Evaluate $ \lim_{n\to\infty} \frac{(n!)^2}{(2n)!} $ I'm completely stuck evaluating $ \lim_{n\to\infty} \frac{(n!)^2}{(2n)!} $ how would I go about solving this?
 A: Hint: for combinatorial reasons (in how many ways can you choose $n$ objects from $n$ pairs of objects?),
$$\binom{2n}n\ge2^n\ ,$$
so
$$\frac{(n!)^2}{(2n)!}=\frac1{\binom{2n}n}\le\frac1{2^n}\ .$$
A: 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:
$$k \leq n \implies 2k \leq k + n \implies \frac{k}{k+n} \leq \frac{1}{2}$$
A: Recalling the Stirling's approximation $$n!\sim\sqrt{2\pi}n^{n+1/2}e^{-n}
 $$ we have $$\sqrt{2\pi}\lim_{n\rightarrow\infty}\frac{n^{2n+1}e^{-2n}}{\left(2n\right)^{2n+1/2}e^{-2n}}=0.
 $$
A: Since each term is bounded above by $2^{-n}$, I guess the answer must be zero.
A: 
Use Stirling's Approximation: 
  $$n! \approx \sqrt{2 \pi n} .n^n e^{-n}$$

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