# Does $\sum^\infty_{n=1} \frac{e^n \cdot n!}{n^n}$ converge?

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?

We have by Stirling's Approximation that

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

• Ah, alright. I wasn't aware of that approximation. Since it would tend to 0, would the reciprocal be convergent or divergent? – Bryden C Mar 14 '18 at 13:49
• @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. – JP McCarthy Mar 14 '18 at 13:52
• Oh my bad I worded my comment poorly. I meant the reciprocal would tend towards 0. Thanks for the answers. – Bryden C Mar 14 '18 at 13:54
• Yes the reciprocal goes to zero but not quickly enough to converge. – JP McCarthy Mar 14 '18 at 13:54

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

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

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