Closed form solution for infinite summation Thomas, Bruckner & Bruckner, Elementary Real Analysis.
Prove that for all r > 1, 
$$\frac{1}{r - 1} = \frac{1}{r+1} + \frac{2}{r^2 + 1} + \frac{4}{r^4 + 1} + \frac{8}{r^8 + 1}  + \cdots$$
So far I have $$ \frac{1}{r-1} -\frac{1}{r+1} = \frac{2}{r^2 -1} $$
$$\sum_{n=1}^\infty \frac{2^n}{r^{2^n} + 1} = \sum_{n=1}^\infty \left(\frac{2^n}{r^{2^n}} - \frac{2^n}{r^{4^n} + r^{2^n}}\right)$$

Rinse and repeat the first line below
$$\begin{align}
{1\over \color{blue}{r}-1}&={1\over \color{blue}{r}+1}+{2\over \color{blue}{r}^2-1}\\
&={1\over r+1}+2\left({1\over\color{blue}{r^2}+1}+{2\over (\color{blue}{r^2})^2-1}\right)\\
&={1\over r+1}+\frac{2}{r^2+1}+\frac{4}{\color{blue}{r^4}-1}\\
&={1\over r+1}+\frac{2}{r^2+1}+4\left(\frac{1}{\color{blue}{r^4}+1}+\frac{2}{(\color{blue}{r^4})^2-1}\right)\\
&=\cdots
\end{align}$$
