Question

Evaluate $\underset{n\to \infty }{lim}{\left[\frac{n!}{n^n}\right]}^{1/n}$.

• A. $\frac{1}{e}$
• B. $\frac{1}{t}$
• C. $\frac{1}{n}$
• D. $noneofabove$

Answer 1

The correct option is A $\frac{1}{e}$

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

$P=\underset{n\to \infty }{lim}{\left(\frac{\mathrm{1.2.3.4...}n}{n.nnn....n}\right)}^{1/n}$

$P=\underset{n\to \infty }{lim}{(\left(\frac{1}{n}\right)\left(\frac{2}{n}\right)\left(\frac{3}{n}\right)...\left(\frac{n}{n}\right))}^{1/n}$

$\Rightarrow lnP=\underset{n\to \infty }{lim}\frac{1}{n}(\log \left(\frac{1}{n}\right)+\log \left(\frac{2}{n}\right)+...\log \left(\frac{n}{n}\right))$

$P=\underset{n\to \infty }{lim}\frac{1}{n}\sum _{r=1}^n\log \frac{r}{n}$

$={\int }_0^{1}lnxdx={[xlnx-x]}_0^1$

$=(0-1)-\underset{x\to 0}{lim}\left(xlnx\right)+0$

$=-1-\underset{x\to 0}{lim}\frac{lnx}{1/x}=-1-\underset{x\to 0}{lim}\frac{1/x}{(-1/x^2)}$

$=-1-\underset{x\to 0}{lim}x=-1+0=-1\Rightarrow lnP=-1$

$P=e^{-1}=1/e$