How to calculate zeta-like function $\sum_{n=k}^\infty \frac{1}{(n+a)^s}$

Question

Here I want to calculate a zeta-like function $\sum_{n=k}^\infty \frac{1}{(n+a)^s}$ where $k \in \mathbb{N}$, $s>1$ and $0<a<1$. I usually calculate the Riemann zeta function by the Poisson summation. The problem is that since there is a small translation and it's not summing over the whole integer, I cannot directly use the Poisson summation for calculation. Is there any way to calculate this function? Or what theory should I refer?

Thank you so much for your help!

Answer 1

For $|a|< 1$ $$\sum_{n\ge 1} (n+a)^{-s} = \sum_{m\ge 0} {-s \choose m} a^m \zeta(s+m)$$ This is rapidly convergent if you know how to calculate $\zeta(s)$ and if you don't take $|s|,|a|$ too large.

Answer 2

This is the definition of the Hurwitz zeta function $$\sum_{n=k}^\infty \frac{1}{(n+a)^s}=\zeta (s,a+k)$$ Its integral representation is $$\zeta (s,a+k)=\frac{1}{\Gamma (s)}\int_0^\infty \frac{t^{s-1} }{1-e^{-t}}e^{-(a+k)t}\,dt$$