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

Question

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?

Answer 1

We have by Stirling's Approximation that

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

Answer 2

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

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