# Is the Hurwitz zeta function in $l^1(\mathbb{N})$?

Let $$Re(s) >1, n = 1,2,\dots$$ and let
$$\zeta(s,n) = \sum_{m=0}^{\infty}\frac{1}{{(n+m)}^{s}}$$ be the Hurwitz zeta function. I want to prove that there is some $$p \geq 1$$ such that for all $$Re(s) >1$$ we have that $$\sum_{n=1}^\infty\lvert \zeta(s,n) \rvert^p < \infty.$$ I have the intuition that $$p = 2$$ works but I am unable to prove this. Is there any literature regarding this question?

I also think this can be proven using formula $$\zeta(s,q) = \frac{1}{\Gamma(s)}\int_0^\infty \frac{t^{s-1}e^{-qt}}{1-e^{-t}}dt.$$

• For $a >0$ and $\Re(s)> 1$ and by analytic continuation for $\Re(s) >0$ $\zeta(s,a) - \frac{a^{1-s}}{s-1}= \sum_{m=0}^\infty (a+m)^{-s}-\int_m^{m+1} (a+x)^{-s}dx$ $= \sum_{m=0}^\infty \int_m^{m+1} \int_m^x s(a+t)^{-s-1}dtdx = \sum_{m=0}^\infty O(\int_m^{m+1} |s| (a+x)^{-\Re(s)-1}dx)=O(|s| \frac{ a^{-\Re(s)}}{\Re(s)})$ Sep 4, 2019 at 3:30

Note that for $$k \ge 1, a>1, \frac{1}{k^a}+\frac{1}{(k+1)^a}+...> \int_{k}^{\infty}\frac{1}{x^a}dx=\frac{1}{(a-1)k^{a-1}}$$, so $$\zeta(a,k) >\frac{1}{(a-1)k^{a-1}}$$. In particular unless you impose a condition $$\Re s \ge a_0 >1$$ there is no such $$p$$, while for $$\Re s \ge a_0 >1$$, you need to take $$p > \frac{1}{a_0-1}$$
(which works since for $$k \ge 2, a>1, \frac{1}{k^a}+\frac{1}{(k+1)^a}+...< \int_{k-1}^{\infty}\frac{1}{x^a}dx=\frac{1}{(a-1)(k-1)^{a-1}}$$, hence for all $$k \ge 2$$, we have $$|\zeta(s,k)| \le \zeta(\Re{s},k) \le \frac{1}{(a_0-1)(k-1)^{a_0-1}}$$ , so $$p(a_0-1) >1$$ insures convergence for the required series as the first term doesn't matter for that)