Question

$\lim_{n \rightarrow \infty}\frac{(n!)^{1/n}}{n}$ or $\lim_{n \rightarrow \infty}\left( \frac{n!}{n^{n}} \right)^{1/n}$ is equal to

• A. e
• B. e^–1
• C. 1
• D. None of these

Answer 1

Answer: (B)

e^–1

Answer 2

Let $A = \lim_{n \rightarrow \infty}\frac{(n!)^{1/n}}{n}$

⇒ $\log A = \lim_{n \rightarrow \infty}{\log\left( \frac{1.2.3......n}{n^{n}} \right)}^{1 ⥂ / ⥂ n}$

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

⇒ $\log A = \lim_{n \rightarrow \infty}\frac{1}{n}\sum_{r = 1}^{n}\left\lbrack \log\left( \frac{r}{n} \right) \right\rbrack$

⇒ $\log A = \int_{0}^{1}{\log xdx} = \lbrack x\log x - x\rbrack_{0}^{1}$ ⇒ $\log A = - 1$ ⇒ $A = e^{- 1}$