<s>[INST] Calculation of a series It seems that we have:
$$\sum_{n\geq 1} \frac{2^n}{3^{2^{n-1}}+1}=1.$$
Please, how can one prove it?
 [/INST] 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 $(*)$.
</s>