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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.

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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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