Question

${lim}_{\pi \to \infty }{\left[\frac{n!}{n^n}\right]}^{\frac{1}{n}}equals$ [Kurukshetra CEE 1998]

• A. $e$
• B. $\frac{1}{e}$
• C. $\frac{\pi }{4}$
• D. $\frac{4}{\pi }$

Answer 1

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

$LetP={lim}_{x\to \infty }{\left(\frac{n!}{n^n}\right)}^{\frac{1}{n}}={lim}_{x\to \infty }{(\frac{1}{n}.\frac{2}{n}.\frac{3}{n}.\frac{4}{n}\cdots \cdots \frac{n}{n})}^{\frac{1}{n}}$
$\therefore logP=\frac{1}{n}{lim}_{\pi \to \infty }(log\frac{1}{n}+log\frac{2}{n}+\cdots \cdots +log\frac{n}{n})$
$logP{lim}_{x\to \infty }{\sum }_{r=1}^n\frac{1}{n}log\frac{r}{n}$
$logP={\int }_0^{1}logxdx=(xlogx-x{)}_0^1=(-1)\Rightarrow P=\frac{1}{e}$