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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

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marked as duplicate by Key Flex, José Carlos Santos complex-analysis Jan 11 at 7:08

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    $\begingroup$ Wrong formula. What values do $a_n$ take? $\endgroup$ – xbh Jan 11 at 5:25
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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 $1$.

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By the Cauchy-Hadamard theorem, $r=\frac1{\limsup_{n\to\infty}\sqrt[n^2]{n!}}=1$.

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