Finding the value of $\sum\limits_{k=0}^{\infty}\frac{2^{k}}{2^{2^{k}}+1}$ Does this weighted sum of reciprocals of Fermat numbers,
$$
F=\sum_{k=0}^{\infty}\dfrac{2^{k}}{2^{2^{k}}+1}
$$
have a nice closed form? Wolfram says it's $1$.
Thanks.
 A: Use the identity
$$\frac{1}{t+1} = \frac{1}{t-1} - \frac{2}{t^2 -1}.$$
A: Hint:
Try to guess and prove a formula for partial sums
$$
S(n)=\sum_{k=0}^n\frac{2^k}{2^{2^k}+1}.
$$
Here
$$
S(0)=\frac13,\ S(1)=\frac{11}{15},\ S(2)=\frac{247}{255},\ldots
$$
See a pattern for $1-S(n)$?
A: This might be another way of looking at Jyrki's hint, but here is the way I did this:
$$
\begin{align}
\color{#C0C0C0}{1-\frac1{2-1}+}\frac1{2+1}&=1-\frac2{4-1}\\
1-\frac{2}{4-1}+\frac{2}{4+1}&=1-\frac{4}{16-1}\\
1-\frac{4}{16-1}+\frac{4}{16+1}&=1-\frac8{256-1}\\
1-\frac8{256-1}+\frac8{256+1}&=1-\frac{16}{65536-1}\\
&\vdots
\end{align}
$$

Motivation behind this approach
One trick to try is conjugates. Noting that
$$
\frac{2^k}{2^{2^k}-1}-\frac{2^k}{2^{2^k}+1}=\frac{2^{k+1}}{2^{2^{k+1}}-1}
$$
we see, using Telescoping Series, that
$$
\begin{align}
\sum_{k=0}^{n-1}\frac{2^k}{2^{2^k}+1}
&=\sum_{k=0}^{n-1}\left(\frac{2^k}{2^{2^k}-1}-\frac{2^{k+1}}{2^{2^{k+1}}-1}\right)\\
&=1-\frac{2^n}{2^{2^n}-1}
\end{align}
$$
Therefore, letting $n\to\infty$,
$$
\sum_{k=0}^\infty\frac{2^k}{2^{2^k}+1}=1
$$
