Find $\lim_{n\to\infty} \frac{(n!)^{1/n}}{n}$. 
Find $$\lim_{n\to\infty} \frac{(n!)^{1/n}}{n}.$$

I don't know how to start. Hints are also appreciated.
 A: $$
f_n=\frac{e^{(1/n)\log n!}}{n}
$$
and the use Stirling's approximation $\log n!\sim n\log n -n$
to conclude
$f_n\sim \frac{e^{\log n-1}}{n}\to 1/e$.
A: If you know Stirling
$$n!\sim\sqrt{2\pi n}\left({n\over e}\right)^n$$
You get
$$u_n\sim \left(\sqrt{2\pi n}\right)^{1\over n}\cdot {1\over e}$$
Now at $+\infty$ one has $\left(\sqrt{2\pi}\right)^{1\over n}\to 1$ and $n^{1\over n}\to 1$ (look at the logarithm) and so
$$u_n\to {1\over e}$$
A: let $$y=  \frac{(n!)^{1/n}}{n}.$$
$$\implies y=\left (\frac{n(n-1)(n-2)....3.2.1}{n.n.n....nnn}\right)^\frac{1}{n} $$
Note that we can distribute the $n$ in the denominator and give an $'n'$ to each term
$$\implies \log y= \frac {1}{n}\left(\log\frac{1}{n}+\log\frac{2}{n}+...+\log\frac{n}{n}\right)$$
applying $\lim_{n\to\infty}$ on both sides, we find that R.H.S is of the form 
$$ \lim_{n\to\infty} \frac{1}{n} \sum_{r=0}^{r=n}f \left(\frac{r}{n}\right)$$
which can be evaluated by integration $$=\int_0^1\log(x)dx$$
$$=x\log x-x$$ plugging in the limits (carefully here) we get $-1$.
$$ \log y=-1,\implies y=\frac{1}{e}$$
A: Let $f_n=\frac{n!}{n^n}$ so that $f_{n+1}= \frac{(n+1)!}{(n+!)^n+1}$
So $\frac{f_{n+1}}{f_n}=(1+\frac{1}{n})^{-n}$
Therefore 
$\lim_{n\to\infty}(1+\frac{1}{n})^{-n}=\frac{1}{e}>0 $
We know that if $<f_n>$ is a sequence such that $f_n$ is greater than 0 for all n, and $\lim_{n\to\infty}\frac{f_{n+1}}{f_n}=l, l>0$. Then $\lim_{n\to\infty}{f_n}^{\frac{1}{n}}=l$.
Using the above theorem we have $\lim_{n\to\infty} \frac{(n!)^{\frac{1}{n}}}{n}=\frac{1}{e}$
