Given series $\sum a_n,\sum b_n$ converge absolutely show that $\sum c_n$ where $c_n=a_n\cdot b_n$ converges.

Question

Given series $\sum a_n,\sum b_n$ converge absolutely to $a,b$ show that $\sum c_n$ where $c_n=a_n\cdot b_n$ converges absolutely and converges to $a\cdot b$

I believe I can show that $\sum c_n$ converges by comparison since $\sum \vert c_n\vert=\sum \vert a_n\vert\vert b_n\vert\leq \sum \vert a_n\vert\sum\vert b_n\vert$

But my issue is showing the convergence is $a\cdot b$

I want to show that $\lim_{n\to \infty}\sum_{i=0}^n \vert c_n\vert =\lim_{n\to \infty} \sum_{i=0}^n \vert a_i\vert\vert b_i\vert=ab$

But the same thing I used before no longer seems to work because I get $\lim_{n\to \infty} \sum_{i=0}^n\vert a_i\vert\vert b_i\vert\leq \lim_{n\to \infty} \sum_{i=0}^n \vert a_i\vert \sum_{i=0}^n \vert b_i\vert$

Which I don't believe is what I want, I would need to somehow show that $ab$ is also a lower bound for the limit.

Answer 1

$\sum a_nb_n=ab$ is false. Take $a_n=b_n=0$ for $n \geq 2$ and you will easily get a counter-example.

[Any choice of $a_1,a_2,b_1,b_2$ in $(0,\infty)$ will do].

Answer 2

Have a look at $$a_n=\begin{cases}2^{-n}&n\text{ odd}\\0&n\text{ even}\end{cases}$$ and $$b_n=\begin{cases}2^{-n}&n\text { even}\\0&n\text{ odd}\end{cases},$$ which makes $c_n=0$.