Series of rational fractions I read in Gradshteyn I , Ryzhik I Table Of Integrals, Series And Products (7Ed , Elsevier, 2007)(Isbn 0123736374)(1220S)the following series
$$-\frac{x}{x-1}=\sum _{k=0}^{\infty } -\frac{x^{2^k}}{x^{2^{k+1}}-1}=\sum _{k=1}^{\infty } \frac{2^k x^{2^k-1}}{x^{2^k-1}+1}$$
the serie $$\frac{2^k x^{2^k}}{x^{2^k}+1}$$ it is easy calculating using Van Wijngaarden's trick positive series in alternate serie but when come from the second series??
 A: We show the following is valid for |x|<1:
\begin{align*}
\frac{x}{1-x}&=\sum_{k=0}^\infty\frac{2^kx^{2^{k}}}{1+x^{2^{k}}}\tag{1}\\
&=\sum_{k=0}^{\infty}\frac{x^{2^{k}}}{1-x^{2^{k+1}}}\tag{2}\\
\end{align*}
Since we have
\begin{align*}
\frac{2^{k+1}x^{2^{k+1}}}{1-x^{2^{k+1}}}=\frac{2^kx^{2^k}}{1-x^{2^k}}-\frac{2^kx^{2^k}}{1+x^{2^k}}
\end{align*}

we obtain by telescoping
\begin{align*}
\color{blue}{\sum_{k=0}^\infty}&\color{blue}{\frac{2^kx^{2^k}}{1+x^{2^k}}}\\
&=\lim_{N\to\infty}\sum_{k=0}^N\frac{2^kx^{2^{k}}}{1+x^{2^{k}}}\\
&=\lim_{N\to\infty}\left(\sum_{k=0}^N\frac{2^kx^{2^{k}}}{1-x^{2^{k}}}-\sum_{k=0}^N\frac{2^{k+1}x^{2^{k+1}}}{1-x^{2^{k+1}}}\right)\\
&=\lim_{N\to\infty}\left(\sum_{k=0}^N\frac{2^kx^{2^{k}}}{1-x^{2^{k}}}-\sum_{k=1}^{N+1}\frac{2^kx^{2^{k}}}{1-x^{2^{k}}}\right)\\
&=\lim_{N\to\infty}\left(\frac{x}{1-x}-\frac{2^{N+1}x^{2^{N+1}}}{1-x^{2^{N+1}}}\right)\tag{$2^{N+1}x^{2^{N+1}}\to0$}\\
&\,\,\color{blue}{=\frac{x}{1-x}}
\end{align*}
and the claim (1) follows.

Since we have
\begin{align*}
\frac{x^{2^{k}}}{1-x^{2^{k+1}}}&=\frac{\left(1+x^{2^{k}}\right)-1}{1-x^{2^{k+1}}}=\frac{1}{1-x^{2^{k}}}-\frac{1}{1-x^{2^{k+1}}}
\end{align*}

we obtain again by telescoping
\begin{align*}
\color{blue}{\sum_{k=0}^\infty}&\color{blue}{\frac{x^{2^k}}{1+x^{2^{k+1}}}}\\
&=\lim_{N\to\infty}\sum_{k=0}^N\frac{x^{2^{k}}}{1+x^{2^{k+1}}}\\
&=\lim_{N\to\infty}\left(\sum_{k=0}^N\frac{1}{1-x^{2^{k}}}-\sum_{k=0}^N\frac{1}{1-x^{2^{k+1}}}\right)\\
&=\lim_{N\to\infty}\left(\sum_{k=0}^N\frac{1}{1-x^{2^{k}}}-\sum_{k=1}^{N+1}\frac{1}{1-x^{2^{k}}}\right)\\
&=\lim_{N\to\infty}\left(\frac{1}{1-x}-\frac{1}{1-x^{2^{N+1}}}\right)\tag{$x^{2^{N+1}}\to0$}\\
&=\frac{1}{1-x}-1\\
&\,\,\color{blue}{=\frac{x}{1-x}}
\end{align*}
and the claim (2) follows.

Note: We find in edition 07 of Table of integrals, series, and products by I.S. Gradshteyn and I.M. Ryzhik formula 1.121(1.):
\begin{align*}
\frac{x}{1-x}=\sum_{k=1}^\infty\frac{2^{k-1}x^{2^{k-1}}}{1+x^{2^{k-1}}}=\sum_{k=1}^\infty\frac{x^{2^{k-1}}}{1-x^{2^{k}}}\qquad\qquad[x^2<1]
\end{align*}
