Closed form sum of $\sum^{\infty}_{n=1} \frac{1}{3^n-1}$

Question

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?

Answer 1

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.

Answer 2

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 $0.6821535026.$