Question

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

Answer 1

Let $\frac{(n!{)}^{1/n}}{n}=t$

$\log t=\frac{1}{n}(\log n!-n\log n)=\frac{1}{n}\sum _{k=1}^n\log \frac{k}{n}$

Here we have used $\log n!=\log 1+\log 2+\dots \cdots +\log n$

$\underset{n\to \infty }{lim}\frac{1}{n}\sum _{k=0}^n\log \frac{k}{n}$

Above sum is the Riemann sum.

$\therefore \underset{n\to \infty }{lim}\frac{1}{n}\sum _{k=0}^n\log \frac{k}{n}={\int }_0^1\log xdx$

$\underset{n\to \infty }{lim}\log t={\int }_0^1\log xdx=[x\log x-x{]}_0^1=-1$

$\underset{n\to \infty }{lim}t=\frac{1}{e}$