Unless I'm making a mistake, if one of the factors $\sum_{n=1}^\infty a_n$ of a convolution product or Cauchy product converges, and the other $\sum_{n=1}^\infty b_n$ corverges absolutely then the product $$\sum_{n=1}^\infty \sum_{k=1}^{n} a_{k} b_{n-k+1}$$ converges.
For $n\geq 1$ I've computed this product, without formality, for $a_n=b_n=\frac{1}{(\zeta(s))^n}$ where $\zeta(s)$ is the Riemann zeta function and those $s$ such that $\zeta(s)\neq 0$ (for example I know that for $\Re s>1$ the Riemann Zeta funciton doesn't vanish). Then I've computed without formality $$ \left( \sum_{n=1}^\infty\frac{1}{ \left( \zeta(s) \right) ^n}\right)^2= \frac{1}{\zeta(s)}\sum_{n=1}^\infty\frac{1}{ \left( \zeta(s) \right) ^n} \sum_{i=1}^{n} 1=\frac{1}{\zeta(s)}\sum_{n=1}^\infty\frac{n}{ \left( \zeta(s) \right) ^n}.$$
Question. For wich $s$ does make sense the previous product? I say for some easy region in the whole plane, if my problem is well posed you can study the more advance topic, but to me is enough for easy regions. Thanks in advance.
My only computation was that I belive that for those $s\neq 0$ with $|\Re\zeta(s)|>1$ I can write $$ \left| \sum_{n=1}^\infty\frac{1}{ \left( \zeta(s) \right) ^n}\right| \leq \frac{1}{1-|\zeta(s)|}-1=\frac{|\zeta(s)|}{|\zeta(s)|-1}.$$