# Find the radius of convergence of $\sum_{n=0}^{\infty} n!x^{n^{2}}$ [duplicate]

I have to find find the radius of convergence of the series $$\sum_{n=0}^{\infty} n!x^{n^{2}}$$.

Here $$\lim_{n\to\infty} \frac{(n+1)!}{n!} = \infty$$. Then radius of convergence is 0. Where I'm doing wrong? Please help

## marked as duplicate by Key Flex, José Carlos Santos complex-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jan 11 at 7:08

• Wrong formula. What values do $a_n$ take? – xbh Jan 11 at 5:25
You have found the radius of convergence of $$\sum n! x^{n}$$ not that of $$\sum n! x^{n^{2}}$$. Here $$a_n=0$$ of $$n$$ is not a square and $$a_n=k!$$ if $$n=k^{2}$$. To show that the series $$\sum n! |x^{n^{2}}|$$converges only for $$|x|<1$$ use Stirling's approximation. You have to observe that $$e^{-n} e^{n^{2} \ln\, x} e^{(n+\frac 1 2) \ln\, n} \to \infty$$ for $$|x|\geq 1$$ whereas the series is dominated by a series of the type $$\sum Ce^{-\delta |x|}$$ for $$|x| <1$$. Hence the radius of convergence is $$1$$.
By the Cauchy-Hadamard theorem, $$r=\frac1{\limsup_{n\to\infty}\sqrt[n^2]{n!}}=1$$.