Count $ \lim_{n\rightarrow\infty}\frac{n!}{n^n}$

Question

How to count this limit: $\lim_{n\rightarrow\infty}\dfrac{n!}{n^n}$ ? I have no idea how to even start. I will be glad for any tips or help.

Hint: $${n!\over n^n}=\underbrace{{n\over n}\cdot{n-1\over n}\cdot{n-2\over n}\cdots {2\over n}}_{\le 1}\cdot{1\over n}\le {1\over n}.$$ — David Mitra, Jan 26 '14 at 19:49

score 5 · Accepted Answer · answered Jan 26 '14 at 19:49

5

What are you allowed to use? The simplest approach is Stirling expansion of $n!=(\frac{n}{e})^n \sqrt{2 \pi n}(1+o(1))$. The limit is of course $0$.

answered Jan 26 '14 at 19:49

Alex

19,262

score 5 · Answer 2 · answered Jan 26 '14 at 19:59

5

David Mitra's suggestion is the simplest one. Just notice that

$$\frac{n!}{n^n}=\frac n n \frac{n-1} n \frac {n-2}n ...\frac 1 n < \frac 1 n$$

The inequality happenning because every term on the product is smaller or equal to 1. But you know the limit for $\frac 1 n$ (in most books, it's one of the most basic concepts), and $\frac {n!} n$ is always bigger or equal to zero, so you have your answer.

answered Jan 26 '14 at 19:59

Leo Azevedo

385

Sorry, this was done before David editted his comment to LaTeX. – Leo Azevedo Jan 26 '14 at 20:00
Stirling in this case is using a bazooka to get an ant. – Leo Azevedo Jan 26 '14 at 20:05
Knee-jerk reaction... – Alex Jan 26 '14 at 20:15

Count $ \lim_{n\rightarrow\infty}\frac{n!}{n^n}$

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