# 2.4 Properties of convergent sequences

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A discussion of the properties of convergent sequences, such as when they form a bounded set, the squeeze theorem, and the properties of convergence when sequences are combined algebraically.

Often, our goal is to show that a given sequence is convergent. However, as we study convergent sequences, we would liketo establish various properties that they have in common. The first theorem of this section is just such a result.

Suppose $\left\{{a}_{n}\right\}$ is a convergent sequence of real or complex numbers. Then the sequence $\left\{{a}_{n}\right\}$ forms a bounded set.

Write $L=lim{a}_{n}.$ Let $ϵ$ be the positive number $1.$ Then, there exists a natural number $N$ such that $|{a}_{n}-L|<1$ for all $n\ge N.$ By the backward triangle inequality, this implies that $||{a}_{n}|-|L||<1$ for all $n\ge N,$ which implies that $|{a}_{n}|\le |L|+1$ for all $n\ge N.$ This shows that at least the tail of the sequence is bounded by the constant $|L|+1.$

Next, let $K$ be a number larger than the finitely many numbers $|{a}_{1}|,...,|{a}_{N-1}|.$ Then, for any $n,$ $|{a}_{n}|$ is either less than $K$ or $|L|+1.$ Let $M$ be the larger of the two numbers $K$ and $|L|+1.$ Then $|{a}_{n}| for all $n.$ Hence, the sequence $\left\{{a}_{n}\right\}$ is bounded.

Note that the preceding theorem is a partial converse to [link] ; i.e., a convergent sequence is necessarily bounded. Of course, not every convergent sequence must beeither nondecreasing or nonincreasing, so that a full converse to [link] is not true. For instance, take $z=-1/2$ in part (1) of [link] . It converges to 0 all right, but it is neither nondecreasing nor nonincreasing.

1. Suppose $\left\{{a}_{n}\right\}$ is a sequence of real numbers that converges to a number $a,$ and assume that ${a}_{n}\ge c$ for all $n.$ Prove that $a\ge c.$ HINT: Suppose not, and let $ϵ$ be the positive number $c-a.$ Let $N$ be a natural number corresponding to this choice of $ϵ,$ and derive a contradiction.
2. If $\left\{{a}_{n}\right\}$ is a sequence of real numbers for which $lim{a}_{n}=a,$ and if $a\ne 0,$ then prove that ${a}_{n}\ne 0$ for all large enough $n.$ Show in fact that there exists an $N$ such that $|{a}_{n}|>|a|/2$ for all $n\ge N.$ HINT: Make use of the positive number $ϵ=|a|/2.$
1. If $\left\{{a}_{n}\right\}$ is a sequence of positive real numbers for which $lim{a}_{n}=a>0,$ prove that $lim\sqrt{{a}_{n}}=\sqrt{a}.$ HINT: Multiply the expression $\sqrt{{a}_{n}}-\sqrt{a}$ above and below by $\sqrt{{a}_{n}}+\sqrt{a}.$
2. If $\left\{{a}_{n}\right\}$ is a sequence of complex numbers, and $lim{a}_{n}=a,$ prove that $lim|{a}_{n}|=|a|.$ HINT: Use the backward triangle inequality.

Suppose $\left\{{a}_{n}\right\}$ is a sequence of real numbers and that $L=lim{a}_{n}.$ Let ${M}_{1}$ and ${M}_{2}$ be real numbers such that ${M}_{1}\le {a}_{n}\le {M}_{2}$ for all $n.$ Prove that ${M}_{1}\le L\le {M}_{2}.$

HINT: Suppose, for instance, that $L>{M}_{2}.$ Make use of the positive number $L-{M}_{2}$ to derive a contradiction.

We are often able to show that a sequence converges by comparing it to another sequence that we already know converges.The following exercise demonstrates some of these techniques.

Let $\left\{{a}_{n}\right\}$ be a sequence of complex numbers.

1. Suppose that, for each $n,$ $|{a}_{n}|<1/n.$ Prove that $0=lim{a}_{n}.$
2. Suppose $\left\{{b}_{n}\right\}$ is a sequence that converges to $0,$ and suppose that, for each $n,$ $|{a}_{n}|<|{b}_{n}|.$ Prove that $0=lim{a}_{n}.$

The next result is perhaps the most powerful techniquewe have for showing that a given sequence converges to a given number.

## Squeeze theorem

Suppose that $\left\{{a}_{n}\right\}$ is a sequence of real numbers and that $\left\{{b}_{n}\right\}$ and $\left\{{c}_{n}\right\}$ are two sequences ofreal numbers for which ${b}_{n}\le {a}_{n}\le {c}_{n}$ for all $n.$ Suppose further that $lim{b}_{n}=lim{c}_{n}=L.$ Then the sequence $\left\{{a}_{n}\right\}$ also converges to $L.$

We examine the quantity $|{a}_{n}-L,|$ employ some add and subtract tricks, and make the following computations:

$\begin{array}{ccc}\hfill rcl|{a}_{n}-L|& \le & |{a}_{n}-{b}_{n}+{b}_{n}-L|\hfill \\ & \le & |{a}_{n}-{b}_{n}|+|{b}_{n}-L|\hfill \\ & =& {a}_{n}-{b}_{n}+|{b}_{n}-L|\hfill \\ & \le & {c}_{n}-{b}_{n}+|{b}_{n}-L|\hfill \\ & =& |{c}_{n}-{b}_{n}|+|{b}_{n}-L|\hfill \\ & \le & |{c}_{n}-L|+|L-{b}_{n}|+|{b}_{n}-L|.\hfill \end{array}$

So, we can make $|{a}_{n}-L|<ϵ$ by making $|{c}_{n}-L|<ϵ/3$ and $|{b}_{n}-L|<ϵ/3.$ So, let ${N}_{1}$ be a positive integer such that $|{c}_{n}-L|<ϵ/3$ if $n\ge {N}_{1},$ and let ${N}_{2}$ be a positive integer so that $|{b}_{n}-L|<ϵ/3$ if $n\ge {N}_{2}.$ Then set $N=max\left({N}_{1},{N}_{2}\right).$ Clearly, if $n\ge N,$ then both inequalities $|{c}_{n}-L|<ϵ/3$ and $|{b}_{n}-L|<ϵ/3,$ and hence $|{a}_{n}-L|<ϵ.$ This finishes the proof.

The next result establishes what are frequently called the “limit theorems.” Basically, these results show how convergenceinteracts with algebraic operations.

Let $\left\{{a}_{n}\right\}$ and $\left\{{b}_{n}\right\}$ be two sequences of complex numbers with $a=lim{a}_{n}$ and $b=lim{b}_{n}.$ Then

1. The sequence $\left\{{a}_{n}+{b}_{n}\right\}$ converges, and
$lim\left({a}_{n}+{b}_{n}\right)=lim{a}_{n}+lim{b}_{n}=a+b.$
2. The sequence $\left\{{a}_{n}{b}_{n}\right\}$ is convergent, and
$lim\left({a}_{n}{b}_{n}\right)=lim{a}_{n}lim{b}_{n}=ab.$
3. If all the ${b}_{n}$ 's as well as $b$ are nonzero, then the sequence $\left\{{a}_{n}/{b}_{n}\right\}$ is convergent, and
$lim\left(\frac{{a}_{n}}{{b}_{n}}=\frac{lim{a}_{n}}{lim{b}_{n}}=\frac{a}{b}.$

Part (1) is exactly the same as [link] . Let us prove part (2).

By [link] , both sequences $\left\{{a}_{n}\right\}$ and $\left\{{b}_{n}\right\}$ are bounded. Therefore, let $M$ be a number such that $|{a}_{n}|\le M$ and $|{b}_{n}|\le M$ for all $n.$ Now, let $ϵ>0$ be given. There exists an ${N}_{1}$ such that $|{a}_{n}-a|<ϵ/\left(2M\right)$ whenever $n\ge {N}_{1},$ and there exists an ${N}_{2}$ such that $|{b}_{n}-b|<ϵ/\left(2M\right)$ whenever $n\ge {N}_{2}.$ Let $N$ be the maximum of ${N}_{1}$ and ${N}_{2}.$ Here comes the add and subtract trick again.

$\begin{array}{ccc}\hfill rcl|{a}_{n}{b}_{n}-ab|& =& |{a}_{n}{b}_{n}-a{b}_{n}+a{b}_{n}-ab|\hfill \\ & \le & |{a}_{n}{b}_{n}-a{b}_{n}|+|a{b}_{n}-ab|\hfill \\ & =& |{a}_{n}-a||{b}_{n}|+|a||b-{b}_{n}|\hfill \\ & \le & |{a}_{n}-a|M+M|{b}_{n}-b|\hfill \\ & <& ϵ\hfill \end{array}$

if $n\ge N,$ which shows that $lim\left({a}_{n}{b}_{n}\right)=ab.$

To prove part (3), let $M$ be as in the previous paragraph, and let $ϵ>0$ be given. There exists an ${N}_{1}$ such that $|{a}_{n}-{a|<\left(ϵ|b|}^{2}\right)/\left(4M\right)$ whenever $n\ge {N}_{1};$ there also exists an ${N}_{2}$ such that $|{b}_{n}-{b|<\left(ϵ|b|}^{2}\right)/\left(4M\right)$ whenever $n\ge {N}_{2};$ and there exists an ${N}_{3}$ such that $|{b}_{n}|>|b|/2$ whenever $n\ge {N}_{3}.$ (See [link] .) Let $N$ be the maximum of the three numbers ${N}_{1},{N}_{2}$ and ${N}_{3}.$ Then:

$\begin{array}{ccc}\hfill rcl|\frac{{a}_{n}}{{b}_{n}}-\frac{a}{b}|& =& |\frac{{a}_{n}b-{b}_{n}a}{{b}_{n}b}|\hfill \\ & =& |{a}_{n}b-{b}_{n}a|\frac{1}{|{b}_{n}b|}\hfill \\ & <& |{a}_{n}b-{b}_{n}a|\frac{1}{{|b|}^{2}/2}\hfill \\ & \le & \left(|{a}_{n}-a||b|+|a||{b}_{n}-b|\right)\frac{2}{{|b|}^{2}}\hfill \\ & <& \left(M|{a}_{n}-a|+M|{b}_{n}-b|\right)\frac{2}{{|b|}^{2}}\hfill \\ & <& ϵ\hfill \end{array}$

if $n\ge N.$ This completes the proof.

REMARK The proof of part (3) of the preceding theorem may look mysterious. Where, for instance, does this number ${ϵ|b|}^{2}/4M$ come from? The answer is that one begins such a proof by examining the quantity $|{a}_{n}/{b}_{n}-a/b|$ to see if by some algebraic manipulation one can discover how to control its size by using the quantities $|{a}_{n}-a|$ and $|{b}_{n}-b|.$ The assumption that $a=lim{a}_{n}$ and $b=lim{b}_{n}$ mean exactly that the quantities $|{a}_{n}-a|$ and $|{b}_{n}-b|$ can be controlled by requiring $n$ to be large enough. The algebraic computation in the proof above shows that

$|\frac{{a}_{n}}{{b}_{n}}-\frac{a}{b}|\le \left(M|{a}_{n}-a|+M|{b}_{n}-b|\right)\frac{2}{{|b|}^{2}},$

and one can then see exactly how small to make $|{a}_{n}-a|$ and $|{b}_{n}-b|$ so that $|{a}_{n}/{b}_{n}-a/b|<ϵ.$ Indeed, this is the way most limit proofs work.

If possible, determine the limits of the following sequences by using [link] , [link] , [link] , and the squeeze theorem [link] .

1. $\left\{{n}^{1/{n}^{2}}\right\}.$
2. $\left\{{\left({n}^{2}\right)}^{1/n}\right\}.$
3. $\left\{{\left(1+n\right)}^{1/n}\right\}.$
4. $\left\{{\left(1+{n}^{2}\right)}^{1/{n}^{3}}\right\}.$
5. $\left\{{\left(1+1/n\right)}^{2/n}\right\}.$
6. $\left\{{\left(1+1/n\right)}^{2n}\right\}.$
7. $\left\{{\left(1+1/n\right)}^{{n}^{2}}\right\}.$
8. $\left\{{\left(1-1/n\right)}^{n}\right\}.$ HINT: Note that
$1-1/n=\frac{n-1}{n}=\frac{1}{\frac{n}{n-1}}=\frac{1}{\frac{n-1+1}{n-1}}=\frac{1}{1+\frac{1}{n-1}}.$
9. $\left\{{\left(1-1/\left(2n\right)\right)}^{3n}\right\}.$
10. $\left\{{\left(n!\right)}^{1/n}\right\}.$

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