Question

$\lim_{n \rightarrow \infty}\left\lbrack \frac{n!}{n^{n}} \right\rbrack^{1/n}$equals

• A. e
• B. $1/e$
• C. $\pi/4$
• D. $4/\pi$

Answer 1

Answer: (B)

$1/e$

Answer 2

Let $P = \lim_{x \rightarrow \infty}\left( \frac{n!}{n^{n}} \right)^{1/n}$

$= \lim_{n \rightarrow \infty}\left( \frac{1}{n}.\frac{2}{n}.\frac{3}{n}.\frac{4}{n}..........\frac{n}{n} \right)^{1/n}$

$\therefore\log{}P = \frac{1}{n}\lim_{n \rightarrow \infty}\left( \log\frac{1}{n} + \log\frac{2}{n} + ...... + \log\frac{n}{n} \right)$

$\log{}P = \lim_{n \rightarrow \infty}\sum_{r = 1}^{n}{}\frac{1}{n}\log\frac{r}{n}$

$\log{}P = \int_{0}^{1}{}\log xdx = (x\log x - x)_{0}^{1} = ( - 1)$ ⇒ $P = \frac{1}{e}$ .