# Is there any generic way to obtain $\sum_{n=0}^\infty \frac{1}{(kn+q)^s}$ from Riemann zeta function?

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?

• Do you intend that $k$ and $q$ are integers, or can they be more general numbers? Commented Nov 11, 2019 at 2:58
• @EricTowers Right now I only consider integers, but if there's a formula it should work for reals. Commented Nov 11, 2019 at 3:07

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 $$\zeta (s,a)=a^{-s}+s\sum_{n=0}^\infty (-1)^n\frac{ (s+1)_{n-1} }{n!}\zeta (n+s)\, a^n$$
For $$k$$ and $$q$$ real and positive, we should expect $$\sum_{n=0}^\infty \frac{1}{(kn+q)^s} = \frac{1}{k^s} \sum_{n=0}^\infty \frac{1}{\left( n+\frac{q}{k} \right)^s} = \frac{1}{k^s} \zeta(s,q/k) \text{.}$$
There's the multiplication theorem. But this is probably exactly backwards from what you want (it starts with the zeta series, then groups the terms with congruent $$n \pmod{k}$$). It is $$k^s \zeta(s) = \sum_{q=1}^k \zeta \left(s, \frac{q}{k} \right) \text{.}$$