Find all complex values of $|z|$ for which series converges

Question

I want to find all $z \in \mathbb{C}$ for which $\sum_{n = 1}^\infty2^{-n}z^{2^n}$ converges. I've already figured out that it converges for $|z| \leq 1$ by comparison test, and I also know that for real $z > 1$, the series diverges since $2^{-n}z^{2^n}\rightarrow \infty$. What about for $z > 1$ where $z \not \in \mathbb{R}$?

Answer 1

Hint. Is it true that for $|z|>1$, $\displaystyle \lim_{n\to \infty}2^{-n}z^{2^n}=0?$ (necessary condition for convergence).

Note that by letting $m=2^n$ then $$|2^{-n}z^{2^n}|=\frac{|z|^m}{m}=\frac{(1+(|z|-1))^m}{m}\\\geq \frac{1+m(|z|-1)+\binom{m}{2}(|z|-1)^2}{m}\geq \frac{(m-1)(|z|-1)^2}{2}$$ and the right-hand side goes to infinity as $m=2^n\to +\infty$.

Answer 2

For $|z| \le 1$ we have $2^{-n}|z|^{2^n} \le 2^{-n}$, thus the series converges.

Now let $|z|>1$. Then there is $s>0$ such that $|z|=1+s$. The Bernoulli iequality gives

$$|z|^{2^n}=(1+s)^{2^n} \ge 1+2^ns.$$

Hence

$$ 2^{-n}|z|^{2^n} \ge 2^{-n}(1+2^ns)=2^{-n}+s>s$$ for all $n$.

This shows that $(2^{-n}|z|^{2^n})$ does not converge to $0$.