Question

The value of $\underset{n\to \infty }{lim}\frac{(n!{)}^{\frac{1}{n}}}{n}$ is?

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

Answer 1

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

$\underset{n\to \infty }{lim}\frac{(n!{)}^{\frac{1}{n}}}{n}=\underset{n\to \infty }{lim}{\left(\frac{n!}{n^n}\right)}^{\frac{1}{n}}$

We have,

$\frac{n!}{n}=\frac{1\cdot 2\cdot 3n}{n\cdot n\cdot nn}$

$\therefore {\left\{\frac{n!}{n^n}\right\}}^{\frac{1}{n}}={\{\frac{1}{n}\cdot \frac{2}{n}\cdot \frac{3}{n}\cdot \cdot \cdot \frac{r}{n}\cdot \cdot \cdot \frac{n}{n}\}}^{\frac{1}{n}}$

$\Rightarrow \underset{n\to \infty }{lim}{\left\{\frac{n!}{n^n}\right\}}^{\frac{1}{n}}=\underset{n\to \infty }{lim}{\{\frac{1}{n}\cdot \frac{2}{n}\cdot \frac{3}{n}\cdot \cdot \cdot \frac{r}{n}\cdot \cdot \cdot \frac{n}{n}\}}^{\frac{1}{n}}$

Let $A=\underset{n\to \infty }{lim}{\left\{\frac{n!}{n^n}\right\}}^{\frac{1}{n}}$

Then, $A=\underset{n\to \infty }{lim}{\{\frac{1}{n}\cdot \frac{2}{n}\cdot \frac{3}{n}\cdot \cdot \cdot \frac{r}{n}\cdot \cdot \cdot \frac{n}{n}\}}^{\frac{1}{n}}$

$\Rightarrow \log A=\underset{n\to \infty }{lim}\frac{1}{n}\sum \log \left(\frac{r}{n}\right)={\int }_0^1\log xdx$

$={[x\log x-\int \frac{1}{x}\cdot xdx]}_0^1$

Integrating by parts, we get

$=[x\log x-x{]}_0^1=-1$

$\Rightarrow A=e^{-1}=\frac{1}{e}$