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 $$ \frac{1}{r-1} -\frac{1}{r+1} = \frac{2}{r^2 -1} $$
$$\sum_{n=1}^\infty \frac{2^n}{r^{2^n} + 1} = \sum_{n=1}^\infty \left(\frac{2^n}{r^{2^n}} - \frac{2^n}{r^{4^n} + r^{2^n}}\right)$$

For any $x\in(0,1)$ we have
$$\frac{1}{1-x}= (1+x)(1+x^2)(1+x^4)(1+x^8)\cdots \tag{1} $$
that can be read as every positive integer has a unique representation in base-$2$.
By taking $\frac{d}{dx}\log(\cdot)$ of both sides, we get:
$$ \frac{1}{1-x}=\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\frac{8x^7}{1+x^8}+\cdots \tag{2} $$
and it is enough to evaluate the last identity at $x=\frac{1}{r}$ to prove the given claim, after trivial simplifications.
