# Finding limit with approximation

I'm trying to find the limit of the following, where $$m$$ is a constant

$$\lim_{n \rightarrow \infty}\frac{(n-1)!}{2}\bigg(\frac{m}{n}\bigg)^{n}.$$

$$(n-1)! \approx \sqrt{2\pi (n-1)}\bigg(\frac{n-1}{e}\bigg)^{(n-1)}$$

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

• seems like if $-1<m<1$ the sequence converge Mar 13, 2019 at 15:37

$$\lim_{n\to \infty}\dfrac{n!}{2n}\left(\dfrac{m}{n}\right)^n=\lim_{n\to \infty}\dfrac{\sqrt{2\pi n}}{2n}\left(\dfrac{n}{e}\right)^n\left(\dfrac{m}{n}\right)^n=\lim_{n\to \infty}\dfrac{\sqrt{2\pi n}}{2n}\left(\dfrac{m}{e}\right)^n$$
• I obtain $\sqrt{2\pi}\lim_{n \rightarrow \infty} \frac{1}{n^.5}\bigg(\frac{m}{e}\bigg)^{n}$ which will equal 0 for $|m|<1$ ? This is not the result I was expecting. Mar 13, 2019 at 15:44