12 - Algebra of Limits - part 2

Algebra of Limits - part 2

Continuing off Algebra of Limits - part 1
Let $(a_n)$ and $(b_n)$ be sequences with

$$a_n\to L\text{ and }{b}_n\to M\text{ as }n\to \infty$$

6. Product: Sequence $(a_n{b}_n)$ converges with

$$a_n{b}_n\to LM\text{ as }n\to \infty$$

7. Reciprocal: If $M\ne 0$, sequence $\left(\frac{1}{b_n}\right)$ converges with

$$\frac{1}{b_n}\to \frac{1}{M}\text{ as }n\to \infty$$

8. Quotient: Sequence $\left(\frac{a_n}{b_n}\right)$ converges with

$$\frac{a_n}{b_n}\to \frac{L}{M}\text{ as }n\to \infty$$

Proof

6. Take $ϵ>0$ (and assume $ϵ<1$)
   Since $a_n\to L$, there is $N_1$ such that

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

Since $b_n\to M$, there is $N_2$ such that

$$\text{if }n\ge N_2\text{ then }|b_n-M|<ϵ$$

Let $N=\max \{N_1,N_2\}$
If $n\ge N$, then $|a_n-L|<ϵ$ and $|b_n-M|<ϵ$ then

$$|a_n|<|L|+ϵ$$

So

$$\begin{array}{rl}|a_n{b}_n-LM|& =|a_n(b_n-M)+M(a_n-L)|\\ & \le |a_n||b_n-M|+|M||a_n-L|\\ & <(|L|+ϵ)ϵ+|M|ϵ\\ & <ϵ(1+|L|+|M|)\end{array}$$

Since $1+|L|+|M|$ is constant, then

$$(a_n{b}_n)\text{ converges to }LM$$

7. Assume that $M\ne 0$
   Take $ϵ>0$
   Since $b_n\to M$ and $|M|\ge 0$ then there is $N_1$ such that

$$\text{if }n\ge N_1\text{ then }|b_n-M|<\frac{M}{2}\text{ so }|b_n|>\frac{|M|}{2}$$

So by Reverse Triangle Inequality

$$\begin{array}{rl}|b_n|& =|(b_n+(M-b_n))+(-(M-b_n))|\\ & \ge ||b_n+(M-b_n)|-|M-b_n||>\frac{|M|}{2}>0\end{array}$$

Hence there the tail $(b_n{)}_{n\ge N_1}$ has all terms non-zero so then

$${\left(\frac{1}{b_n}\right)}_{n\ge N_1}\text{is well defined}$$

Also, There is $N_2$ such that if $n\ge N_2$ then $|b_n-M|<ϵ$

​​

Let $N=\max \{N_1,N_2\}$. If $n\ge N$

$$\left|\frac{1}{b_n}-\frac{1}{M}\right|=\frac{|M-b_n|}{|M||b_n|}<\frac{ϵ}{|M|}\cdot \frac{2}{|M|}$$

Since $\frac{2}{|M{|}^2}$ is a positive constant then ${\left(\frac{1}{b_n}\right)}_{n\ge N_1}$ converges with

$$\frac{1}{b_n}\to \frac{1}{M}\text{ as }n\to \infty$$

8. Follows from $(6)$ and $(7)$