How can we show that $\frac{x}{1-x} = \sum \frac{x^{2^n}}{1-x^{2^{n+1}}}$ for $0

Question

How can we show that $\frac{x}{1-x} = \sum \frac{x^{2^n}}{1-x^{2^{n+1}}}$

I've been stuck on this question. I've tried using the geometric series but to no avail.

Answer 1

Here we start like this : \begin{align*} s(x)& :=\sum_{n=0}^{\infty}\frac{x^{2^n}}{1-x^{2^{n+1}}}\\ & =\sum_{n=0}^{\infty}\frac{x^{2^n}+x^{2^{n+1}}-x^{2^{n+1}}}{1-x^{2^{n+1}}}\\ & =\sum_{n=0}^{\infty}\left(\frac{x^{2^n}}{1-x^{2^n}}-\frac{x^{2^{n+1}}}{1-x^{2^{n+1}}}\right)\\ & =\frac{x}{1-x}. \end{align*}