Theorem Suppose
(a) $\sum_{n=0}^{\infty}a_n$ converges absolutely,
(b) $\sum_{n=0}^{\infty}a_n=A$,
(c)$\sum_{n=0}^{\infty}b_n=B$,
(d)$c_n=\sum_{k=0}^{n}a_kb_{n-k}$ $(n=0,1,2,\dots)$.
Then $$\sum_{n=0}^{\infty}c_n=AB.$$
(Rudin's Principles of Mathematical Analysis, 3rd Edition, page 74)
My question is: This theorem guarantees that, if at least one of the series $\sum_{n=0}^{\infty}a_n$ or $\sum_{n=0}^{\infty}b_n$ converges absolutely, then the product series $\sum_{n=0}^{\infty}c_n$ converges (and only this). Is there an example such that this $\sum_{n=0}^{\infty}c_n$ converges but nonabsolutely?
If one of the two series is only conditionally convergent, you cannot expect more than conditional convergence of the Cauchy product.
A class of series where it is easy to see that $\sum c_n$ can only be conditionally convergent is obtained by letting both series be alternating, $a_n = (-1)^n\lvert a_n\rvert$ and $b_n = (-1)^n\lvert b_n\rvert$. Then for the terms of the Cauchy product we have
$$c_n = \sum_{k=0}^n a_k b_{n-k} = \sum_{k=0}^n (-1)^k\lvert a_k\rvert (-1)^{n-k}\lvert b_{n-k}\rvert = (-1)^n \sum_{k=0}^n \lvert a_k\rvert\cdot\lvert b_{n-k}\rvert,$$
which yields for any given $\kappa\in \mathbb{N}$ and all $n\geqslant \kappa$ the estimate
$$\lvert c_n\rvert \geqslant \lvert a_{\kappa}\rvert \cdot \lvert b_{n-\kappa}\rvert,$$
so unless $a_n = 0$ for all $n$, the Cauchy product is never absolutely convergent if $\sum b_n$ is only conditionally convergent when both series are alternating.