What can be said about the radius of convergence of the poower series $$\sum_0^{\infty} n!x^{n^2}$$
I know that $\limsup_{n\to\infty}(n!)^{\frac1{n}}\to\infty$. Is that of any use here? Should I use ratio test or root test? Any hints? Thanks beforehand.
With the ratio test: let $a_n=n!x^{n^2}$. For $x \ne 0$ we have
$|\frac{a_{n+1}}{a_n}|= (n+1)|x|^{2n+1}$.
If $|x|<1$, then $|\frac{a_{n+1}}{a_n}| \to 0 <1$ and the series is convergent.
If $|x| \ge 1$, then $|\frac{a_{n+1}}{a_n}| \to \infty$ and the series is divergent.
Consequence: the radius of convergence $=1$.
The ratio test is the best! We have $$\lim_{n\to \infty} {a_{n+1}\over a_n}=\lim_{n\to \infty}{(n+1)!\over n!}{x^{(n+1)^2}\over x^{n^2}}=\lim_{n\to \infty}{(n+1)}{x^{2n+1}}$$which means that for $|x|<1$ the series converges and diverges elsewhere. Therefore the radius of convergence is 1.
For a power series $\sum_{n=0}^{\infty} a_nx^n$ radius of convergence is given by $\frac{1}{\rho}$ where $\rho=lim\ sup_{n\rightarrow \infty} |a_n|^{\frac{1}{n}}$. Here $a_n=k!$ if $n=k^2$ and $=0$ ,otherwise. Hence $\rho=lim_{n\rightarrow \infty} (n!)^{\frac{1}{n^2}}=1$. Hence radius of convergence is $1$.
