# Limit of Stirling's approximation as n goes to infinity. [duplicate]

I would like to see some detailed solution for $$\frac{n!}{\sqrt{2\pi n} \left(\frac{n}{e}\right)^n}$$ as $n\to\infty$. I know that the answer is 1 but i am not sure why? Here is what is tried:
I rewrote the stirling's formula like this. $$\frac{(e/n)\cdot (2e/n)\cdot (3e/n)\cdots (ne/n)}{\sqrt{2\pi n}}\to 0$$ as $n\to \infty$. I am not sure where I went wrong.
• How do conclude that $\frac{(e/n)\cdot (2e/n)\cdot (3e/n)\cdots (ne/n)}{\sqrt{2\pi n}}\to 0$? – leonbloy Jan 26 '16 at 21:43