Thursday, October 24, 2019

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 $-1$.




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


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