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
 A: The first term is $z/(1-z^2)$ which is a series where every exponent is odd.
What are the exponents in the series of the second term?
A: Hint: Let $\displaystyle F(z)=\sum_{n\geq 0}\frac{z^{2^n}}{1-z^{2^{n+1}}}$. Put $\displaystyle G(z)=F(z)-\frac{z}{1-z}$. Compare $G(z)$ and $G(z^2)$.
A: $$\sum_{n=0}^{\infty}\frac{z^{2^n}}{1-z^{2^{n+1}}}=\frac{z}{1-z^2}+\frac{z^2}{1-z^4}+\frac{z^4}{1-z^8}+\cdots$$
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 $|z|<1$):
$$\lim_{m \to \infty}\frac{-z^{2^{m}}}{1-z^{2^{m}}}=0$$
A: Hint: 

Every integer $k\geqslant1$ can be written in one and only one way as $k=2^n+i2^{n+1}=2^n(1+2i)$, for some $n\geqslant0$ and $i\geqslant0$.

To apply the hint, note that, for every $n\geqslant0$, $$\frac{z^{2^n}}{1-z^{2^{n+1}}}=z^{2^n}\sum_{i=0}^\infty\left(z^{2^{n+1}}\right)^i=\sum_{i=0}^\infty z^{2^n+i2^{n+1}},$$ hence the sum on $n\geqslant0$ of the LHS is the sum of $z^k$ on $k\geqslant1$, that is, $$\sum_{n=0}^\infty \frac{z^{2^n}}{1-z^{2^{n+1}}}=\sum_{k=1}^\infty z^k=\frac{z}{1-z}.$$
This only uses (but twice) the expansion $$\frac1{1-x}=\sum_{k=0}^\infty x^k.$$
To prove the hint, note that each integer $k\geqslant1$ is $k=2^np$ where $p$ is odd and positive and $2^n$ is the highest power of $2$ dividing $k$, that is, $k=2^n(1+2i)$ for some $i\geqslant0$, which proves simultaneously the existence and the uniqueness of this factorization.
