calculus - Find $lim_{ntoinfty} frac{(n!)^{1/n}}{n}$.

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

I don't know how to start. Hints are also appreciated.

Answer

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 $-1$.

$$ \log y=-1,\implies y=\frac{1}{e}$$