I know there are a couple of proofs of this result on this website, but I would like to verify if my proof is correct, as it is slightly different.
Let $\sum\limits_{n=0}^\infty a_n$ and $\sum\limits_{n=0}^\infty b_n$ be two absolutely convergent series of complex numbers, let $c_n = \sum\limits_{k=0}^n a_k b_{n-k}$. Prove that $\sum\limits_{n=0}^\infty c_n$ is absolutely convergent and that $$\sum\limits_{n=0}^\infty c_n = \left(\sum\limits_{n=0}^\infty a_n\right)\left(\sum\limits_{n=0}^\infty b_n\right) \text{ (*) }$$
Proof:
(I'm not sure why it is asked to prove a definition, because (*) is a Cauchy product of infinite series, by definition).
Let $\sum\limits_{n=0}^\infty a_n = L$ and $\sum\limits_{n=0}^\infty b_n=M$. By definition of the Cauchy product, $\left(\sum\limits_{n=0}^\infty a_n\right)\left(\sum\limits_{n=0}^\infty b_n\right)=\sum\limits_{k=0}^\infty\sum\limits_{l=0}^k a_l b_{k-l}=\sum\limits_{k=0}^\infty c_k$.
Now, $\left\|\left(\sum\limits_{n=0}^\infty a_n\right)\left(\sum\limits_{n=0}^\infty b_n\right)\right\|\le \sum\limits_{n=0}^\infty \|a_n\|\sum\limits_{n=0}^\infty \|b_n\|=LM<\infty$, so $\sum\limits_{n=0}^\infty c_n$ converges absolutely.
If we didn't use the Cauchy product, then
$\sum\limits_{n=0}^N \|c_n\|=\sum\limits_{n=0}^N \|\sum\limits_{k=0}^n a_k b_{n-k} \|\le \sum\limits_{n=0}^N \sum\limits_{k=0}^n \|a_k b_{n-k} \|=\|a_0b_0\|+\|a_0b_1\|+...+\|a_Nb_0\|+...+\|a_Nb_N\|=\|a_0\|\sum\limits_{n=0}^N \| b_n\|+...+\|a_N\|\sum\limits_{n=0}^N\|b_n\|=(\|a_0\|+...+\|a_N\|)\sum\limits_{n=0}^N\|b_n\|=\sum\limits_{n=0}^N\|a_n\|\sum\limits_{n=0}^N\|b_n\|$. Now $\lim\limits_{N\to\infty}\sum\limits_{n=0}^N\|a_n\|\sum\limits_{n=0}^N\|b_n\|=LM$.
Do you think the above is correct? And which approach is better - the first or the second? It seems to me that the first is better since it is shorter.