TITLE: Radius of convergence in power series $\sum_{n=0}^{\infty}(-1)^nx^{2^n}$
QUESTION [5 upvotes]: 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 $-R$?


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?
By the ratio test, I get that this series converges when $|x|<1$, diverges when $|x|>1$, so $R=1$, is that right?
When $x=1$ or $-1$, series both becomes $$\sum_{n=0}^{\infty}(-1)^n,$$
then obviously, series diverges. Right?

REPLY [3 votes]: (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 $\rho=1$.