I'm trying to find the limit of the following, where $m$ is a constant
$$\lim_{n \rightarrow \infty}\frac{(n-1)!}{2}\bigg(\frac{m}{n}\bigg)^{n}.$$
I start with using Stirling's approximation
$$(n-1)! \approx \sqrt{2\pi (n-1)}\bigg(\frac{n-1}{e}\bigg)^{(n-1)}$$
So I obtain
$$ \frac{\sqrt{2\pi}}{2}\lim_{n \rightarrow \infty} = (n-1)^{n-\frac12}e^{1-n}\bigg(\frac{m}{n}\bigg)^{n}$$
to which I can't see any simplification. Please suggest some hints. Thanks
Best Answer
Hint:
By Stirling's Approximation
$$\lim_{n\to \infty}\dfrac{n!}{2n}\left(\dfrac{m}{n}\right)^n=\lim_{n\to \infty}\dfrac{\sqrt{2\pi n}}{2n}\left(\dfrac{n}{e}\right)^n\left(\dfrac{m}{n}\right)^n=\lim_{n\to \infty}\dfrac{\sqrt{2\pi n}}{2n}\left(\dfrac{m}{e}\right)^n$$
Can you proceed?