17 - Limits - first key results

Sandwiching (first version)

Let $(a_n)$ and $(b_n)$ be real sequences with

$$0\le a_n\le b_n\text{ for all }n\ge 1$$

If $b_n\to 0$ as $n\to \infty$, then $a_n\to \infty$ as $n\to \infty$

Proof

Assume that $0\le a_n\le b_n$ for all $n$, and that $b_n\to 0$ as $n\to \infty$
Take $ϵ>0$
Since $b_n\to 0$, there exists $N$ such that

$$\text{if }n\ge N\text{ then }|b_n|<ϵ$$

Now if $n\ge N$ then $0\le a_n\le b_n<ϵ$, so $|a_n|<ϵ$
So $a_n\to 0$ as $n\to \infty$

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Useful Convergence lemma

1. Take $c\in \mathbb{R}$ with $|c|<1$ then

$$c^n\to 0\text{ as }n\to \infty$$

2. Let $a_n=\frac{n}{2^n}$ for $n\ge 1$ then

$$a_n\to 0\text{ as }n\to \infty$$

Proof

1. Write $|c|=\frac{1}{1+y}$ where $y>0$
   Take $ϵ>0$
   Let $N=\lceil \frac{1}{y\cdot ϵ}\rceil +1$
   Take $n\ge N$
   By Bernoulli’s Inequality (since $y>0$ and $n\ge 1$ we have

$$(1+y{)}^n\ge 1+ny$$

Then

$$|c^n|=\frac{1}{(1+y{)}^n}\le \frac{1}{1+ny}\le \frac{1}{Ny}<ϵ$$

So $c^n\to 0$ as $n\to \infty$
2) If $n\ge 2$ then $2^n=(1+1{)}^n\ge (\genfrac{}{}{0ex}{}{n}{2})$ (by the binomial theorem)
Take $ϵ>0$
Let $N=\lceil 2+\frac{2}{ϵ}\rceil$
For $n\ge N$ then

$$|a_n-0|=\frac{n}{2^n}\le \frac{n}{\left(\genfrac{}{}{0ex}{}{n}{2}\right)}=\frac{2}{n-1}\le \frac{2}{N-1}<ϵ$$

Hence $a_n\to 0$ as $n\to \infty$

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Uniqueness of limits

Let $(a_n)$ be a convergent sequence then the limit is unique

Proof

Assume that $a_n\to L_1$ and $a_n\to L_2$ as $n\to \infty$
Suppose for a contradiction that $L_1\ne L_2$
Let $ϵ=\frac{|L_1-L_2|}{2}>0$
Since $a_n\to L_1$ as $n\to \infty$, there $N_1$ such that

$$\text{if }n\ge N_1\text{ then }|a_n-L_1|<ϵ$$

Also, since $a_n\to L_2$ as $n\to \infty$, there is $N_2$ such that

$$\text{if }n\ge N_2\text{ then }|a_n-L_2|<ϵ$$

For $n\ge \max \{N_1,N_2\}$ we have $|a_n-L_1|<ϵ$ and $|a_n-L_2|<ϵ$

$$\begin{array}{rl}|L_1-L_2|& =|(L_1-a_n)+(a_n-L_2)|\\ & \le |L_1-a_n|+|a_n-L_2|(\text{by the triangle inequality})\\ & <2ϵ=|L_1-L_2|\end{array}$$

This is a contradiction hence $L_1=L_2$

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