# Find the radius of convergence of the power series $\sum_{n=1}^{\infty}\frac{n!}{n^n}(x+3)^n$ [duplicate]

Find the radius of convergence of the power series $$\sum_{n=1}^{\infty}\frac{n!}{n^n}(x+3)^n$$

We easily get that $$R=\lim_{n\to\infty}\frac{n!}{n^n}.\frac{(n+1)^n}{(n+1)!}=e$$ so the radius is $$x+3 \in (-R,R)$$ or $$x\in (-e-3,e-3)$$. I have a problem at the end points ($$x=-e-3$$ and $$x=e-3$$.

Trying to see if the series $$\sum_{n=1}^{\infty}\frac{n!}{n^n}e^n$$, I get nowhere with the usual criteria (they all wield 1). What should I do?

• use the Hadamard's formula for the radius of convergence of a power series and the Stirling approximation for the factorial – Masacroso Jun 19 '19 at 10:45

Use Stirling's formula $$n!\sim\sqrt{2\pi n}(n/e)^n$$, we get at the end-points the series behaves like $$\sum(\pm 1)^n\sqrt{n}$$, so diverges at both ends.
• @Masacroso No, we are testing the end-points, so an $e^n$ factor from $(x-3)^n$ cancels with the $e^n$ in $(n/e)^n$ from Stirling. Similarly the $n^n$ is cancelled in the coefficient. – user10354138 Jun 19 '19 at 10:51