Evaluate $ \lim_{n\to\infty} \frac{(n!)^2}{(2n)!} $

Question

I'm completely stuck evaluating $ \lim_{n\to\infty} \frac{(n!)^2}{(2n)!} $ how would I go about solving this?

Answer 1

Recalling the Stirling's approximation $$n!\sim\sqrt{2\pi}n^{n+1/2}e^{-n} $$ we have $$\sqrt{2\pi}\lim_{n\rightarrow\infty}\frac{n^{2n+1}e^{-2n}}{\left(2n\right)^{2n+1/2}e^{-2n}}=0. $$

Answer 2

Since each term is bounded above by $2^{-n}$, I guess the answer must be zero.

Answer 3

Hint: for combinatorial reasons (in how many ways can you choose $n$ objects from $n$ pairs of objects?), $$\binom{2n}n\ge2^n\ ,$$ so $$\frac{(n!)^2}{(2n)!}=\frac1{\binom{2n}n}\le\frac1{2^n}\ .$$

Answer 4

As all terms are positive, we have $$0 \leq \frac{(n!)^2}{(2n)!} = \frac{n!}{2n \cdot \dots \cdot (n+1)} = \prod_{k=1}^n \frac{k}{k+n} \leq \prod_{k=1}^n \frac{1}{2} = \left(\frac{1}{2}\right)^n$$

So then as

$$\lim\limits_{n\rightarrow\infty} \left(\frac{1}{2}\right)^n = 0$$

It follows that

$$\lim\limits_{n\rightarrow\infty} \frac{(n!)^2}{(2n)!} = 0$$

Edit:

For what it's worth if you want quick justification of the second inequality step:

$$k \leq n \implies 2k \leq k + n \implies \frac{k}{k+n} \leq \frac{1}{2}$$

Answer 5

Use Stirling's Approximation: $$n! \approx \sqrt{2 \pi n} .n^n e^{-n}$$

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