Question

$\underset{n\to \infty }{\lim }{\left(\frac{n!}{n^n}\right)}^{1/n}$ equals

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

Answer 1

Answer: (B)

Let $y=\underset{n\to \infty }{\lim }{\left(\frac{n!}{n^n}\right)}^{\frac{1}{n}}$
$\ell ny=\underset{n\to \infty }{\lim }\frac{1}{n}\ell n\left(\frac{n!}{n^n}\right)=\underset{n\to \infty }{\lim }\frac{1}{n}\ell n\left(\frac{\mathrm{1.2.3}\dots \dots n}{n^n}\right)$
$=\underset{n\to \infty }{\lim }\frac{1}{n}\left[\ell n\left(\frac{1}{n}\right)+\ell n\left(\frac{2}{n}\right)+\ell n\left(\frac{3}{n}\right)+\dots \dots +\ell n\left(\frac{n}{n}\right)\right]$
${=\underset{n\to \infty }{\lim }\frac{1}{n}\sum _{r=1}^n\ell n\left(\frac{r}{n}\right)=\underset{0}{\overset{1}{\int }}\ell \ln xdx=x\ell nx-x]}_0^1$
$=(0-1)-\underset{x\to 0^{+}}{\mathrm{Lt}}x\ell nx+0$
$=-1-0=-1$
$\Rightarrow y=\frac{1}{e}$