Lecture 11 - Bounded Sequences, Limit Laws

Note that this mean $\{a_n|n\in \mathbb{N}\}$ is bounded.

Proof
Let $a_n\to a$. We need to show our $M$. Notice that since $a_n\to a$ then use $\epsilon =1>0$ to get that $\mathrm{\exists }N$ such that for all $n\ge N$ we have:
$|a_n-a|<1$

Notice (as described in the figure) that:
$|a_n|<|a|+1$
for every $n\ge N$. This algebraically comes from that:
$\begin{array}{r}-1<a_n-a<1\\ a-1<a_n<a+1\\ a_n<a+1\le |a|+1\\ \Rightarrow |a_n|<|a|+1\end{array}$
Now then choose $M=max(|a|+1,|a_1|,...,|a_{N-1}|)\in \mathbb{R}$. Then:
$|a_n|<|a|+1=M$
when $n\ge N$. For when $n<N$ then by construction then our $|a_n|\le M$ as desired.

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Scratch Work:

1. We want, for some $\epsilon >0$:
   $|c{a}_n-ca|<\epsilon \Rightarrow |c||a_n-a|<\epsilon$
   Consider the cases. If $c=0$ then the case is trivial. If $c\ne 0$ then:
   $|a_n-a|<\frac{\epsilon }{|c|}$
   We know that $\mathrm{\exists }N$ such that for all $n\ge N$.

2. We want, for some $\epsilon >0$:
   $|(a_n+b_n)-(a+b)|<\epsilon$
   Notice then:
   $\begin{array}{rl}|(a_n-a)+(b_n-b)|& \le \underset{<\frac{\epsilon }{2}}{\underbrace{|a_n-a|}}+\underset{<\frac{\epsilon }{2}}{\underbrace{|b_n-b|}}\end{array}$

3. We want, for some $\epsilon >0$:
   $|a_n{b}_n-ab|<\epsilon$
   Notice that, ignoring the absolute values for a moment:
   $\begin{array}{rl}{a}_n{b}_n-ab-a{b}_n+a{b}_n& =(a_n-a)b_n+a(b_n-b)\end{array}$
   So putting the absolute values back in:
   $\begin{array}{rl}|(a_n{b}_n)-ab|& =|(a_n-a)b_n+a(b_n-b)|\\ & \le |b_n|\underset{<\frac{\epsilon }{2M}}{\underbrace{|a_n-a|}}+|a|\underset{<\frac{\epsilon }{2(|a|)}\vee \frac{\epsilon }{2(|a|+1)}}{\underbrace{|b_n-b|}}& & \text{Triangle Ineq.}\end{array}$
   where since $b_n$ is convergent, then it has a bound $M$, so then $|b_n|\le M$. Notice further that $M>0$ since we consider the absolute value of $b_n$.

4. Say $b\ne 0$. Then we want:
   $\frac{1}{b_n}\to \frac{1}{b}$
   Because then we can apply (3) by using the new sequence $1/b_n$ instead. This means we need to show:
   $|\frac{1}{b_n}-\frac{1}{b}|=\frac{|b-b_n|}{|b||b_n|}$
   Note that because $(b_n)\to b$, then we can make the numerator as small we we like by making $n$ large. On the other hand, for the denominator we want some $|b_n|\ge \delta >0$. This will lead to a bound on the size of $\frac{1}{|b||b_n|}$.

The trick is to look far enough out into the sequence $(b_n)$ so that the terms are closer to $b$ than they are to $0$. Consider ${\epsilon }_0=\frac{|b|}{2}$. Because $(b_n)\to b$ then there exists and $N_1$ such that $|b_n-b|<\frac{|b|}{2}$ for all $n\ge N_1$. This implies that $|b_n|>\frac{|b|}{2}$. So then choose $N_2$ such that $N\ge N_2$ implies:
$|b_n-b|<\frac{\epsilon |b{|}^2}{2}$
and choose $N=max(N_1,N_2)$.

Proof

1. Let $\epsilon >0$. Choose $N\in \mathbb{N}$ such that:
   a. If $c=0$ then we are done (trivial) case.
   b. Since $c\ne 0$ then because $\frac{\epsilon }{|c|}>0$ then $n\ge N\Rightarrow |a_n-a|<\frac{\epsilon }{|c|}$ . Then:
   $\begin{array}{rl}|a_n-a|& <\frac{\epsilon }{|c|}\\ |c||a_n-a|& <\epsilon \\ |c{a}_n-ca|& <\epsilon \end{array}$
2. Let $\epsilon >0$. Since $\frac{\epsilon }{2}>0$ then $\mathrm{\exists }{N}_1$ where $n\ge N_1$ implies $|a_n-a|<\frac{\epsilon }{2}$. Similarly then $\mathrm{\exists }{N}_2$ where $n\ge N_2$ implies $|b_n-b|<\frac{\epsilon }{2}$. Choose $N=max(N_1,N_2)\in \mathbb{N}$, since then letting $n\ge N$:
   $\begin{array}{rl}|(a_n+b_n)-(a+b)& =|(a_n-a)+(b_n-b)|\\ & \le |a_n-a|+|b_n-b|& & \mathrm{\Delta }\text{Inequality}\\ & <\frac{\epsilon }{2}+\frac{\epsilon }{2}\\ & =\epsilon \end{array}$
3. Let $\epsilon >0$. Since $b_n\to b$ then $\mathrm{\exists }M\ge 0$ where $|b_n|\le M$. Notice that:
   $\epsilon >0\Rightarrow \frac{\epsilon }{2M}>0$
   And further:
   $\epsilon >0\Rightarrow \frac{\epsilon }{2(|a|+1)}>0$
   (or choose $\epsilon /2(|a|)$ if $|a|\ne 0$. Either way choosing either doesn't affect the later arguments). Then since $a_n\to a$ then $\mathrm{\exists }{N}_1$ where for all $n\ge N_1$ then $|a_n-a|<\frac{\epsilon }{2(|a|+1)}$. Similarly since $b_n\to b$ then $\mathrm{\exists }{N}_2$ where for all $n\ge N_2$ when $|b_n-b|<\frac{\epsilon }{2M}$.

Now choose $N=max(N_1,N_2)$. Let $n\ge N$. Then:
$\begin{array}{rl}|a_n{b}_n-ab|& =|a_n{b}_n-a{b}_n+a{b}_n-ab|\\ & =|(a_n-a)b_n+a(b_n-b)|\\ & \le |b_n||a_n-a|+|a||b_n-b|\\ & <\frac{|b_n|\epsilon }{2M}+\frac{|a|\epsilon }{2(|a|+1)}\\ & <\underset{|b_n|\le M}{\underbrace{\frac{\epsilon }{2}}}+\underset{|a|<\frac{\epsilon }{2(|a|+1)}}{\underbrace{\frac{\epsilon }{2}}}\\ & =\epsilon \end{array}$
4. If we can prove that $\frac{1}{b_n}\to \frac{1}{b}$ then we can apply (3) and we are done (assuming $b\ne 0$). Let $\epsilon >0$. First note:
$|\frac{1}{b_n}-\frac{1}{b}|=\frac{|b-b_n|}{|b||b_n|}$
Now because $(b_n)\to b$ then $|b_n-b|<\frac{\epsilon |b{|}^2}{2}$ since it's $>0$. There's some $N_2$ such that $n\ge N_2$ allows for this inequality to hold.

Now notice also that because $(b_n)\to b$ then notice for a specific ${\epsilon }_0=\frac{|b|}{2}$ that then $\mathrm{\exists }{N}_1$ where $|b_n-b|<\frac{|b|}{2}$ for all $n\ge N_1$. Then this implies that $|b_n|>\frac{|b|}{2}$.

Thus choose $N=max(N_1,N_2)$ so then for all $n\ge N$:
$|\frac{1}{b_n}-\frac{1}{b}|=|b-b_n|\frac{1}{|b||b_n|}<\frac{\epsilon |b{|}^2}{2}\cdot \frac{1}{|b|\frac{|b|}{2}}=\epsilon$
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