<s>[INST] Show that $\sum_0^{\infty}\frac{z^{2^n}}{z^{2^{n+1}}-1}$ converges to $\frac{z}{z-1}$ or $\frac{-1}{1-z}$ How do we show that the sum below converges to $\frac{z}{z-1}$ if $|z|<1$ and $\frac{-1}{1-z}$ if $|z|>1$?
$$\sum_0^{\infty}\frac{z^{2^n}}{z^{2^{n+1}}-1}=\frac{z}{z^2-1}+\frac{z^2}{z^4-1}+\frac{z^4}{z^8-1}+...$$
Any help is appreciated, thanks!
 [/INST] Note
\begin{align}\frac{z^{2^n}}{z^{2^{n+1}}-1} &= \frac{z^{2^n}(z^{2^n}-1)}{(z^{2^n}-1)(z^{2^{n+1}}-1)} = \frac{z^{2^{n+1}}-z^{2^n}}{(z^{2^n}-1)(z^{2^{n+1}}-1)} \\
&= \frac{(z^{2^{n+1}}-1) - (z^{2^n}-1)}{(z^{2^n}-1)(z^{2^{n+1}}-1)}\\
& = \frac{1}{z^{2^n} - 1} - \frac{1}{z^{2^{n+1}}-1}.
\end{align}
So for every positive integer $N$, 
$$\sum_{n = 0}^N \frac{z^{2^n}}{z^{2^{n+1}}-1} = \frac{1}{z-1} - \frac{1}{z^{2^{N+1}}-1}.\tag{*}$$
by telescoping. Now evaluate the limit as $N\to \infty$ of the right-hand side (*) depending on the cases $|z| < 1$ and $|z| > 1$.
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