Is it possible to express this sum in a closed form?

Question

Consider the sum $$\sum_{n=0}^\infty\frac1{2^n+3^n},$$ which clearly converges (by comparison with $\sum\frac1{2^n}$, say), and Mathematica approximates the limit to be $0.821354$.

Is there any way to write this in a closed form?

Answer 1

The closest thing I've found so far, with reference to Dieckman's tables, is that \begin{align*} \sum_{n=0}^\infty\frac{1}{2^n+3^n} &=\frac12\sum_{n=0}^\infty\big(\tfrac1{\sqrt6}\big)^n\operatorname{sech}(\tfrac n2\log\tfrac32)\\[5pt] &=\frac12\,{_2\phi_1}(\tfrac23,-1;-\tfrac23;\tfrac23;\tfrac13), \end{align*} where $_r\phi_s(a;b;q;z)$ is the basic hypergeometric series.