Summation of a geometric like series I wonder if there is a formula to calculate a series as following:
$\sum_{n=0}^{\infty}\frac{1}{q^n-x},$
where $q>1$ and $x<1$. Obviously, this series converges. If $x=0$, it is actually a geometric series. But when $x\neq0$, it seems much difficult. I just wonder if there is an analytic solution? 
Thanks!
 A: The result given by Wolfram Alpha is not bad
$$\sum_{n=0}^{\infty}\frac{1}{q^n-x}=-\frac{2 \psi _q^{(0)}\left(-\frac{\log (x)}{\log (q)}\right)+\log \left(q(q-1)^2 
   x^2\right)}{2 x \log (q)}$$ where appears the q-digamma function.
What you could also to is to write
$$\frac{1}{q^n-x}=\sum_{p=0}^\infty q^{-n(1+p)} x^p$$ and reversing the order of summation
$$\sum_{n=0}^\infty q^{-n(1+p)} x^p=\frac{x^p}{1-\left(\frac{1}{q}\right)^{p+1}}$$making
$$\sum_{n=0}^{\infty}\frac{1}{q^n-x}=\sum_{p=0}^{\infty}\frac{x^p}{1-\left(\frac{1}{q}\right)^{p+1}}=\sum_{p=0}^{\infty}\frac{q (qx)^p}{q^{p+1}-1}$$ which would converge quite fast is $x \ll q$.
For illustration purposes, let us try for $x=\frac 12$ and $q=2$ and consider the sumation up to $p=10$. This would lead to
$$\frac{111324238097962}{34654391537535}\approx 3.21241$$ while the exact value would be $\approx 3.21339$ which is in a relative error of $3\times 10^{-2}$%.
Pushing the summation up to $p=20$ reduces the relative error to $3\times 10^{-5}$%.
