty} b_{n}=\ell$.

Proof

Fix any $\varepsilon>0$. Since $\lim _{n \rightarrow \infty} a_{n}=\ell$, there exists $N_{1} \in \mathbb{N}$ such that

$$\ell-\varepsilon<a_{n}<\ell+\varepsilon \nonumber$$

for all $n \geq N_{1}$. Similarly, since $\displaystyle \lim _{n \rightarrow \infty} c_{n}=\ell$, there exists $N_{2} \in \mathbb{N}$ such that

$$\ell-\varepsilon<c_{n}<\ell+\varepsilon \nonumber$$

for all $n \geq N_{2}$. Let $N=\max \left\{N_{1}, N_{2}\right\}$. Then, for $n \geq N$, we have

$$\ell-\varepsilon<a_{n} \leq b_{n} \leq c_{n}<\ell+\varepsilon, \nonumber$$

which implies $\left|b_{n}-\ell\right|<\varepsilon$. Therefore, $\lim _{n \rightarrow \infty} b_{n}=\ell$. $\square$

Definition $\PageIndex{2}$

A sequence $\left\{a_{n}\right\}$ is bounded above if the set $\left\{a_{n}: n \in \mathbb{N}\right\}$ is bounded above. Similarly, the sequence \left\{a_{n}: n \in \mathbb{N}\right\} is bounded below if the set $\left\{a_{n}: n \in \mathbb{N}\right\}$ is bounded below. We say that the sequence $\left\{a_{n}\right\}$ is bounded if the set $\left\{a_{n}: n \in \mathbb{N}\right\}$ is bounded, that is, if it is both bounded above and bounded below.

It follows from the observation after Definition 1.5.1 that the sequence $\left\{a_{n}\right\}$ is bounded if and only if there is $M \in \mathbb{R}$ such that $\left|a_{n}\right| \leq M$ for all $n \in \mathbb{N}$.

Theorem $\PageIndex{7}$

A convergent sequence is bounded.

Proof

Suppose the sequence $\left\{a_{n}\right\}$ converges to $a$. Then, for $\varepsilon=1$, there exists $N \in \mathbb{N}$ such that

$$\left|a_{n}-a\right|<1 \text { for all } n \geq N.$$

Since $\left|a_{n}\right|-|a| \leq \| a_{n}|-| a|| \leq\left|a_{n}-a\right|$, this implies $\left|a_{n}\right|<1+|a|$ for all $n \geq N$. Set

$$M=\max \left\{\left|a_{1}\right|, \ldots,\left|a_{N-1}\right|,|a|+1\right\}$$

Then $\left|a_{n}\right| \leq M$ for all $n \in \mathbb{N}$. Therefore, $\left\{a_{n}\right\}$ is bounded. $\square$

Definition $\PageIndex{3}$

Let $\left\{a_{n}\right\}_{n=1}^{\infty}$ be a sequence of real numbers. The sequence $\left\{b_{n}\right\}_{n=1}^{\infty}$ is called a subsequence of $\left\{a_{n}\right\}_{n=1}^{\infty}$ if there exists a sequence of increasing positive integers

$$n_{1}<n_{2}<n_{3}<\cdots,$$

such that $b_{k}=a_{n_{k}}$ for each $k \in \mathbb{N}$.

Example $\PageIndex{6}$

Consider the sequence $a_{n}=(-1)^{n}$ for $n \in \mathbb{N}$.

Solution

Then $\left\{a_{2 k}\right\}$ is a subsequence of $\left\{a_{n}\right\}$ and $a_{2 k}=1$ for all $k$ (here $n_{k}=2 k$ for all $k$). Similarly, $\left\{a_{2 k+1}\right\}$ is also a subsequence of $\left\{a_{n}\right\}$ and $a_{2 k+1}=-1$ for all $k$ (here $n_{k}=2 k +1$ for all $k$).

Lemma $\PageIndex{8}$

Let $\left\{n_{k}\right\}_{k}$ be a sequence of positive integers with

$$n_{1}<n_{2}<n_{3}<\cdots$$

Then $n_{k} \geq k$ for all $k \in \mathbb{N}$.

Proof

We use mathematical induction. When $k=1$, it is clear that $n_{1} \geq 1$ since $n_{1}$ is a positive integer. Assume $n_{k} \geq k$ for some $k$. Now $n_{k+1}>n_{k}$ and, since $n_{k}$ and $n_{k+1}$ are integers, this implies, $n_{k+1} \geq n_{k}+1$. Therefore, $n_{k+1} \geq k+1$ by the inductive hyposthesis. The conclusion now follows by the principle of mathematical induction. $\square$

Theorem $\PageIndex{9}$

If a sequence $\left\{a_{n}\right\}$ converges to $a$, then any subsequence $\left\{a_{n_{k}}\right\}$ of $\left\{a_{n}\right\}$ also converges to $a$.

Proof

Suppose $\left\{a_{n}\right\}$ converges to $a$ and let $\varepsilon>0$ be given. Then there exists $N$ such that

$$\left|a_{n}-a\right|<\varepsilon \text { for all } n \geq N.$$

For any $k \geq N$, since $n_{k} \geq k$, we also have

$$\left|a_{n_{k}}-a\right|<\varepsilon.$$

Thus, $\left\{a_{n_{k}}\right\}$ converges to $a$ as $k \rightarrow \infty$. $\square$

Example $\PageIndex{1}$

Let $a_{n}=(-1)^{n}$ for $n \in \mathbb{N}$.

Solution

Then the sequence $\left\{a_{n}\right\}$ is divergent. Indeed suppose by contradiction that

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

Then every subsequence of $\left\{a_{n}\right\}$ converges to a number $\ell \in \mathbb{R}$. From the previous theorem, it follows, in particular, that

$\ell=\lim _{k \rightarrow \infty} a_{2 k}=1 \text { and } \ell=\lim _{k \rightarrow \infty} a_{2 k+1}=-1$.

This contradiction shows that the sequence is divergent.

Since the sequence $\left\{a_{n}\right\}$ is bounded but not convergent, this example illustrates the fact that the converse of theorem 2.1.7 is not true.

Remark $\PageIndex{10}$

Given a positive integer $k_{0}$, it will be convenient to also talk about the sequence $\left\{a_{n}\right\}_{n \geq k_{0}}$, that is, a function defined only for the integers greater than or equal to $k_{0}$. For simplicity of notation, we may also denote this sequence by $\left\{a_{n}\right\}$ whenever the integer $k_{0}$ is clear form the context. For instance, we talk of the sequence $\left\{a_{n}\right\}$ given by

$$a_{n}=\frac{n+1}{(n-1)(n-2)}.$$

although $a_{1}$ and $a_{2}$ are not defined. In all cases, the sequence must be defined from some integer onwards.

Exercise $\PageIndex{1}$

Prove the following directly from the definition of limit.

1. $\lim _{n \rightarrow \infty} \frac{2 n^{2}+2}{3 n^{3}+1}=0$.
2. $\lim _{n \rightarrow \infty} \frac{n^{2}+1}{5 n^{2}+n+1}=\frac{1}{5}$.
3. $\lim _{n \rightarrow \infty} \frac{2 n^{3}+1}{4 n^{3}-n}=\frac{1}{2}$.
4. $\lim _{n \rightarrow \infty} \frac{3 n^{2}+5}{6 n^{2}+n}=\frac{1}{2}$.
5. $\lim _{n \rightarrow \infty} \frac{4 n^{2}-1}{n^{2}-n}=4$.

Answer

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Exercise $\PageIndex{2}$

Prove that if $\left\{a_{n}\right\}$ is a convergent sequence, then $\left\{|a_{n}|\right\}$ is a convergent sequence. Is the converse true?

Answer

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Exercise $\PageIndex{3}$

Let $\left\{a_{n}\right\}$ be a sequence. Prove that if the $\left\{|a_{n}|\right\}$ converges to 0, then $\left\{a_{n}\right\}$ also converges to 0.

Answer

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Exercise $\PageIndex{4}$

Prove that $\lim _{n \rightarrow \infty} \frac{\sin n}{n}=0$.

Answer

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Exercise $\PageIndex{5}$

Let $\left\{x_{n}\right\}$ be a bounded sequence and let $\left\{y_{n}\right\}$ be a sequence that converges to 0. Prove that the sequence $\left\{x_{n} y_{n}\right\}$ converges to 0.

Answer

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Exercise $\PageIndex{6}$

Prove that the following limits are 0. (Hint: use Theorem 2.1.6.)

1. $\lim _{n \rightarrow \infty} \frac{n+\cos \left(n^{2}-3\right)}{2 n^{2}+1}$.
2. $\lim _{n \rightarrow \infty} \frac{3^{n}}{n !}$.
3. $\lim _{n \rightarrow \infty} \frac{n !}{n^{n}}$.
4. $\lim _{n \rightarrow \infty} \frac{n^{2}}{3^{n}}$. (Hint: see Exercise 1.3.4(c)).

Answer

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Exercise $\PageIndex{7}$

Prove that if $\lim _{n \rightarrow \infty} a_{n}=\ell>0$, then there exists $N \in \mathbb{N}$ such that $a_{n}>0$ for all $n \geq N$.

Answer

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Exercise $\PageIndex{8}$

Prove that if $\lim _{n \rightarrow \infty} a_{n}=\ell \neq 0$, then $\lim _{n \rightarrow \infty} \frac{a_{n+1}}{a_{n}}=1$. Is the conclusion still true if $\ell =0$?

Answer

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Exercise $\PageIndex{9}$

Let $\left\{a_{n}\right\}$ be a sequence of real numbers such that $\lim _{n \rightarrow \infty} a_{n}=3$. Use Definition 2.1.1 to prove the following

1. $\lim _{n \rightarrow \infty} 3 a_{n}-7=2$;
2. $\lim _{n \rightarrow \infty} \frac{a_{n}+1}{a_{n}}=\frac{4}{3}$; (Hint: prove first that there is $N$ such that $a_{n}>1$ for $n \geq N$.)

Answer

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Exercise $\PageIndex{10}$

Let $a_{n} \geq 0$ for all $n \in \mathbb{N}$. Prove that if $\lim _{n \rightarrow \infty} a_{n}=\ell$, then $\lim _{n \rightarrow \infty} \sqrt{a_{n}}=\sqrt{\ell}$.

Answer

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Exercise $\PageIndex{11}$

Prove that the sequence $\left\{a_{n}\right\}$ with $a_{n}=\sin (n \pi / 2)$ is divergent.

Answer

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Exercise $\PageIndex{12}$

Consider a sequence $\left\{a_{n}\right\}$.

1. Prove that $\lim _{n \rightarrow \infty} a_{n}=\ell$ if and only if $\lim _{k \rightarrow \infty} a_{2 k}=\ell$ and $\lim _{k \rightarrow \infty} a_{2 k+1}=\ell$.
2. Prove that $\lim _{n \rightarrow \infty} a_{n}=\ell$ if and only if $\lim _{k \rightarrow \infty} a_{3 k}=\ell$, $\lim _{k \rightarrow \infty} a_{3 k+1}=\ell$, and $\lim _{k \rightarrow \infty} a_{3 k+2}=\ell$.

Answer

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Exercise $\PageIndex{13}$

Given a sequence $\left\{a_{n}\right\}$, define a new sequence $\left\{b_{n}\right\}$ by

$$b_{n}=\frac{a_{1}+a_{2}+\ldots+a_{n}}{n}.$$

1. Prove that if $\lim _{n \rightarrow \infty} a_{n}=\ell$, then $\lim _{n \rightarrow \infty} b_{n}=\ell$
2. Find a counterexample to show that the converse does not hold in general.

Answer

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