# 5.3: Real Analysis - Convergent Sequences

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## Notation $$5.3.1$$.

For $$x \in \mathbb{R}$$, $$|x|$$ denotes the absolute value of $$x$$: |x|=\left\{\begin{aligned} x & \text { if } x \geq 0 , \\ -x & \text { if } x<0 . \end{aligned}\right.

You may assume the following basic properties of absolute value (without proof):

## Lemma $$5.3.2$$.

For $$x, y, z \in \mathbb{R}$$, we have:

1. $$|x| \geq 0$$ (and $$|x|=0 \Leftrightarrow x=0$$).
2. $$|x| = |−x|$$.
3. $$|x+y| \leq |x|+|y|$$. (“triangle inequality”)
4. $$|x y|=|x| \cdot|y|$$.
5. $$−|x| \leq x \leq |x|$$.
6. $$\exists N \in \mathbb{N}, N>|x|$$.
7. If $$|x| < |y|$$ and $$z \neq 0$$, then $$|xz| < |yz|$$.
8. If $$|x| > |y| \neq 0$$, then $$1/|x| < 1/|y|$$.

## Definition $$5.3.3$$.

Assume $$a_{1}, a_{2}, a_{3}, \ldots$$ is an infinite sequence of real numbers, and $$L \in \mathbb{R}$$. We say that the sequence converges to $$L$$ (and write $$a_{n} \rightarrow L$$) iff $\forall \epsilon>0, \exists N \in \mathbb{N}, \forall n>N,\left|a_{n}-L\right|<\epsilon .$

## Other Terminology.

When $$a_{n} \rightarrow L$$, we can also say that the limit of the sequence is $$L$$.

## Example $$5.3.4$$.

Let $$t \in \mathbb{R}$$. If $$a_{n} = t$$ for all $$n$$, then $$a_{n} \rightarrow t$$.

Solution

PROOF.

Given$$\epsilon>0$$, let $$N = 0$$. Given $$n > N$$, we have $$\left|a_{n}-t\right|=|t-t|=|0|=0<\epsilon$$.

## Example $$5.3.5$$.

If $$a_{n} = 1/n$$ for all $$n$$, then $$a_{n} \rightarrow 0$$.

Scratchwork. To prove $$a_{n} \rightarrow 0$$, we want: $\left|a_{n}-0\right| \stackrel{?}{<} \epsilon \quad 1 / n \stackrel{?}{<} \epsilon \quad 1 / \epsilon \stackrel{?}{<} n$
Since $$n > N$$, it suffices to choose $$N > 1 / \epsilon$$.

Solution

PROOF.

Given $$\epsilon > 0$$, Lemma $$5.3.2(6)$$ tells us there exists $$N \in \mathbb{N}$$, such that $$N > 1 / \epsilon$$. Given $$n > N$$, we have \begin{aligned} \left|a_{n}-0\right| &=1 / n & &\left(a_{n}=1 / n>0\right) \\ &<1 / N & &(n>N \text { and Lemma 5.3.2(8)) }\\ &<\epsilon & &(N>1 / \epsilon \text { and Lemma 5.3.2(8)) } \end{aligned}

## Exercise $$5.3.6$$.

Show that if $$a_{n} = n/(n + 1)$$ for all $$n$$, then $$a_{n} \rightarrow 1$$.

## Proposition $$5.3.7$$.

If $$a_{n} \rightarrow L$$ and $$b_{n} \rightarrow M$$, then $$a_{n} + b_{n} \rightarrow L + M$$.

Scratchwork. To prove $$a_{n} + b_{n} \rightarrow L + M$$, $\text { we want to make }\left|\left(a_{n}+b_{n}\right)-(L+M)\right| \text { small (less than } \epsilon \text { ). }$
What we know is that we can make $$\left|a_{n}-L\right|$$ and $$\left|b_{n}-M\right|$$ as small as we like. By the triangle inequality, we have $\left|\left(a_{n}-L\right)+\left(b_{n}-M\right)\right|<\left|a_{n}-L\right|+\left|b_{n}-M\right|$
By simple algebra, the left-hand side is equal to $$\left|\left(a_{n}+b_{n}\right)-(L+M)\right|$$, so we just need to make the right-hand side less than $$\epsilon$$. This will be true if $$\left|a_{n}-L\right|$$ and $$\left|b_{n}-M\right|$$ are both less than $$\epsilon / 2$$.

Since $$a_{n} \rightarrow L$$, there is some large $$N_{a}$$, such that $$\left|a_{n}-L\right|<\epsilon / 2$$ for all $$n > N_{a}$$. Similarly, since $$b_{n} \rightarrow M$$, there is some large $$N_{b}$$, such that $$\left|b_{n}-M\right|<\epsilon / 2$$ for all $$n > N_{b}$$. Now, we just need know that $$n$$ will be larger than both $$N_{a}$$ and $$N_{b}$$ whenever $$n > N$$. So we should choose $$N$$ to be whichever of $$N_{a}$$ and $$N_{b}$$ is larger. That is, we let $$N$$ be the maximum of $$N_{a}$$ and $$N_{b}$$, which is denoted max(Na, Nb).

Proof

Given $$\epsilon > 0$$, we know that $$\epsilon / 2 > 0$$. Hence:

• Since $$a_{n} \rightarrow L$$, we know $$\exists N_{a} \in \mathbb{N}, \forall n>N_{a},\left|a_{n}-L\right|<\epsilon / 2$$.
• Since $$b_{n} \rightarrow M$$, we know $$\exists N_{b} \in \mathbb{N}, \forall n>N_{b},\left|b_{n}-M\right|<\epsilon / 2$$.

Let $$N=\max \left(N_{a}, N_{b}\right) \in \mathbb{N}$$, so $$N \geq N_{a}$$ and $$N \geq N_{b}$$.

Given $$n > N$$:

1. We have $$n > N \geq N_{a}$$, so $$\left|a_{n}-L\right|<\epsilon / 2$$.
2. We have $$n > N \geq N_{b}$$, so $$\left|b_{n}-M\right|<\epsilon / 2$$.

Therefore \begin{aligned} \left|\left(a_{n}+b_{n}\right)-(L+M)\right| &=\left|\left(a_{n}-L\right)+\left(b_{n}-M\right)\right| & & \text { (high-school algebra) } \\ & \leq\left|a_{n}-L\right|+\left|b_{n}-M\right| & & \text { (triangle inequality) } \\ &<\epsilon / 2+\epsilon / 2 & &((*) \text { and }(* *)) \\ &=\epsilon . & & \end{aligned}

## Exercise $$5.3.8$$.

Assume $$a_{n} \rightarrow L$$, and $$c \in \mathbb{R}$$. Do these proofs directly from the definition of convergence.

1. Show $$−a_{n} \rightarrow −L$$.
2. Show $$a_{n} + c \rightarrow L + c$$.
3. Show $$2a_{n} \rightarrow 2L$$.
4. Show $$ca_{n} \rightarrow cL$$ if $$c \neq 0$$.
5. Show that if $$L > 0$$, then $$\exists N \in \mathbb{N}$$, such that $$a_{n} > 0$$ for all $$n > N$$.
6. (harder) Show that if $$L = 1$$, then $$1/a_{n} \rightarrow 1$$.

## Exercise $$5.3.9$$.

Assume $$a_{n} \rightarrow L$$ and $$b_{n} \rightarrow M$$.

1. Show that if $$M = 0$$, and $$|a_{n}| \leq 2$$ for all $$n$$, then $$a_{n}b_{n} \rightarrow 0$$.
2. (harder) Show $$a_{n}b_{n} \rightarrow LM$$.

This page titled 5.3: Real Analysis - Convergent Sequences is shared under a CC BY-NC-SA 2.0 license and was authored, remixed, and/or curated by Dave Witte Morris & Joy Morris.