It seems that we have:
$$\sum_{n\geq 1} \frac{2^n}{3^{2^{n-1}}+1}=1.$$
Please, how can one prove it?
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13$\begingroup$ It may help to know how this question arises. $\endgroup$– Iosif PinelisCommented Apr 4, 2021 at 16:14
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1 Answer
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This is the special case $q=3$ of a formula $$ \qquad\qquad \sum_{n=1}^\infty \frac{2^n}{q^{2^{n-1}}+1} = \frac{2}{q-1} \qquad\qquad(*) $$ which holds for all $q$ such that the sum converges, i.e. such that $|q|>1$. This follows from the identity $$ \frac{1}{x-1} - \frac{2}{x^2-1} = \frac{1}{x+1}. $$ Substitute $q^{2^{n-1}}$ for $x$, multiply by $2^n$, and sum from $n=1$ to $n=N$ to obtain the telescoping series $$ \frac{2}{q-1} - \frac{2^{N+1}}{q^{2^N}-1} = \sum_{n=1}^N \left( \frac{2^n}{q^{2^{n-1}}-1} - \frac{2^{n+1}}{q^{2^n}-1} \right) = \sum_{n=1}^N \frac{2^{n-1}} {q^{2^n} + 1}. $$ Taking the limit as $N \to \infty$ yields the claimed formula $(*)$.
answered Apr 4, 2021 at 16:22
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3$\begingroup$ (+1) I had, only two days prior, essentially answered this question in chat. I suggested a similar approach. $\endgroup$– robjohnCommented Apr 6, 2021 at 3:53
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$\begingroup$ Maybe I misread something -- but shouldn't the last term in the chain of equalities read $\sum_{n=1}^N \frac{2^n}{q^{2^{n-1}}+1}$? $\endgroup$– Stefan Kohl ♦Commented May 8, 2021 at 18:38