## watchmath 5 years ago Compute the limit $\lim_{n\to\infty}\frac{n!}{n^n}$

1. anonymous

0

2. anonymous

i got 1

3. watchmath

why?

4. anonymous

sorry... it is 0 :)

5. anonymous

let y=n!/n^n lny=$\sum_{r=0}^{n}\ln(n-r)/n$ now ln((n-r)/n)=ln(1-r/n) as n->infinity so r/n=0 so ln(1-r/n)=ln(1)=0; lny=0 y=1 so the limit is 1

6. watchmath

Something wrong with your argument dipankar. I found the answer $0<\frac{n!}{n^n}<\frac{1\cdot n^{n-1}}{n^n}=\frac{1}{n}$ By squeeze theorem, then the limit is zero.