Question

Compute the limit
\[ \lim_{n\to\infty}\frac{n!}{n^n}\]

Answer 1

0

Answer 2

i got 1

Answer 3

why?

Answer 4

sorry... it is 0 :)

Answer 5

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

Answer 6

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.