Closed form sum of $\sum^{\infty}_{n=1} \frac{1}{3^n-1}$ Wolframalpha uses $q$-Polygamma function to represent the sum, hence essentially does nothing. Here I wonder if this sum can be represented by elementary function.
The summation is like a infinite summation of geometric series:
$$
\begin{aligned}
\sum^{\infty}_{n=1} \frac{1}{3^n-1} =& \sum^{\infty}_{n=1} \frac{(\frac{1}{3})^n}{1-(\frac{1}{3})^n} 
\\
=& \left(\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \cdots\right)
\\
+& \left(\frac{1}{3^2} + \frac{1}{(3^2)^2} + \frac{1}{(3^2)^3} + \cdots\right)
\\
+& \left(\frac{1}{3^3} + \frac{1}{(3^3)^2} + \frac{1}{(3^3)^3} + \cdots\right)
\\
+& \cdots
\end{aligned}
$$
Is there any special technique to sum this kind of series?

Generically, constants of this form ($\displaystyle\sum_n\dfrac{1}{a^n-b}$) are known to be irrational, but AFAIK closed forms aren't known; see http://mathworld.wolfram.com/Erdos-BorweinConstant.html for some of the details.  Note that the Erdos-Borwein constant itself (which is your sum with $2^n-1$ in place of $3^n-1$) arises in the analysis of selection sort; see the end of section 5.2.3 of The Art Of Computer Programming, and particularly exercise 27, for the details on how it arises in that analysis.
