0
$\begingroup$

By ratio test,

$$ \sum^\infty_{n=1} \frac{n!}{n^n}$$

converges as the limit will become $e^{-1}$. If I add $e^n$ to it, the ratio test then fails:

$$ \sum^\infty_{n=1} \frac{e^n \cdot n!}{n^n}$$

How would I be able to show that this series is convergent/divergent?

$\endgroup$
7
$\begingroup$

We have by Stirling's Approximation that

$$\frac{e^n\cdot n!}{n^n}\geq \frac{e^n}{n^n}\cdot \sqrt{2\pi}\cdot n^n\cdot \sqrt{n} \cdot e^{-n}=\sqrt{2\pi }\cdot \sqrt{n}\underset{n\rightarrow \infty}{\not\rightarrow} 0.$$

$\endgroup$
  • $\begingroup$ Ah, alright. I wasn't aware of that approximation. Since it would tend to 0, would the reciprocal be convergent or divergent? $\endgroup$ – Bryden C Mar 14 '18 at 13:49
  • $\begingroup$ @BrydenC Sorry the terms do not go to zero so the series is divergent. The reciprocal $\displaystyle \frac{n^n}{e^n\cdot n!}\geq \frac{1}{\sqrt{n}}$ so diverges as a $p$-series. $\endgroup$ – JP McCarthy Mar 14 '18 at 13:52
  • $\begingroup$ Oh my bad I worded my comment poorly. I meant the reciprocal would tend towards 0. Thanks for the answers. $\endgroup$ – Bryden C Mar 14 '18 at 13:54
  • $\begingroup$ Yes the reciprocal goes to zero but not quickly enough to converge. $\endgroup$ – JP McCarthy Mar 14 '18 at 13:54
5
$\begingroup$

Note that

$$\frac{e^n \cdot n!}{n^n}\to \infty$$

but it is sufficient to prove that

$$\frac{e^n \cdot n!}{n^n}\ge 1$$

indeed

base case

  • $n=1 \implies e \ge 1$

induction step

  • assume $\frac{e^n \cdot n!}{n^n}\ge 1$

  • $\frac{e^{n+1} (n+1)! }{ (n+1)^{n+1} }=\frac{e (n+1)n^n }{ (n+1)^{n+1} }\frac{e^n \cdot n!}{n^n}\ge e\frac{n^n }{ (n+1)^{n} }=\frac{e }{\left(1+\frac1n\right)^n }\ge 1$

$\endgroup$
  • $\begingroup$ Is it trivial ? $\endgroup$ – Gabriel Romon Mar 14 '18 at 13:52
  • $\begingroup$ @GabrielRomon No it was not without Stirling, I've presented a simpler result by induction which suffices to prove divergence. $\endgroup$ – gimusi Mar 14 '18 at 14:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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