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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?

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    $\begingroup$ Good question. If instead of $2$, $3$ we take the golden ratio and its negative inverse, we end up with the reciprocal fibonacci constant, for which no closed form exists. This makes it unlikely that a closed form exists in this case, but still not impossible. $\endgroup$ – Josef E. Greilhuber Dec 2 at 11:00
  • $\begingroup$ "... the reciprocal fibonacci constant, for which no closed form exists. ..." This is not proved. You should say "no closed form is known" $\endgroup$ – jjagmath Dec 2 at 11:15
  • $\begingroup$ Is there any reason that you suspect that this sum should have a closed form? $\endgroup$ – Omnomnomnom Dec 2 at 11:42
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    $\begingroup$ This is of the form $\sum_{n=0}^\infty a^n\,\text{sech}(nx)$. According to Dieckman's tables, even with $a=1$ it involves hypergeometric functions. $\endgroup$ – Conifold Dec 2 at 11:54
  • $\begingroup$ @Omnomnomnom None really, only the fact that is a naturally arising question given the very familiar limit of $\sum\frac{1}{2^n}$. $\endgroup$ – Luke Collins Dec 2 at 12:44
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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.

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