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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

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  • $\begingroup$ seems like if $-1<m<1$ the sequence converge $\endgroup$ – sango Mar 13 '19 at 15:37
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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?

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  • $\begingroup$ 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. $\endgroup$ – rami_salazar Mar 13 '19 at 15:44
  • $\begingroup$ @rodger_kicks Yes, as also shown on this Desmos graph. $\endgroup$ – Toby Mak Mar 15 '19 at 7:51

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