How to compute the radius of convergence of $\sum_{n=0}^{\infty}{4^n\cdot x^{2^n}}$

Question

How to compute the radius of convergence of $\sum_{n=0}^{\infty}{4^n\cdot x^{2^n}}$.

I was wondering wheter it even matters that it isn't $x^n$ but $x^{2n}$. According to the root test $4^n$ converges to $4$, so we have a radius of convergence of $x\in(-1/4,1/4)$. Is that true?

Answer 1

$$\frac{4^{n+1}x^{2^{n+1}}}{4^nx^{2^n}}=4x^{2^n}$$ ensures convergence for all $$|x|<1$$ and divergence for $$|x|\ge1.$$

Answer 2

The coefficients $a_k$ are given by $a_k=4^{n}$ if $k =2^{n}$. Hence $a_k^{1/k}=4^{\log_2 k/k}$ which tends to $1$. Hence the radius of convergence is $1$.