Radius of convergence in power series $\sum_{n=0}^{\infty}(-1)^nx^{2^n}$

Question

Given the series $$\sum_{n=0}^{\infty}(-1)^nx^{2^n}$$ determine the radius of convergence, and what can we say when $x=R$ and $-R$?

1. Is it a power series? Power series should have the form of $$\sum_{n=0}^{\infty}a_nx^n$$ but the given series does not match this form. If not a power series, why can we say about its radius of convergence?

2. By the ratio test, I get that this series converges when $|x|<1$, diverges when $|x|>1$, so $R=1$, is that right?

3. When $x=1$ or $-1$, series both becomes $$\sum_{n=0}^{\infty}(-1)^n,$$ then obviously, series diverges. Right?

Answer 1

(1): see the comments.

(2): the radius of convergence $\rho$ of $\sum a_nx^n$ is defined to be:

$$ \rho:=\frac{1}{\alpha} $$

where

$$ \alpha = \lim_{n\rightarrow \infty} \sup \sqrt[n]{|a_n|} $$.

Here we have:

$|a_n| = 1$ if $n$ is a power of 2, $|a_n|=0$ otherwise. Thus $\rho=1$.