1
$\begingroup$

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 $$ \zeta(s,q) = \frac{1}{\Gamma(s)}\int_0^\infty \frac{t^{s-1}e^{-qt}}{1-e^{-t}}dt. $$

$\endgroup$
1
  • $\begingroup$ 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)})$ $\endgroup$
    – reuns
    Sep 4, 2019 at 3:30

1 Answer 1

2
$\begingroup$

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)

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .