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?
 A: 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 $|z|=1$.
A: If $|z| < 1$, then by geometric series, $$\sum_{n = 0}^\infty \frac{z^{2^n}}{1-z^{2^{n+1}}} = \sum_{n = 0}^\infty \sum_{k: v_2(k) = n} z^k = \sum_{n = 1}^\infty z^n = \frac{z}{1-z}$$ where $v_2(k)$ denotes the largest power of $2$ dividing $k$. 
If $|z| > 1$, then $$\sum_{n=0}^{\infty} \frac{z^{2^n}}{1-z^{2^{n+1}}} = -\sum_{n=0}^{\infty} \frac{z^{-2^n}}{1-{z^{-2^{n+1}}}} = -\sum_{n = 0}^\infty \sum_{k: v_2(k) = n} z^{-k} = -\sum_{n=1}^\infty z^{-n} = \frac{1}{1-z}$$
A: Let $f(z)=\dfrac z{1-z^2}$. Then your series is $\displaystyle\sum_{n=0}^\infty f(z^{2^n})$. So, let\begin{align}f_n(z)&=f(z^{2^n})\\&=z^{2^n}+z^{3\times2^n}+z^{5\times2^n}+\cdots\end{align}If $K$ is a compact subset of $D(0,1)$ and if $M=\sup_{z\in J}\lvert z\rvert$, then$$\bigl(\forall z\in D(0,1)\bigr)(\forall n\in\mathbb{Z}_+):|f_n(z)|=\left|f\left(
 z^{2^n}\right)\right|=\left|\frac{z^{2^n}}{1-z^{2^{n+1}}}\right|=
 \frac{|z|^{2^n}}{\left|1-z^{2^{n+1}}\right|}\leq\frac{M^{2^n}}{
 1-M^{2^{n+1}}}$$and so $\displaystyle\sum_{n=0}^\infty f_n$ converges uniformly on each compact subset of $D(0,1)$. It follows from Weierstrass's Double Series Theorem that, if $\displaystyle\sum_{k=0}^\infty c_{n,k}z^k$ is the Taylor series of $f_n$ centered at $0$, then$$(\forall z\in D(0,1)):\sum_{n=0}^\infty f_n(z)=\sum_{k=0}^\infty
  \sum_{n=0}^\infty c_{n,k}z^k,$$which means that$$\bigl(\forall z\in D(0,1)\bigr):\sum_{n=0}^\infty\frac{z^{2^n}}{1+z^{2^{n+1}}}=1+z+z^2+z^3+\cdots=\frac1{1-z}.$$
In order to see what happens when $\lvert z\rvert>1$, replace $z$ with $\frac1z$.
