<s>[INST] Convergence and divergence of a Complex Series I've been given the following series:
$$\frac{z}{1-z^2} + \frac{z^2}{1-z^4} + \frac{z^4}{1-z^8} + ...$$
and been told to investigate the convergence. Clearly this diverges if $z=1$ (possibly if $\vert{z}\vert = 1$?), but other than that I am at a loss as to how to proceed. Wolfram Alpha tells me that this converges to $\frac{z}{1-z}$ if $\vert{z}\vert<1$ and to $\frac{1}{1-z}$ if $\vert{z}\vert>1$ but how would one go about showing this?
 [/INST] We can show inductively that
$$\begin{align*}
\sum_{j=0}^n \frac{z^{2^j}}{1-z^{2^{j+1}}}&=\frac{\sum\limits_{k=1}^{2^{n+1}-1}z^k}{1-z^{2^{n+1}}}\\&=\frac{z-z^{2^{n+1}}}{(1-z)(1-z^{2^{n+1}})}\\&=\frac{z-1+1-z^{2^{n+1}}}{(1-z)(1-z^{2^{n+1}})}\\&=-\frac{1}{1-z^{2^{n+1}}}+\frac1{1-z}.
\end{align*}$$ The base case $n=0$ is quite obvious. We find that
$$\begin{align*}
-\frac{1}{1-z^{2^{n+1}}}+\frac1{1-z}+\frac{z^{2^{n+1}}}{1-z^{2^{n+2}}}&=\frac{-1-z^{2^{n+1}}}{1-z^{2^{n+2}}}+\frac1{1-z}+\frac{z^{2^{n+1}}}{1-z^{2^{n+2}}}\\&=-\frac{1}{1-z^{2^{n+2}}}+\frac1{1-z}.
\end{align*}$$ By induction, the claim is proved. 
We can easily see that the given series converges for $|z|<1$ and $|z|>1$, to $\frac{z}{1-z}$ and $\frac1{1-z}$, respectively.
 If $|z|=1$, then the given series converges if and only if
$$
\exists \lim_{n\to\infty} z^{2^n} \ne 1.
$$ If we denote such limit by $L$, then it should satisfy 
$$
L^2 =\lim_{n\to\infty} z^{2^{n+1}}=\lim_{n\to\infty} z^{2^n}=L
$$ giving $L=1$ or $L=0$. Since $L\ne 0,1$, the series does not converge for all $|z|=1$.
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