What is the Convergence Radius and Domain of $\sum\limits_{n=0}^\infty \frac{1}{2^{n}} z^{2^n}$? So I need to find the Convergence Radius and Convergence Domain of the complex power series:
$\sum\limits_{n=0}^\infty \frac{1}{2^{n}} z^{2^n}$
I've tried looking at the individual parts of the series and rewriting it somehow but couldn't find a way.
It can probably be solved, after a step or two, by the formula $R=\frac{1}{\displaystyle\limsup_{n\rightarrow \infty}\sqrt[n]{|a_n|}}$ .
Thanks.
 A: For $n \ge 0$, let $a_n = \begin{cases} \frac1n, & n = 2^k\text{ for } k \in \mathbb{N}\\ 0,&\text{otherwise}\end{cases}$. 
The radius of convergence $R$ of the power series
$$\sum_{k=0}^\infty \frac{1}{2^k} z^{2^k} = \sum_{n=1}^\infty a_n z^n$$
can be computed using the root test.
$$\begin{array}{rll}
\frac1R &= \limsup_{n\to\infty} |a_n|^{1/n}
& \color{blue}{\leftarrow \text{root text}}\\
&= \lim_{n\to\infty} \sup_{m\ge n} |a_m|^{1/m}
& \color{blue}{\leftarrow \text{definition of "limsup"}}\\
&= \lim_{n\to\infty}\sup_{2^k\ge n}  2^{-\frac{k}{2^k}}
& \color{blue}{\leftarrow \text{only $m$ of the from $2^k$ matter}}\\
&= \lim_{n\to\infty}\sup_{k \ge \lceil \log_2n\rceil} 2^{-\frac{k}{2^k}}\\
&= \lim_{\ell\to\infty}\sup_{k \ge \ell}2^{-\frac{k}{2^k}}
& \color{blue}{\leftarrow \text{ $\ell = \lceil \log_2 n \rceil \to \infty$ as $n \to \infty$}}\\
&= \limsup_{\ell\to\infty} 2^{-\frac{\ell}{2^\ell}}
& \color{blue}{\leftarrow \text{definition of "limsup" again}}\\
&= \lim_{\ell\to\infty} 2^{-\frac{\ell}{2^\ell}}
& \color{blue}{\leftarrow\text{ limsup = lim whenvever limit exists}} \\
&= 2^{-\lim_{\ell\to\infty} \frac{\ell}{2^\ell}}
& \color{blue}{\leftarrow 2^{-x} \text{ is a continuous function in } x}
\\
&= 2^0 = 1
\end{array}
$$
The radius of convergence of is $1$. The power series converges to some function $f(z)$ analytic over the open unit disk $|z| < 1$.
For any point $z$ on the unit circle, we have
$$\left| \sum_{n\to 0}^\infty a_n z^n \right| \le \sum_{n=0}^\infty |a_n z^n|
= \sum_{n\to 0}^\infty |a_n| = \sum_{k=0}^\infty \frac{1}{2^k} = 1$$
The power series converges absolutely and hence converges over the unit circle.  This means the power series converges over the whole closed unit disk $|z| \le 1$.
Notice the indices where $a_n \ne 0$ is $n = 2^k$ and $\frac{2^k}{k}$ diverges to $\infty$ as $k \to \infty$. By Fabry gap theorem, the unit circle is a natural boundary for the function $f$.
There is no way to analytic continue $f(z)$ outside the closed unit disk.
This means the domain of the power series is exactly the closed unit disk (even if one
allow analytic continuation).
A: Using the Cauchy Condensation test, which states that the series of the sequence $f(n)$ converges if and only if the series of the sequence $2^n f(2^n)$ converges, the convergence of $a_n=z^{2^n}/2^n$ is equivalent to the convergence of $b_n=z^n/n^2$. Applying the root test and checking the boundary case shows that the series converges whenever $|z|\le 1$.
A: i) If $|z|<1,$ then $|z^{2^n}/2^n| \le 2^{-n}.$ Since $\sum 2^{-n}<\infty,$ the radius of convergence of the series is at least $1.$ ii) If $|z|>1,$ then $|z|^m/m\to \infty$  as $m\to \infty.$ Thus $|z^{2^n}/2^n|\to \infty.$ It follows that the series diverges for such $z,$ which implies the radius of convergence is at most $1.$ Putting i) and ii) together shows the radius of convergence is $1.$
