Finding the limit of $\frac{(2n)!}{n!}$ So I've tried to use Stirling's approximation and got that $\lim\frac{2n!}{n!}=\frac{1}{1}=1$
Any thoughts?

$\frac{2n!}{n!} = \frac{2n(2n-1)(2n-2)!}{n(n-1)!} \ge 2.\frac{2(n-1)!}{(n-1)!}$
So, $\frac{2n!}{n!} \ge 2^n$
$\lim\limits_{n \to +\infty} \frac{2n!}{n!} \ge \lim\limits_{n \to +\infty} 2^n = + \infty$
