complex power series Sorry for taking this from another question, but the second part was never answered, and I'm not sure how to get there. From: Prove the following equation of complex power series.
Show that for $|z| \lt 1$ with $z \in \Bbb C$, we have
$$ \sum_{k=0}^\infty \frac{{z^2}^k}{1-{z^2}^{k+1}} = \frac{z}{1-z} $$
$$ \sum_{k=0}^\infty \frac{2^k{z^2}^k}{1+{z^2}^{k}} = \frac{z}{1-z} $$
My guess is that the second one is obtained by differentiating the first one or something like that, but I can't manage to prove the first one.

For the first identity, you're trying to prove that
$$
0 = \frac{z}{1-z} - \sum_{k=0}^\infty \frac{{z^2}^k}{1-{z^2}^{k+1}} = \lim_{K\to\infty} \bigg( \frac{z}{1-z} - \sum_{k=0}^K\frac{{z^2}^k}{1-{z^2}^{k+1}} \bigg)
$$
for $|z|<1$. If you work out the exact value of the expression on the right-hand side for small values of $K$, you get
\begin{align*}
\frac{z}{1-z} - \sum_{k=0}^0 \frac{{z^2}^k}{1-{z^2}^{k+1}} &= \frac{z}{1-z} - \frac{z}{1-z^2} = \frac{z^2}{1-z^2} \\
\frac{z}{1-z} - \sum_{k=0}^1 \frac{{z^2}^k}{1-{z^2}^{k+1}} &= \frac{z^2}{1-z^2} - \frac{z^2}{1-z^4} = \frac{z^4}{1-z^4} \\
\frac{z}{1-z} - \sum_{k=0}^2 \frac{{z^2}^k}{1-{z^2}^{k+1}} &= \frac{z^4}{1-z^4} - \frac{z^4}{1-z^8} = \frac{z^8}{1-z^8}
\end{align*}
and so on, suggesting a pattern that successfully be proved by induction. Then you only need to show that
$$
\lim_{K\to\infty} \frac{z^{2^{K+1}}}{1-z^{2^{K+1}}} = 0
$$
for $|z|<1$. The second identity can be proved similarly.
