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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.

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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.

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  • $\begingroup$ Thank you. Appreciate it. That's very elaborate. It's all about coming up with two different general functions for $f_n(z)$. Then the convergence is clear enough. $\endgroup$ – user358174 Aug 15 '16 at 4:48
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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 $f.$

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  • $\begingroup$ Thank you. The first explanation with W-M test is pretty clear. May be for $|z|>1$ we can do the inequality the other way round: I'm not quite understand how you get the limit and the natural boundary. For the latter I haven't seen such an argument before. The two limiting functions being not analytic continuation is clear because one couldn't be obtained from the other. I was thinking of showing that the unit circle has a singularity at every point on it. Or else the unit circle is dense with the set of singularities. $\endgroup$ – user358174 Aug 15 '16 at 22:10

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