Show that $\sum_{n=1}^{\infty} \frac{n^n}{n!}$ diverges

Question

So far I have $\lim_{n\rightarrow \infty}\frac{\sqrt[n]{n^n}}{\sqrt[n]{n!}}=\lim_{n\rightarrow \infty}\frac{n}{\sqrt[n]{n}\sqrt[n]{n-1}...\sqrt[n]1}$

I know that $\lim_{n\rightarrow\infty}\sqrt[n]{n}=\lim_{n\rightarrow\infty}\sqrt[n]{1}=1$.

How do I have the argue when using the sandwichtheorem, i.e. how do I know that

$\lim_{n\rightarrow\infty}\sqrt[n]{n}\leq\lim_{n\rightarrow\infty}\sqrt[n]{k}\leq\lim_{n\rightarrow\infty}\sqrt[n]{1},\forall 1<k<n$

?

I think the excercise was designed to apply the rootcriteria for the first time — RM777, Dec 15 '18 at 19:32
This question shows something that implies yours https://math.stackexchange.com/questions/2787512/prove-that-the-series-sum-n-1-infty-fracennnn-diverges?rq=1 — Yanko, Dec 15 '18 at 19:42

score 2 · Answer 1 · answered Dec 15 '18 at 19:31

2

Since each number $\dfrac{n^n}{n!}$ is greater than or equal to $1$, you don't have $\lim_{n\to\infty}\dfrac{n^n}{n!}=0.$

answered Dec 15 '18 at 19:31

score 0 · Answer 2 · answered Dec 15 '18 at 19:37

0

Note that

$$ \bigg| \frac{ a_{n+1}}{a_n} \bigg| = \frac{(n+1)^{n+1} n!}{(n+1)! n^n} = \frac{(n+1)^n}{n^n} = \left( 1 + \frac{1}{n} \right)^n \to \mathrm{e}$$

answered Dec 15 '18 at 19:37

James

score 0 · Answer 3 · answered Dec 15 '18 at 19:37

0

Abridged proof. Notice $n! \leq n^n.$ Q.E.D.

answered Dec 15 '18 at 19:37

William M.

3 Answers3