TITLE: How to calculate $\sum\limits_{n=1}^{\infty}\frac{1}{3^n - 2^n}$
QUESTION [7 upvotes]: By the test of reason, this series converges. 
The problem is figuring out which technique to use to calculate your sum.
Thanks for any help.

REPLY [5 votes]: You can express the sum in terms of the $q$-digamma function $\psi_q(z)$.  This isn't a profound simplification, however, because $\psi_q(z)$ is defined as a sum of a similar form:
$$
\psi_q(z) = \frac{\partial \log \Gamma_q(z)}{\partial z} = -\log(1-q) + \log(q) \sum_{n=0}^\infty \frac{q^{n+z}}{1-q^{n+z}}.
$$
But, for what it's worth,
$$
\sum_{n=1}^\infty \frac{1}{3^n-2^n} = \frac{\log3-\psi_{2/3}(\log_{3/2}3)}{\log(3/2)}.
$$