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After looking at the question here Computing summation I wondered if it might be possible to evaluate the following summation with a similar-looking summand term but with $2n$ instead of $2^n$:

$$\sum_{n=1}^{\infty}\frac 1{3^{2n}-3^{-2n}}$$

Although the summand has fewer levels of exponentiation, it may not be as easy to sum. I simulated this on Excel and found that it converges numerically to 0.126391. Would anyone be able to evaluate the summation algebraically?

Community
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Hypergeometricx
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2 Answers2

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We have: $$\sum_{n=1}^{+\infty}\frac{1}{9^n-9^{-n}}=\sum_{n=1}^{\infty}\frac{1}{9^n}\cdot\frac{1}{1-\frac{1}{9^{2n}}}=\sum_{n=1}^{+\infty}\sum_{j=0}^{+\infty}\frac{1}{9^{(2j+1)n}}=\sum_{j=0}^{+\infty}\frac{1}{9^{2j+1}-1}=\sum_{m=1}^{+\infty}\frac{d_1(m)}{9^m}$$ where $d_1(m)$ accounts for the number of odd positive divisors of $m$.

I doubt that a "closer" form exists.

Jack D'Aurizio
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  • Thank you for your answer. A closed form solution would be good though rather than a summation. Have upvoted your answer anyway. – Hypergeometricx Oct 01 '14 at 01:46
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I am surprised but there is a closed form (don't ask me any question - a CAS made it) $$\sum_{n=1}^{+\infty}\frac{1}{9^n-9^{-n}}=\frac{-\psi _9^{(0)}(1)+\psi _9^{(0)}\left(1-\frac{i \pi }{\log (9)}\right)+i \pi }{\log (81)}$$ where appears the q-digamma function. Its numerical value is approximately $$0.126390773436878184483281022628894471168769898542838$$

Claude Leibovici
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  • Thank you for your answer. The closed form does indeed give the same numerical limit as found on Excel (to 15 decimal places). It would be good to know how to get to the closed form though. I have upvoted your answer. – Hypergeometricx Oct 01 '14 at 01:42