While searching the internet for the Hurwitz zeta function I found this Mathematics stack exchange question:
The multiplication formula for the Hurwitz zeta function
$$\zeta(s,mz) = \frac{1}{m^{s}} \sum_{k=0}^{m-1} \zeta \left(s,z+\frac{k}{m} \right) \tag{1}$$
This is almost identical to the part $$\frac{1}{k^s}\sum _{n=1}^{k-1} \zeta \left(s,\frac{n}{k}\right)n \tag{2}$$ in the Dirichlet generating function for:
$$\sum _{k=1}^{\infty} \frac{\sum _{n=1}^k \sigma _0(n)}{k^s}=-\sum _{k=1}^{\infty} \frac{1}{k} \frac{1}{k^s}\sum _{n=1}^{k-1} \zeta \left(s,\frac{n}{k}\right)n+\sum _{k=1}^{\infty} \frac{1}{k}\zeta (s-1) \tag{3}$$
where $\sigma _0(n) = 1, 2, 2, 3, 2, 4, 2, 4, 3, 4, 2, 6,...$ is the number of divisors function, and $(3)$ is valid for $s>2$ I believe.
What I would like to do is to extend the relation $(3)$ to hold for $s=0$ (and $z=0$), but I don't know if that is possible.
The first step towards that would be to ask if: $$\frac{1}{m^{s}}\sum_{k=0}^{m-1} \zeta \left(s,z+\frac{k}{m} \right)k \tag{4}$$ can be in written in terms of the zeta function $\zeta(s,mz)$ as in the left hand side of $(1)$? I suspect that it is not possible.
