<s>[INST] How to prove $\sum_{k=0}^\infty \frac{2^k}{2^{2^k}+1}=1$? How can I prove this result?
$$\sum_{k=0}^\infty \frac{2^k}{2^{2^k}+1}=1$$
The sum converges very quickly: The term at $k=4$ is already smaller than $2^{-12}$ and each further term is much smaller than its predecessor.
I've tried replacing the $2$'s by $x$'s and looking for power series, as well as seeing if the sum telescopes, but nothing so far has worked. 
 [/INST] Make it into a telescopic sum as follows
\begin{align*}
\sum_{k=0}^n \frac{2^k}{2^{2^k}+1}&=\sum_{k=0}^n \frac{2^k(2^{2^k}-1)}{(2^{2^{k+1}}-1)}\\
&=\sum_{k=0}^n \frac{2^k(2^{2^k}+1-2)}{(2^{2^{k+1}}-1)}\\
&=\sum_{k=0}^n \frac{2^k(2^{2^k}+1)}{(2^{2^{k+1}}-1)}-\frac{2^{k+1}}{(2^{2^{k+1}}-1)}\\
&=\sum_{k=0}^n \frac{2^k}{(2^{2^{k}}-1)}-\frac{2^{k+1}}{(2^{2^{k+1}}-1)}\\
&=1-\frac{2^{n+1}}{(2^{2^{n+1}}-1)}\rightarrow 1
\end{align*}
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