Show $\sum_n \frac{z^{2^n}}{1-z^{2^{n+1}}} = \frac{z}{1-z}$ Show $\displaystyle\sum_{n=0}^\infty \frac{z^{2^n}}{1-z^{2^{n+1}}} = \frac{z}{1-z}$ for $|z|<1$.
This is an additional problem for my complex analysis class and I've attempted it for a few hours but ended up taking wrong routes. All of my attempts I haven't used complex analysis at all and I don't see how I could here.
edit: this is meant for a BEGINNER complex analysis course so please try keep the solutions to that (if you could)
Any help would be great

$$\sum_{n=0}^{\infty}\frac{z^{2^n}}{1-z^{2^{n+1}}}=\frac{z}{1-z^2}+\frac{z^2}{1-z^4}+\frac{z^4}{1-z^8}+\cdots$$
Add $\displaystyle \frac{-z}{1-z}$ to both sides. It's Telescoping series:
$$\frac{-z}{1-z}+\sum_{n=0}^{\infty}\frac{z^{2^n}}{1-z^{2^{n+1}}}=\frac{-z}{1-z}+\frac{z}{1-z^2}+\frac{z^2}{1-z^4}+\frac{z^4}{1-z^8}+\cdots=\\ =\frac{-z-z^2}{1-z^2}+\frac{z}{1-z^2}+\frac{z^2}{1-z^4}+\frac{z^4}{1-z^8}+\cdots= \\ = \frac{-z^2}{1-z^2}+\frac{z^2}{1-z^4}+\frac{z^4}{1-z^8}+\cdots= \\ = \frac{-z^4-z^2}{1-z^4}+\frac{z^2}{1-z^4}+\frac{z^4}{1-z^8}+\cdots=\frac{-z^4}{1-z^4}+\frac{z^4}{1-z^8}+ \cdots$$
$\textbf{Edit:}$ If you want to have finite sum rather than infinity. You can show that (like above - by cancelling terms) :
$$\frac{-z}{1-z}+\sum_{n=0}^{m}\frac{z^{2^n}}{1-z^{2^{n+1}}}=\frac{-z^{2^{m}}}{1-z^{2^{m}}}$$
Next calculate limit for $m \to \infty$ both sides.
$$\lim_{m \to \infty}\frac{-z}{1-z}+\sum_{n=0}^{m}\frac{z^{2^n}}{1-z^{2^{n+1}}}=\frac{-z}{1-z}+\sum_{n=0}^{\infty}\frac{z^{2^n}}{1-z^{2^{n+1}}}$$
Finally (because $|z|<1$):
$$\lim_{m \to \infty}\frac{-z^{2^{m}}}{1-z^{2^{m}}}=0$$
