Can anyone provide me a hint for finding the limit involving a factorial function?

Question

I have to find the limit of this function: $\lim _{n\rightarrow \infty }\frac{(n!)^{2}}{(2n)!}$

Can anyone provide a hint?

Answer 1

Hint: $$\frac{\frac{(n+1)!^2}{(2n+2)!}}{\frac{n!^2}{(2n)!}}=\frac{(n+1)^2}{(2n+2)(2n+1)}=\frac{n+1}{4n+2}\xrightarrow[n\to +\infty]{}\frac{1}{4}. $$

Answer 2

When you have factorials, Stirling approximation $$n!\approx \sqrt{2\pi n}\,n^n \,e^{-n}$$ is very useful. So for your case $$\frac{(n!)^{2}}{(2n)!}\approx \frac{\sqrt{\pi n}}{2^{2n}} $$

Answer 3

This answer was inspired by a remark of @Soke: $\frac{(n!)^2}{(2n)!}$ is the reciprocal of the binomial coefficient $\binom{2n}{n}$.

There are $n$ couples in a room. We wish to choose $n$ people. There are $\binom{2n}{n}$ ways to do this.

There are $2^n$ ways to choose $1$ person from each couple. It follows that $\binom{2n}{n}\ge 2^n$.

Answer 4

You could use Stirling's approximation, which makes the limit

$$ \lim_{n \to \infty} \frac {{2 \pi n}(\frac{n}{e})^{2n}}{\sqrt{4 \pi n}(\frac{2n}{e})^{2n}}=\lim_{n\to \infty}\frac{\sqrt{\pi n}}{2^{2n}}=0$$