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?
On
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 $$0.126390773436878184483281022628894471168769898542838$$
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.