MCQ
$\mathop {\lim }\limits_{n \to \infty } {\left[ {\frac{{n!}}{{{n^n}}}} \right]^{1/n}}$ equals
  • A
    $e$
  • $1/e$
  • C
    $\pi /4$
  • D
    $4/\pi $

Answer

Correct option: B.
$1/e$
b
(b) Let $P = \mathop {\lim }\limits_{x \to \infty } \,{\left( {\frac{{n\,\,!}}{{{n^n}}}} \right)^{1/n}}$

$ = \mathop {\lim }\limits_{n \to \infty } \,{\left( {\frac{1}{n}\,.\,\frac{2}{n}\,.\,\frac{3}{n}\,.\,\frac{4}{n}\,..........\frac{n}{n}} \right)^{1/n}}$

$\therefore \,\,\,\log \,\,P = \frac{1}{n}\,\mathop {\lim }\limits_{n \to \infty } \,\left( {\log \frac{1}{n} + \log \frac{2}{n} + ...... + \log \frac{n}{n}} \right)$

$\log \,\,P = \mathop {\lim }\limits_{n \to \infty } \,\sum\limits_{r = 1}^n {} \frac{1}{n}\log \frac{r}{n}$

$\log \,\,P = \int_0^1 {} \log x\,dx = (x\,\log x - x)_0^1 = ( - 1)$

==> $P = \frac{1}{e}$ .

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