Jtdylktuy

What can be said about the radius of convergence of the poower series
$$sum_0^{infty} n!x^{n^2}$$

I know that $limsup_{ntoinfty}(n!)^{frac1{n}}toinfty$. 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 $$lim_{nto infty} {a_{n+1}over a_n}=lim_{nto infty}{(n+1)!over n!}{x^{(n+1)^2}over x^{n^2}}=lim_{nto infty}{(n+1)}{x^{2n+1}}$$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 $rho=lim sup_{nrightarrow infty} |a_n|^{frac{1}{n}}$. Here $a_n=k!$
if $n=k^2$ and $=0$ ,otherwise. Hence $rho=lim_{nrightarrow infty} (n!)^{frac{1}{n^2}}=1$. Hence radius of convergence is $1$.