Question

$\underset{n\to \infty }{\lim }\frac{(n!{)}^{1/n}}{n}$ equals

• A. $e$
• B. $e^{-1}$
• C. $1$
• D. None of these

Answer 1

Answer: (B)

$\frac{n!}{n^n}=\frac{1}{n}\cdot \frac{2}{n}\cdot \frac{3}{n}\cdot \dots \cdot \frac{n}{n}$
Let $A=\underset{n\to \infty }{\lim }\frac{(n!{)}^{1/n}}{n}=\underset{n\to \infty }{\lim }{\left(\frac{n!}{n^n}\right)}^{1/n}$
$\Rightarrow A=\underset{n\to \infty }{\lim }{\left[\frac{1}{n}\cdot \frac{2}{n}\cdot \frac{3}{n}\dots \cdot \frac{n}{n}\right]}^{1/n}$
$\Rightarrow \log A=\underset{n\to \infty }{\lim }\frac{1}{n}[\log \left(\frac{1}{n}\right)+\log \left(\frac{2}{n}\right)+\log \left(\frac{3}{n}\right)+\dots +\log \left(\frac{n}{n}\right)]$
$=\underset{n\to \infty }{\lim }\frac{1}{n}\sum _{t=1}^n\log \left(\frac{r}{n}\right)$
$=\underset{0}{\overset{1}{\int }}\log xdx=|x\log x-x{|}_0^1$
$=[1\log 1-1]-\left[\underset{x\to 0}{\lim }x\log x-0\right]=-1$
$\therefore A=e^{-1}=\frac{1}{e}$