# Prove the following equation of complex power series.

Show that for $|z| \lt 1$ with $z \in \Bbb C$, we have

$$\sum_0^\infty \frac{{z^2}^k}{1-{z^2}^{k+1}} = \frac{z}{1-z}$$

$$\sum_0^\infty \frac{2^k{z^2}^k}{1+{z^2}^{k}} = \frac{z}{1-z}$$

My guess is that the second one is obtained by derivating the first one or something like that, but I can't manage to prove the first one.

-
Stop doing stupid edits! – user26857 Feb 4 '13 at 23:56
What YACP means is: stop vandalizing the post. – Cameron Buie Feb 5 '13 at 0:47

## 1 Answer

We have $$\sum_{k=0}^{\infty}\dfrac{z^{2^k}}{1-z^{2^{k+1}}} = \sum_{k=0}^{\infty}\sum_{l=0}^{\infty} z^{2^k + l \cdot 2^{k+1}} = \sum_{l=0}^{\infty} \sum_{k=0}^{\infty}z^{2^k(2l+1)} = \sum_{m=1}^{\infty} z^m$$ since there is a unique decomposition of any number $m \in \mathbb{Z}^+$ as $2^k(2l+1)$. But $$\sum_{m=1}^{\infty} z^m = \dfrac{z}{1-z}$$ Hence, we get that $$\sum_{k=0}^{\infty}\dfrac{z^{2^k}}{1-z^{2^{k+1}}} = \dfrac{z}{1-z}$$

-