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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$\begingroup$ I think there is a closed form for such things using the digamma function, if that qualifies as closed form for you. $\endgroup$– André NicolasCommented Sep 26, 2012 at 23:27
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$\begingroup$ @AndréNicolas well, I suppose it's better than nothing. How do you express it in terms of the digamma function? $\endgroup$– NavinCommented Sep 27, 2012 at 0:49
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$\begingroup$ 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. $\endgroup$– Gerry MyersonCommented Sep 27, 2012 at 0:51
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$\begingroup$ @Navin: It was sort of a guess. Go to Maple, or Mathematica, or Alpha and it will tell you. $\endgroup$– André NicolasCommented Sep 27, 2012 at 1:22
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1$\begingroup$ I think all comments together make up an answer ... $\endgroup$– mickCommented Dec 13, 2013 at 23:33
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1 Answer
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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.
answered Dec 28, 2015 at 18:51