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?

  • $\begingroup$ use the Hadamard's formula for the radius of convergence of a power series and the Stirling approximation for the factorial $\endgroup$ – 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.

| cite | improve this answer | |
  • 1
    $\begingroup$ @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. $\endgroup$ – user10354138 Jun 19 '19 at 10:51

Not the answer you're looking for? Browse other questions tagged or ask your own question.