# For which $s$ is defined $\frac{1}{\zeta(s)}\sum_{n=1}^\infty\frac{n}{ \left( \zeta(s) \right) ^n}$, where $\zeta(s)$ is the Riemann Zeta function?

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 $$\left| \sum_{n=1}^\infty\frac{1}{ \left( \zeta(s) \right) ^n}\right| \leq \frac{1}{1-|\zeta(s)|}-1=\frac{|\zeta(s)|}{|\zeta(s)|-1}.$$

• IN my last computation was a typo the right is $|\zeta(s)|>1$ intead of $|\Re \zeta(s)|>1$. – user243301 Apr 29 '16 at 10:16
• of course for considering $\sum_n z^n$ as a series (and not only a formal series) you need $|z| < 1$, hence here $|\zeta(s)| > 1$, in particular this is true for $s$ in the neighborhood of $]1,\infty[$. and here you don't need your first assumption : if $\sum_n |a_n|$ and $\sum_n |b_n|$ both converge then $\sum_{n,m} a_n b_m$ converges absolutely, hence the order of summation doesn't matter and $(\sum_n a_n)(\sum_m b_m) = \sum_n \sum_k a_k b_{n-k}$ – reuns Apr 29 '16 at 18:13
• Very thanks much for your contribution, you are welcome to add an answer @user1952009 – user243301 Apr 29 '16 at 18:22
• Many thanks @PreservedFruit – user243301 Jul 9 '18 at 9:42