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I have known that $\displaystyle\sum_{n=0}^\infty\dfrac{2^n}{x^{2^{n}}+1}=\dfrac{1}{x-1}$ for $x>1$.

Taking the derivative of both sides , $\displaystyle\sum_{n=0}^\infty\dfrac{4^nx^{2^n-1}}{(x^{2^{n}}+1)^2}=\dfrac{1}{(x-1)^2}$ .

This impiles $\displaystyle\sum_{n=0}^\infty\dfrac{4^n}{x^{2^{n}}+1}-\sum_{n=0}^\infty\dfrac{4^n}{(x^{2^{n}}+1)^2}=\dfrac{x}{(x-1)^2}$ .

Replacing $x$ by $2$ will get $\displaystyle\sum_{n=0}^\infty\dfrac{4^n}{2^{2^{n}}+1}-\sum_{n=0}^\infty\dfrac{4^n}{(2^{2^{n}}+1)^2}=2$ .

But I have no idea to find $\sum_{n=0}^\infty\dfrac{4^n}{2^{2^{n}}+1}$ . Thank for helping.

Bless
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  • Also used \displaystyle in title, another no-no. I've fixed. @NormalHuman – Thomas Andrews Oct 23 '15 at 17:18
  • Your title series converges to something like $0.327443470\ldots$ so it doesn't look like a simple fraction. BTW: your summations should start at $n=0$. – Intelligenti pauca Oct 23 '15 at 21:19
  • Do you know of a proof for $\displaystyle\sum_{n=0}^\infty\dfrac{2^n}{x^{2^{n}}+1}=\dfrac{1}{x-1}$ ? – Lucian Oct 24 '15 at 09:26
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    From $\sum_{n=0}^m\ln{(x^{2^n}+y^{2^n})}=\ln{\prod_{n=0}^m(x^{2^n}+y^{2^n})} = \ln{\frac{(x-y)}{x-y}\prod_{n=0}^m(x^{2^n}+y^{2^n})}= \ln{\frac{x^{2^{m+1}}-y^{2^{m+1}}}{x-y}}$. Then take derivative with respect to y and substitue $y=1$ and take limit $m\to\infty$ will get the result. – Bless Oct 24 '15 at 11:25

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