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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?

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If Wolfram Alpha can't do it, there probably isn't a more elementary form. –  Potato Mar 24 '13 at 20:14
    
@MetinY. Yes, each term $\frac{(1/3)^n}{(1-(1/3)^n)}$ is the sum of a geometric series. –  Shuhao Cao Mar 24 '13 at 20:19
    
The value is known to be irrational, but in general no closed forms are known for these numbers; see mathworld.wolfram.com/Erdos-BorweinConstant.html for more details. –  Steven Stadnicki Mar 24 '13 at 20:40
    
@Steven: This is an answer, not a comment. –  Martin Brandenburg Mar 24 '13 at 20:53
    
@MartinBrandenburg I feel almost guilty about making it an answer because of how little content is going into it! But it's perhaps better that the question receive some answer, so I've gone ahead and converted it (along with a little bit of additional information to at least flesh it out). –  Steven Stadnicki Mar 24 '13 at 21:15

2 Answers 2

up vote 4 down vote accepted

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.

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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 $0.6821535026.$

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Of course to be precise it does not make sense to say that we compute an asymptotic expansion of the sum, as it is a number. What I mean and what I hope is clear is that we can compute the asymptotic expansion of a harmonic sum of which the sum in question is a special value. –  Marko Riedel Mar 25 '13 at 1:00

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