Radius of convergence of power series $\sum_{n=0}^{\infty} n!x^{n^2}$

Question

Radius of convergence of power series $\sum_{n=0}^{\infty} n!x^{n^2}$

$\sum_{n=0}^{\infty} n!x^{n^2} = 1 + x + 2x^4 + 6x^9\ldots$

Comparing this with $\sum_{n=0}^{\infty} a_nx^n=$

$a_n= n! $ or $0$

$|a_n|^{\frac1n} ={n!}^{\frac1n} $ or $0^{\frac1n}$

We need $L= \lim \sup |a_n|^{\frac1n}$ and the radius of convergence $R$ will be $R=\frac1L$

$0^{\frac1n} \to 0 $ as $n\to \infty$

$\lim {n!}^{\frac{1}{n}}= \lim \frac{(n+1)!}{n!} \to \infty$ as $n \to \infty$

So $\lim |a_n|^{\frac1n} = 0 $ or $ \infty$

But I am not sure what $\lim \sup |a_n|^{\frac1n}$ is ??
My doubt is that : We are looking for the greatest limit point of $|a_n|^{\frac1n}. $ Clearly $\infty$ seems the correct answer but $\infty$ does not belong in R and can't be a limit point. So what is the correct answer?

Answer 1

The radius of convergence is $1$, because, if $x>0$, then$$\frac{(n+1)!x^{(n+1)^2}}{n!x^{n^2}}=(n+1)x^{2n+1}$$and the sequence $\bigl((n+1)x^{2n+1}\bigr)_{n\in\mathbb N}$ converges to $0$ if $x\in(0,1)$ and diverges if $x\geqslant1$.