Cauchy product of an absolutely convergent series and a conditionally convergent series

Question

We know by Mertens' theorem, Let $(a_n)_{n≥0}$ and $(b_n)_{n≥0}$ be real or complex sequences, then if $\sum_{n=0}^{+\infty}a_n$ converges absolutely and $\sum_{n=0}^{+\infty}b_n$ converges only conditionally, then their Cauchy product $\sum_{n=0}^{+\infty}c_n$ (with $c_n:=\sum_{i+j=n}a_i\cdot b_j$) converges to $AB$. ($A:=\sum_{n=0}^{+\infty}a_n,B:=\sum_{n=0}^{+\infty}b_n$)

But can we say that $\sum_{n=0}^{+\infty}c_n$ converges absolutely? I searched through the Internet and found nothing about it.

Answer 1

No, we cannot say that. Take$$a_n=\begin{cases}1&\text{ if }n=0\\-1&\text{ if }n=1\\0&\text{ otherwise.}\end{cases}$$Then $\sum_{n=0}^\infty a_n$ converges absolutely. Now, let $b_n=\frac{(-1)^n}{n+1}$. Then $\sum_{n=0}^\infty b_n$ converges conditionally. And if $(c_n)_{n\in\Bbb Z_+}$ is their Cauchy product, then$$(\forall n\in\Bbb N):c_n=(-1)^n\frac{2n+1}{n(n+1)},$$and therefore the series $\sum_{n=0}^\infty c_n$ does not converge absolutely.