Given series of complex terms converges locally uniformly Prove that the series 
$$f(z)=\frac{z}{1-z^2}+\frac{z^2}{1-z^4}+\frac{z^4}{1-z^8}+ \cdots$$
conveges locally uniformly to $\frac{z}{1-z}$ when $z$ is in the unit disc $D$ and to $\frac{1}{1-z}$ when $z \notin \overline{D}.$ Are these two limiting functions analytic continuations of each other? If not, is $\partial D$ the natural boundary of analyticity for $f(z)$ ? 
I came across this problem while studying for my complex preliminaries. I tried to factor out the term $\frac{z}{1-z}$ and consider the difference $|f(z)-\frac{z}{1-z}|$ and go by to show that it's less than $\epsilon$ for all $z.$ But it wasn't successful. And any help in this part and other two parts are much appreciated. I wouldn't ask if I was able to solve the problem. Thank you for your time. 
 A: Consider the partial sums
$$
  f_n(z)=\sum_{k=0}^{n-1}\frac{z^{(2^k)}}{1-z^{(2^{k+1})}}.
$$
By induction on $n$, we have
$$
  f_n(z)=\frac1{1-z}-\frac1{1-z^{(2^n)}}.
$$
Indeed
$$
  f_1(z)=\frac{z}{1-z^2}=\frac{1+z}{1-z^2}-\frac{1}{1-z^2}
    =\frac1{1-z}-\frac{1}{1-z^2}.
$$
Assuming the formula holds for $n$, we have
$$\begin{eqnarray*}
  f_{n+1}(z)
    &=&\frac1{1-z}-\frac1{1-z^{(2^n)}}+\frac{z^{(2^n)}}{1-z^{(2^{n+1})}}\\
    &=&\frac1{1-z}-\frac{1+z^{(2^n)}}{1-z^{(2^{n+1})}}+\frac{z^{(2^n)}}{1-z^{(2^{n+1})}}\\
    &=&\frac1{1-z}-\frac{1}{1-z^{(2^{n+1})}},
\end{eqnarray*}$$
as required. If $|z|>r>1$ then $|z^{(2^n)}|\geq r^{(2^n)}$, so $f_n(z)\rightarrow\frac1{1-z}$ uniformly. On the other hand
$$
  f(z)=\frac{z}{1-z}-\frac{1}{z^{(-2^n)}-1}.
$$
If $|z|<r<1$ then $|z^{(-2^n)}|\geq r^{(-2^n)}$, so $f_n(z)\rightarrow\frac{z}{1-z}$ uniformly.
A: The local uniform convergence can be seen (without knowing the limit) pretty easily: Let's take the $|z|<1$ case. If $|z| \le r < 1,$ then
$$\left |\frac{z^{2^n}}{1-z^{2^n+1} }\right | \le \frac{|z|^{2^n}}{1-|z|^{2^n+1} }\le \frac{r^{2^n}}{1-r^2 }.$$
Since $\sum r^{2^n}/(1-r^2) < \infty,$ we get uniform convergence on $\{|z|\le r\}$ by Weierstrass M.
As for what the sum converges to, note the sum equals
$$\sum_{n=0}^{\infty}\sum_{m=0}^{\infty} z^{(2m+1)2^n}.$$
Because every positive integer power of $z$ occurs once and only once here, and $|z|<1$ gives us absolute convergence, the above is just the sum $\sum_{k=1}^{\infty}z^k = z/(1-z).$
I assume similar shenanigans will work for $|z|>1.$
The fact that $\{|z|=1\}$ is a natural boundary for $f$ follows from this, but actually there is a more direct way. Note that set $E=\{e^{j\pi i/2^k}: j,k\in \mathbb N\}$ is dense in the unit circle. If you fix any $\zeta \in E,$ you'll see that along the radius terminating at $\zeta,$ the terms of the series defining $f$ in the disc equal $r^{2^n}/(1-r^{2^{n+1}})$ for large $n.$ That shows $f$ blows up at each $\zeta\in E,$ which implies the unit circle is a natural boundary for $f.$
