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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?

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

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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 $$\zeta (s,a)=a^{-s}+s\sum_{n=0}^\infty (-1)^n\frac{ (s+1)_{n-1} }{n!}\zeta (n+s)\, a^n$$

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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 $$ k^s \zeta(s) = \sum_{q=1}^k \zeta \left(s, \frac{q}{k} \right) \text{.} $$

If the relation you want were known, I imagine

Ashton, A.C.L, and A.S. Fokas, "Relations among the Riemann Zeta and Hurwitz Zeta Functions, as Well as Their Products", Symmetry 2019, 11, 754; doi:10.3390/sym11060754

from June of this year would have been much, much shorter.

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