For $q>1$ and $s\ge{}1$, I'm trying to express
$$\sum_{n=1}^\infty \dfrac{1}{(1-q^n)^s}$$
in terms of the Qpolygamma function.
Just from its definition, it's clear to see that for $s$ the sum is
$$\sum_{n=1}^\infty \dfrac{1}{(1-q^n)^1} = \frac{1}{\log (q)} \left( \psi_{ \frac{1}{q} }^{ (0) }(1) + \log ( 1-\tfrac{1}{q} ) \right)$$
and, fixing a $q$ and using Mathematica, for higher $s$, there are also relatively nice expressions involving logs and phi's in that way.