1. ## Limit

Prove that $\displaystyle \lim_{n \to \infty} \left( \frac{n!}{n^n}\right)^{1 \over n}=\frac{1}{e}$.

I wrote the limit as $\displaystyle \exp{\lim_{n \to \infty} {1 \over n}\ln\left( \frac{n!}{n^n}\right)}$

What should I do next?? Can I use L hospital's rule (in any way)?

2. ## Re: Limit

1.
$\displaystyle k=\lim_{n \to \infty}\left( \frac{n!}{n^n}\right)^{\frac{1}{n}}= \exp{\left[ \lim_{n \to \infty}{\frac{1}{n}}}\ln\left( \frac{n!}{n^n}\right)\right]= \exp{\left[ \lim_{n \to \infty}{\frac{1}{n}}}\ln\left( \frac{1\cdot2\cdot3\cdots (n-1)\cdot n}{n\cdot n \cdot n \cdots n \cdot n}\right)\right]$

$\displaystyle =\exp{\left[ \lim_{n \to \infty} {1 \over n}\sum_{r=1}^{n}\ln\left( \frac{r}{n}\right)\right]}$

2. It is known that : $\displaystyle \lim_{n \to \infty}\frac{1}{n}\sum_{r=1}^{n}f\left( \frac{r}{n}\right)=\int_{0}^{1} f(x) \ dx$. Therefore :

$\displaystyle k=\exp{\left( \int_{0}^{1}\ln(x) dx \right) }=\exp[{\left( x\cdot\ln(x) -x\right)^1_0 }]=\exp{\left(-1-\lim_{x \to 0}x\ln(x) \right)}$

Finish it...