# Proving an inequality involving the summation of Riemann zeta function.

If $0\lt a \le 1, s\gt 1,$ define $\zeta(s,a)=\sum (n+a)^{-s}$.

Show that this series converges absolutely for $s \gt 1$ and prove that

$$\sum_{h=1}^k \zeta(s,\frac{h}{k})=k^s\zeta (s)\ \ \ \text{if} \ k=1,2,...,$$

where $\zeta(s)=\zeta(s,1)$ is the Riemann zeta function.

I know that the series converges absolutely for $s \gt 1$ from the p-series test.

$\sum_{h=1}^k \zeta(s,\frac{h}{k})=\sum_{h=1}^k \sum_{n=0}^\infty \frac{k^s}{(kn+h)^s}=k^s \sum_{n=0}^\infty \sum_{h=1}^k \frac{1}{(kn+h)^s}$.

I am stuck here. How can I show that this last summation equals $k^s \sum_{n=0}^\infty \frac{1}{(n+1)^s}$?

I would greatly appreciate any help.

We have $$\sum_{h=1}^{k}\zeta\left(s,\frac{h}{k}\right)=\sum_{h=1}^{k}\sum_{n\geq0}\frac{k^{s}}{\left(nk+h\right)^{s}}=k^{s}\sum_{n\geq0}\sum_{h=0}^{k-1}\frac{1}{\left(nk+h+1\right)^{s}}.$$ Now note that every positive integer $a$ can be expressed in the form $a=nk+h$ for some $n\in\mathbb{N}$ and $h\in\left[0,k-1\right]$. It's simply $a\textrm{ mod }k$. Then $$k^{s}\sum_{n\geq0}\sum_{h=0}^{k-1}\frac{1}{\left(nk+h+1\right)^{s}}=k^{s}\sum_{n\geq0}\frac{1}{\left(n+1\right)^{s}}=k^{s}\zeta\left(s\right).$$