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

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.

• The question doesn't match the title... certainly $\sum a_n b_n$ converges, but it rarely if ever converges to $(\sum a_n) (\sum b_n)$. Commented Feb 24, 2021 at 7:01
• Note that $\sum c_n$ converges absolutely already when $\sum a_n$ and $\sum b_n$ converge and just one of them absolutely Commented Feb 24, 2021 at 7:03

$$\sum a_nb_n=ab$$ 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].
• Its not absolutely convergent or it doesnt converge to $ab$? Commented Feb 24, 2021 at 6:59
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 $$c_n=0$$.