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Can this series be expressed in closed form, and if so, what is it? $$ \sum_{n=1}^\infty\frac{1}{9^{n+1}-1} $$

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I think there is a closed form for such things using the digamma function, if that qualifies as closed form for you. – André Nicolas Sep 26 '12 at 23:27
@AndréNicolas well, I suppose it's better than nothing. How do you express it in terms of the digamma function? – Navin Sep 27 '12 at 0:49
Erdos asked whether numbers of that sort are irrational. Peter Borwein settled that question in a paper in the Journal of Number Theory, volume 37 (1991), no. 3, pages 253–259. – Gerry Myerson Sep 27 '12 at 0:51
@Navin: It was sort of a guess. Go to Maple, or Mathematica, or Alpha and it will tell you. – André Nicolas Sep 27 '12 at 1:22
I think all comments together make up an answer ... – mick Dec 13 '13 at 23:33
up vote 1 down vote accepted

Let us denote by $f$ the following function

$$ f(a):=\sum_{n=1}^\infty\frac{1}{a^{n+1}-1}. $$


$$f(a) = \sum_{n=1}^\infty\frac{1}{a^{n+1}-1}= \frac{1}{1-a}-\sum_{n=1}^\infty\frac{1}{1-a^n}\stackrel{\left(\spadesuit\right)}{=} \frac{1}{1-a}-\frac{\psi_{1/a}(1)+\ln(a-1)+\ln(1/a)}{\ln(a)},$$

where $\psi_q(z)$ denotes the $q$-polygamma function, and in $\left(\spadesuit\right)$ we used the equation $(4)$ from here.

$$ f(9) = \frac78 - \frac{\ln(8)}{\ln(9)} - \frac{\psi_{1/9}(1)}{\ln(9)} \approx 0.014045117662188129358728474369089\dots $$ Note that $f(2)=\mathcal{C}_{\textrm{EB}}-1 \approx 0.60669515241529\dots,$ where $\mathcal{C}_{\textrm{EB}}$ is the Erdős–Borwein constant.

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