Similar concepts could be found here:
Riemann zeta function and Hurwitz zeta function, where Riemann zeta function was of the form $\zeta(s)=\sum_{n=1}^\infty \frac{1}{n^s}$.
Is there any generic way to obtain $\sum_{n=0}^\infty \frac{1}{(kn+q)^s}$ from Riemann zeta function directly?
Specifically, how to express $\sum_{n=0}^\infty \frac{1}{(kn+1)^s}$ directly as an expression of Riemann zeta function?
Best Answer
As EricTowers wrote $$\sum_{n=0}^\infty \frac{1}{(kn+q)^s}=k^{-s} \,\zeta \left(s,\frac{q}{k}\right)$$ Let $a=\frac{q}{k}$ an expand as a series $$\zeta (s,a)=a^{-s} +\zeta (s)-a s \zeta (s+1)+\frac{1}{2} a^2 s (s+1) \zeta (s+2)-\frac{1}{6} a^3 s (s+1) (s+2) \zeta (s+3)+O\left(a^4\right)$$ that is to say $$\zeta (s,a)=a^{-s}+s\sum_{n=0}^\infty (-1)^n\frac{ (s+1)_{n-1} }{n!}\zeta (n+s)\, a^n$$