Question

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

• A. $e$
• B. $\frac{1}{ e }$
• C. $2 e$
• D. $- e$

Answer 1

Answer: (B)

Let $ y =\displaystyle\lim _{n \rightarrow \infty}\left(\frac{n !}{n^n}\right)^{\frac{1}{n}} $
$ \ell n y =\displaystyle\lim _{n \rightarrow \infty} \frac{1}{n} \ell n \left(\frac{n !}{n^n}\right)=\displaystyle\lim _{n \rightarrow \infty} \frac{1}{n} \ell n\left(\frac{1.2 .3 \ldots \ldots n}{n^n}\right) $
$ =\displaystyle\lim_{n \rightarrow \infty} \frac{1}{n}\left[\ell n \left(\frac{1}{n}\right)+\ell n \left(\frac{2}{n}\right)+\ell n\left(\frac{3}{n}\right)+\ldots \ldots+\ell n\left(\frac{n}{n}\right)\right] $
$\left.=\displaystyle\lim _{n \rightarrow \infty} \frac{1}{n} \displaystyle\sum_{r=1}^n \ell n\left(\frac{r}{n}\right)=\int\limits_0^1 \ell \ln x d x=x \ell n x-x\right]_0^1 $
$=(0-1)-\underset{x \rightarrow 0^+}{Lt} x \ell n x+0$
$ =-1-0=-1 $
$ \Rightarrow y=\frac{1}{e}$