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Q. $\displaystyle\lim _{n \rightarrow \infty} \frac{(n !)^{1 / n}}{n}$ equals

AMUAMU 2015Limits and Derivatives

Solution:

$\frac{n !}{n^{n}}=\frac{1}{n} \cdot \frac{2}{n} \cdot \frac{3}{n} \cdot \ldots \cdot \frac{n}{n}$
Let $ A=\displaystyle\lim _{n \rightarrow \infty} \frac{(n !)^{1 / n}}{n}=\displaystyle\lim _{n \rightarrow \infty}\left(\frac{n !}{n^{n}}\right)^{1 / n}$
$\Rightarrow A=\displaystyle\lim _{n \rightarrow \infty}\left[\frac{1}{n} \cdot \frac{2}{n} \cdot \frac{3}{n} \ldots \cdot \frac{n}{n}\right]^{1 / n}$
$ \Rightarrow \log A=\displaystyle\lim _{n \rightarrow \infty} \frac{1}{n}\left[\log \left(\frac{1}{n}\right)\right.+\log \left(\frac{2}{n}\right)\left.+\log \left(\frac{3}{n}\right)+\ldots+\log \left(\frac{n}{n}\right)\right]$
$=\displaystyle\lim _{n \rightarrow \infty} \frac{1}{n} \displaystyle\sum_{t=1}^{n} \log \left(\frac{r}{n}\right)$
$=\int\limits_{0}^{1} \log x d x=|x \log x-x|_{0}^{1}$
$=[1 \log 1-1]-\left[\displaystyle\lim _{x \rightarrow 0} x \log x-0\right]=-1$
$\therefore A=e ^{-1}=\frac{1}{e}$