Let $\,{x\in\mathbb{C}}\,$, show that: $$ S_{\small-}=\sum_{n=1}^{\infty}\space\frac{2^{-n}}{{x}^{2^{-n}}+1}\,=\frac{1}{\log{x}}-\frac{1}{x-1}\qquad\qquad\qquad\tag{1} $$ $$ S_{\small+}=\sum_{n=0}^{\infty}\space\frac{2^{+n}}{{x}^{2^{+n}}+1}\,=\frac{1}{x-1}\qquad\qquad\qquad\qquad\qquad\tag{2} $$ $$ S\,\,=\sum_{-\infty}^{+\infty}\space\frac{2^n}{\,{x}^{2^n}\,\,\,+1}\,=S_{\small-}+S_{\small+}=\frac{1}{\log{x}}\qquad\qquad\qquad\tag{3} $$ And find the convergence range of each sum.
In a similar question, the OP start by introducing $\,{(2)}\,$ as a known identity, and gives a turly remarkable proof in the comment:
$$ \sum_{n=0}^m\log{(x^{2^n}+y^{2^n})} =\log{\prod_{n=0}^m(x^{2^n}+y^{2^n})} =\log{\frac{(x-y)}{x-y}\prod_{n=0}^m(x^{2^n}+y^{2^n})} =\log{\frac{x^{2^{m+1}}-y^{2^{m+1}}}{x-y}} $$
Then differentiate with respect to ${y}$, substitute ${y=1}$, and limit $\,{m\to\infty}\,$ give the result.
Could this approach work to proof $\,{(1)}\,$? notice for ${x=1}$:
$$ S_{\small-}=\sum_{n=1}^{\infty}\frac{1}{2^{n+1}}\,=\frac{1}{2} \qquad=\qquad \lim_{\small x\to1}\left[\frac{1}{\log{x}}-\frac{1}{x-1}\right]=\frac{1}{2} $$
New alternative methods are highly welcomed. Thanks.
$$\sum_{n=0}^m \frac{2^n y^{2^n}}{x^{2^n}+y^{2^n}} = \frac{y}{x-y}-\frac{2^{m+1}y^{2^{m+1}}}{x^{2^{m+1}}-y^{2^{m+1}}} $$
For $y=1$ and $m \rightarrow +\infty$ we get what you desire.
– Tolaso Aug 26 '18 at 21:45