5.3: Real Analysis - Convergent Sequences

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 $a_{n}b_{n} \rightarrow LM$.