Well, if $p$ is an integer, you should realize that $frac{1}{(tau+a)^{p+1}}$ can be obtained by integrating $Q(x)e^{-2pi iax}$ against $e^{-2pi itau x}$ where $Q(x)$ is the (unique) polynomial of degree $p$ satisfying $Q^{(m)}(0)=e^{-2pi ia}Q^{(m)}(1)$ for $m<p$ and $Q^{(p)}=frac{(2pi i)^{p+1}}{e^{-2pi ia}-1}$.
So, your function is just $Q(x)e^{-2pi iax}$ on $(0,1)$ extended by periodicity to the entire line. The polynomial $Q$ can be easily found for each particular $p$, so if you need some small range of $p$, you have an exact closed form formula. If you want to consider large $p$, then it is not so useful but the origianal series gives you a high precision approximation if you keep just the first few terms. Either way, you have an "expression one can work with", don't you?
The thing that totally puzzles me is why you think that your series has any relation to the Hurwitz zeta function.
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