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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 all comments together make up an answer ...2013-12-13

1 Answers 1

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Let us denote by $f$ the following function

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

Then

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

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.