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

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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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• Wrong formula. What values do $a_n$ take? – xbh Jan 11 at 5:25

## 2 Answers

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, $$r=\frac1{\limsup_{n\to\infty}\sqrt[n^2]{n!}}=1$$.