2
$\begingroup$

Is the following series convergent?

$$\sum_{n=1}^{\infty}\frac{e^n\,n!}{n^n}$$

I treid the Ratio and Root tests, but both of them failed.

  • 0
    Hint: [Stirling](https://en.wikipedia.org/wiki/Stirling's_approximation) tells us that $n!\sim \left( \frac ne \right)^n\sqrt {2\pi n}$ – 2017-02-26
  • 0
    No. One may prove this using the [Term Test](https://en.m.wikipedia.org/wiki/Term_test). – 2017-02-26

1 Answers 1

6

No, your series is not convergent. Note that $$e^{n}=\sum_{k=0}^{\infty} \frac{n^k}{k!}>\frac{n^{n}}{n!} \implies e^{n}>\frac{n^{n}}{n!}$$ From the series expansion of $e^x$. Multiplying $\dfrac{n!}{n^n}$ on each side, we have that $$\frac{n!e^n}{n^n}>1 $$ Thus, by comparison test we have that the series diverges.

  • 0
    would you say, how by comparison test? – 2017-02-26
  • 0
    @soodehMehboodi Each terms are larger than $1$......But $\sum_{n=1}^{\infty}1 $ diverges... – 2017-02-26
  • 0
    How do you check the necessary condition for the series to converge? – 2017-02-26
  • 1
    @soodehMehboodi I don't think that relates to my answer. If $a_{n}>1$ for all $n$, is a sufficient condition for $\sum\limits_{n=1}^{\infty}a_{n}$ to diverge. So, we just do comparison. We don't need the necessary condition. – 2017-02-27
  • 0
    Wonderfully done! (+1) – 2017-02-28