prove that $\lim_{n\to∞} \frac{n!}{n^n}=0$

Question

any sugestions?, by Stolz–Cesàro?. The trouble says that first try with $k<n$ and then for a particular $k<\frac{n}{2}$ I try by definition of $\epsilon$ but it doesn't help

Are you sure you aren't missing some logarithms or something? Or that the limit should be zero instead of one? As written this doesn't look true at all. — user2566092, Sep 22 '15 at 16:13
I think you will find it very hard to prove, since it is not true. That limit is $0$. — robjohn, Sep 22 '15 at 16:15
The above limit is $0$. This can be done in one short line, minimal machinery. The expression is $\le \frac{1}{n}$. The hint is kind of weird, too much work. — André Nicolas, Sep 22 '15 at 16:16

score 8 · Answer 1 · answered Sep 22 '15 at 16:18

8

Every number in the product $n!$ is less than or equal to $n$. In particular, one of them is $1$. Can you use this to show $n! / n^n \leq 1/n$?

answered Sep 22 '15 at 16:18

user2566092

26,142

score 4 · Answer 2 · answered Sep 22 '15 at 17:15

4

There is a very simple approach: by the AM-GM inequality, $$ n!^2 \leq \left(\frac{n+1}{2}\right)^{2n},$$ hence the wanted limit is clearly $\color{red}{0}$.

answered Sep 22 '15 at 17:15

Jack D'Aurizio

353,855

score 2 · Answer 3 · answered Sep 22 '15 at 17:29

$0\leq{n!}\leq{n^{n-1}}\implies$

$\frac{0}{n^n}\leq\frac{n!}{n^n}\leq\frac{n^{n-1}}{n^n}\implies$

$0\leq\frac{n!}{n^n}\leq\frac{1}{n}\implies$

$\lim\limits_{n\to\infty}0\leq\lim\limits_{n\to\infty}\frac{n!}{n^n}\leq\lim\limits_{n\to\infty}\frac{1}{n}\implies$

$0\leq\lim\limits_{n\to\infty}\frac{n!}{n^n}\leq0\implies$

$\lim\limits_{n\to\infty}\frac{n!}{n^n}=0$

score 1 · Answer 4 · answered Sep 22 '15 at 16:21

1

You can also use Stirling's approximation: $n!\sim \left(\dfrac{n}{e}\right)^n\sqrt{2\pi n}$ as $n\to \infty$

answered Sep 22 '15 at 16:21

RFZ

16,814

prove that $\lim_{n\to∞} \frac{n!}{n^n}=0$

4 Answers4