4.2 Properties of Limits

Now that we have established how to prove the existence of limits, we will consider the uniqueness of them, as well as some algebraic properties of convergent sequences.

Theorem 4.2

If a sequence converges, its limit is unique.

Proof.

Let $(x_{n})_{n}$ be a convergent sequence and suppose that $(x_{n})_{n}\to x$ and $(x_{n})_{n}\to y$ . Let $\varepsilon>0$ . Then there exists $N_{1}\in\mathbb{N}$ such that $n\geq N_{1}\implies\lvert x_{n}-x\rvert<\varepsilon$ , but there also exists $N_{2}\in\mathbb{N}$ such that $n\geq N_{2}\implies\lvert x_{n}-y\rvert<\varepsilon$ . Now,

$\displaystyle\lvert x-y\rvert$	$\displaystyle=\lvert x-x_{n}+x_{n}-y\rvert$	(for $n\geq\max\{N_{1},N_{2}\}$ )
	$\displaystyle\leq\lvert x-x_{n}\rvert+\lvert x_{n}-y\rvert$	(4.14)
	$\displaystyle<\varepsilon+\varepsilon=2\varepsilon.$	(4.15)

Therefore, $\lvert x-y\rvert=0$ , so $x=y$ (since $\forall\varepsilon>0$ , $0\leq a<\varepsilon\implies a=0\ \forall a\in\mathbb{R}$ ). Thus, the two limits are in fact the same. ∎

The next theorem will serve as a stepping stone that will make some more useful theorems much easier to prove. First, recall that a set is bounded if and only if it is bounded both above and below. We extend this definition to the elements of sequences, saying that a sequence is bounded if there exists $M\geq 0$ such that $\lvert x_{n}\rvert\leq M\ \forall n\in\mathbb{N}$ . This is a very quick and easy way to establish the range of a sequence that will come in handy later.

Theorem 4.3

Every convergent sequence is bounded¹¹ 1 Note that the converse of this theorem is obviously not true, $((-1)^{n})_{n}$ is bounded, but does not converge..

Proof.

Let $(x_{n})_{n}$ be a convergent sequence with $x_{n}\to x$ . Then there exists $N\in\mathbb{N}$ such that $n\geq N\implies\lvert x_{n}-x\rvert<1$ . Consider the terms in the sequence $x_{1},x_{2},\dots,x_{N-1}$ and let $M=\max\{\lvert x_{1}\rvert,\lvert x_{2}\rvert,\dots,\lvert x_{N-1}\rvert\}$ . Now, for all $n\geq N$ , $\lvert x_{n}\rvert\leq\max\{M,\lvert x\rvert+1\}$ (since $\lvert x_{n}-x\rvert\geq\lvert x_{n}\rvert-\lvert x\rvert<1\implies\lvert x_{n% }\rvert<1+\lvert x\rvert$ ), which is a constant, hence $(x_{n})_{n}$ is bounded. ∎

4.2.1 Adding and Multiplying Sequences

Now that we have this tool, we can proceed to some more interesting results about how limits of sequences behave when we add and multiply sequences together.

Theorem 4.4

Let $a\in\mathbb{R}$ , $(x_{n})_{n}$ and $(y_{n})_{n}$ converging sequences with $x_{n}\to x$ and $y_{n}\to y$ .

(i)

$x_{n}+y_{n}\to x+y$
(ii)

$x_{n}y_{n}\to xy$
(iii)

If $\forall n,y_{n}\neq 0$ and $y\neq 0$ , then $\frac{1}{y_{n}}\to\frac{1}{y}$
(iv)

If $\forall n,y_{n}\neq 0$ and $y\neq 0$ , then $\frac{x_{n}}{y_{n}}\to\frac{x}{y}$
(v)

$ax_{n}\to ax$
(vi)

If $\forall n,x_{n}\leq y_{n}$ , then $x\leq y$ .

Proof.

For all parts of this proof, begin by letting $\varepsilon>0$ .

(i)

Since $x_{n}\to x,\ \exists N_{1}\in\mathbb{N}$ such that $n\geq N_{1}\implies\lvert x_{n}-x\rvert<\frac{\varepsilon}{2}$ and since $y_{n}\to y,\ \exists N_{2}\in\mathbb{N}$ such that $n\geq N_{2}\implies\lvert y_{n}-y\rvert<\frac{\varepsilon}{2}$ . Choose $N=\max\{N_{1},N_{2}\}$ , now for $n\geq N$ we have

$\displaystyle\lvert(x_{n}+y_{n})-(x+y)\rvert$ $\displaystyle=\lvert(x_{n}-x)+(y_{n}-y)\rvert$ (4.16)

$\displaystyle\leq\lvert x_{n}-x\rvert+\lvert y_{n}-y\rvert$ (4.17)

$\displaystyle<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon.$ (4.18)

(ii)

Since $(x_{n})_{n}$ is bounded, $\exists M\geq 0$ such that $\lvert x_{n}\rvert\leq M\ \forall n\in\mathbb{N}$ . Since $x_{n}\to x,\ \exists N_{1}\in\mathbb{N}$ such that $n\geq N_{1}\implies\lvert x_{n}-x\rvert<\frac{\varepsilon}{2M}$ and since $y_{n}\to y,\ \exists N_{2}\in\mathbb{N}$ such that $n\geq N_{2}\implies\lvert y_{n}-y\rvert<\frac{\varepsilon}{2\lvert y\rvert}$ . Now, choosing $N=\max\{N_{1},N_{2}\}$ , for $n\geq N$ we have

$\displaystyle\lvert x_{n}y_{n}-xy\rvert$	$\displaystyle=\lvert x_{n}y_{n}-x_{n}y+x_{n}y-xy\rvert$	(4.19)
	$\displaystyle=\lvert x_{n}(y_{n}-y)+y(x_{n}-x)\rvert$	(4.20)
	$\displaystyle\leq\lvert x_{n}\rvert\lvert y_{n}-y\rvert+\lvert y\rvert\lvert x% _{n}-x\rvert$	(4.21)
	$\displaystyle<\lvert x_{n}\rvert\frac{\varepsilon}{2M}+\lvert y\rvert\frac{% \varepsilon}{2\lvert y\rvert}$	(4.22)
	$\displaystyle\leq M\frac{\varepsilon}{2M}+\lvert y\rvert\frac{\varepsilon}{2% \lvert y\rvert}=\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon.$	(4.23)

(iii)

Since $y_{n}\to y,\ \exists N_{1}\in\mathbb{N}$ such that $n\geq N_{1}\implies\lvert y_{n}-y\rvert<\frac{\lvert y\rvert}{2}$ . Then for $n\geq N_{1}$

$\lvert y\rvert=\lvert y_{n}+(y-y_{n})\rvert\leq\lvert y_{n}\rvert+\lvert y_{n}% -y\rvert<\lvert y_{n}\rvert+\frac{\lvert y\rvert}{2},$ (4.24)

so $\lvert y_{n}\rvert>\frac{\lvert y\rvert}{2}$ . Now, let $N_{2}$ be such that $n\geq N_{2}\implies\lvert y_{n}-y\rvert<\frac{\varepsilon\lvert y\rvert^{2}}{2}$ , then for $n\geq\max\{N_{1},N_{2}\}$ we have

$\displaystyle\left\lvert\frac{1}{y_{n}}-\frac{1}{y}\right\rvert$ $\displaystyle=\left\lvert\frac{y-y_{n}}{y_{n}y}\right\rvert=\frac{\lvert y_{n}% -y\rvert}{\lvert y_{n}\rvert\lvert y\rvert}$ (4.25)

$\displaystyle<\frac{\varepsilon\lvert y\rvert^{2}}{2\lvert y_{n}\rvert\lvert y% \rvert}<\frac{2\varepsilon\lvert y\rvert^{2}}{2\lvert y\rvert^{2}}=\varepsilon.$ (4.26)
(iv)

Write $\frac{x_{n}}{y_{n}}$ as $x_{n}\cdot\frac{1}{y_{n}}$ , then the result follows from parts (ii) and (iii).
(v)

Let $y_{n}=a$ (so $(y_{n})_{n}$ is the constant sequence $(a,a,\dots)$ with limit $a$ ), then the result follows from part (ii).
(vi)

We will make use of a minor result to assist in proving this.

Lemma

Suppose $(a_{n})_{n}$ is a convergent sequence with $a_{n}\to a$ . If $a_{n}>0\ \forall n\in\mathbb{N}$ then $a\geq 0$ .

Proof.

Suppose by way of contradiction that $a<0$ and let $\varepsilon=-a>0$ . Then since $(a_{n})_{n}$ converges, $\exists N\in\mathbb{N}$ such that $n\geq N\implies\lvert a_{n}-a\rvert<-a$ . This means that

$\displaystyle a<\,$ $\displaystyle a_{n}-a<-a$ (4.27)

$\displaystyle\implies$ $\displaystyle a_{n}-a+a<-a+a$ (4.28)

$\displaystyle\implies$ $\displaystyle a_{n}<0$ (4.29)

which is a contradiction, hence $a$ must be greater than 0. ∎

Now, consider $y_{n}-x_{n}\geq 0\ \forall n\in\mathbb{N}$ . The sequence $(y_{n}-x_{n})_{n}$ converges to $y-x$ by part (i) (and part (v) with $a=-1$ ) and so by our claim $y-x\geq 0$ , which implies $x\leq y$ as required.

∎

4.2.2 Sandwich Principle

A little observation from the last part of this theorem is the sandwich principle, which is that if we have three sequences — $(a_{n})_{n}$ , $(x_{n})_{n}$ and $(b_{n})_{n}$ with limits $a$ , $x$ and $b$ respectively — with $a_{n}\leq x_{n}\leq b_{n}\ \forall n\in\mathbb{N}$ , then $a\leq x\leq b$ . In the special case that $a=b$ , we can use this principle to determine the limit of an unknown sequence $(x_{n})_{n}$ as in this case $a=x=b$ . We will finish off this section with a couple of examples about how to use the arithmetic properties of limits.

Example 4.6.

Let $(x_{n})_{n}=\left(1+\frac{1}{n^{p}}\right)_{n}$ with $p\in\mathbb{N}$ . Show that $x_{n}\to 1$ as $n\to\infty$ .
First, let $a_{n}=1$ and $b_{n}=\frac{1}{n^{p}}$ , then note that we already know that the sequence $\left(\frac{1}{n}\right)_{n}$ converges to 0, so by theorem 4.4 (ii) (applied repeatedly),

\underbrace{\frac{1}{n}\cdot\frac{1}{n}\cdots\frac{1}{n}}_{p\text{ times}}=% \frac{1}{n^{p}}\to 0\text{ as }n\to\infty.

(4.30)

Then $(x_{n})_{n}=(a_{n}+b_{n})_{n}$ which, by theorem 4.4 (i), converges to 1 since $a_{n}\to 1$ and $b_{n}\to 0$ as $n\to\infty$ .

Example 4.7.

Show that the sequence $(x_{n})_{n}=\left(\frac{n+5n^{2}+2n^{3}}{n^{3}+3}\right)$ converges to 2.

\frac{2n^{3}+5n^{2}+n}{n^{3}+3}=\frac{2+5\left(\frac{1}{n}\right)+\left(\frac{% 1}{n}\right)^{2}}{1+3\left(\frac{1}{n}\right)^{3}}\to\frac{2+5(0)+0}{1+3(0)}=2% \text{ as }n\to\infty.

(4.31)

$\displaystyle\lvert(x_{n}+y_{n})-(x+y)\rvert$	$\displaystyle=\lvert(x_{n}-x)+(y_{n}-y)\rvert$	(4.16)
	$\displaystyle\leq\lvert x_{n}-x\rvert+\lvert y_{n}-y\rvert$	(4.17)
	$\displaystyle<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon.$	(4.18)

	$\displaystyle\left\lvert\frac{1}{y_{n}}-\frac{1}{y}\right\rvert$	$\displaystyle=\left\lvert\frac{y-y_{n}}{y_{n}y}\right\rvert=\frac{\lvert y_{n}% -y\rvert}{\lvert y_{n}\rvert\lvert y\rvert}$		(4.25)
		$\displaystyle<\frac{\varepsilon\lvert y\rvert^{2}}{2\lvert y_{n}\rvert\lvert y% \rvert}<\frac{2\varepsilon\lvert y\rvert^{2}}{2\lvert y\rvert^{2}}=\varepsilon.$		(4.26)

$\displaystyle a<\,$	$\displaystyle a_{n}-a<-a$	(4.27)
$\displaystyle\implies$	$\displaystyle a_{n}-a+a<-a+a$	(4.28)
$\displaystyle\implies$	$\displaystyle a_{n}<0$	(4.29)

4.2 Properties of Limits

Theorem 4.2

Proof.

Theorem 4.3

Proof.

4.2.1 Adding and Multiplying Sequences

Theorem 4.4

Proof.

Lemma

Proof.

4.2.2 Sandwich Principle

Example 4.6.

Example 4.7.