Question

The value of $\displaystyle \lim_{n \to \infty}\frac{\left(n!\right)^{\frac{1}{n}}}{n}$ is

• A. 1
• B. $\frac{1}{e^{2}}$
• C. $\frac{1}{2e}$
• D. $\frac{1}{e}$

Answer 1

The Correct Option is D

$\displaystyle \lim _{x \rightarrow \infty} \frac{(n !)^{1 / n}}{n}=\displaystyle \lim _{n \rightarrow \infty}\left(\frac{n !}{n^{n}}\right)^{1 / n}$
We have, $\frac{n !}{n^{n}}=\frac{1 \cdot 2 \cdot 3}{n \cdot n \cdot n} \quad n$
$\therefore \left\{\frac{n !}{n^{n}}\right\}^{1 / n}=\left\{\frac{1}{n} \cdot \frac{2}{n} \cdot \frac{3}{n} \cdots \frac{r}{n} \cdots \frac{n}{n}\right\}^{1 / n}$
$\Rightarrow \displaystyle\lim _{n \rightarrow \infty}\left\{\frac{n !}{n^{n}}\right\}^{1 / n}$
$=\displaystyle\lim _{n \rightarrow \infty}\left\{\frac{1}{n} \cdot \frac{2}{n} \cdot \frac{3}{n} \cdots \frac{r}{n} \cdots \frac{n}{n}\right\}^{1 / n}$
Let $A=\displaystyle\lim _{n \rightarrow \infty}\left\{\frac{n !}{n^{n}}\right\}^{1 / n}$
Then, $A=\displaystyle\lim _{n \rightarrow \infty}\left\{\frac{1}{n} \cdot \frac{2}{n} \cdot \frac{3}{n} \cdots \frac{r}{n} \cdots \frac{n}{n}\right\}^{1 / n}$
$\Rightarrow \log A =\displaystyle\lim _{n \rightarrow \infty} \frac{1}{n} \Sigma \log \left(\frac{r}{n}\right)=\int\limits_{0}^{1} \log x d x$
$=\left[x \log x-\int \frac{1}{x} \cdot x d x\right]_{0}^{1}$
Integrating by parts
$=[x \log x-x]_{0}^{1}=-1$
$\Rightarrow A=e^{-1}=\frac{1}{e}$