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?
 A: We can compute an asymptotic expansion of this sum using Mellin transforms. Let $f(x)$ be the base function $$f(x) = \frac{1}{3^x-1}.$$ The Mellin transform $f^*(s)$ of $f(x)$ is
$$ \mathfrak{M}(f(x); s) = f^*(s) = \int_0^\infty  \frac{x^{s-1}}{3^x-1} dx
= \int_0^\infty \frac{1}{3^x} \frac{1}{1-3^{-x}} x^{s-1} dx \\ =
\int_0^\infty \frac{1}{3^x} \left(\sum_{q\ge 0} 3^{-qx} \right)  x^{s-1} dx =
\sum_{q\ge 0} \int_0^\infty 3^{-(q+1)x}  x^{s-1} dx \\= 
\Gamma(s) \sum_{q\ge 0} \frac{1}{(\log 3)^s (q+1)^s} =
\frac{1}{(\log 3)^s} \Gamma(s) \zeta(s). $$
It follows that the Mellin transform of the harmonic sum
$$ g(x) = \sum_{n\ge 1} \frac{1}{3^{nx}-1}$$ is
$$  \mathfrak{M}(g(x); s) = g^*(s) = \frac{1}{(\log 3)^s} \Gamma(s) \zeta(s)^2.$$
Now invert to get the sum. We list the contributions from the main poles. Sum these to get the aymptotic expansion.
$$\begin{array}
\operatorname{Res}(g^*(x) x^{-s}; s=1) & = & 
 \frac{1}{\log 3} \left( (\gamma - \log \log 3) \frac{1}{x} - \frac{\log x}{x} \right)\\
\operatorname{Res}(g^*(x) x^{-s}; s=0) & = & \frac{1}{4} \\
\operatorname{Res}(g^*(x) x^{-s}; s=-1) & = & -{\frac {1}{144}}\,\log  \left( 3 \right) x \\
\operatorname{Res}(g^*(x) x^{-s}; s=-3) & = & 
-{\frac {1}{86400}}\,{x}^{3} \left( \log  3  \right) ^{3} \\
\operatorname{Res}(g^*(x) x^{-s}; s=-5) & = & 
-{\frac {1}{7620480}}\,{x}^{5} \left( \log 3 \right)^5.
\end{array}$$
This partial expansion is already quite good, it gives $0.6821535092$ at $x=1$ whereas the precise value is $0.6821535026.$
A: 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.
