Can someone help me do this limit? $ \lim_{n\to\infty} \frac{n!\times(2n)!}{(3n)!}$ can someone help me with this limit?
I don't know how to expand that factorial multiplication, so what I've done so far is substitute what given:

$$ \lim_{n\to\infty} \frac{n!\times(2n)!}{(3n)!}$$
  $$ \lim_{n\to\infty} \frac{\infty\times\infty}{\infty}$$

And with this I can apply the Cauchy or L'Hôpital's theorem by deriving both sides of the fraction independently, but my problem also stars here because, I don't how to derive a factorial term.
Can someone help me please?
Thanks

This is going to be a horrible overkill, but a funny one. Since
$$ \sum_{n\geq 0}\frac{n!(2n)!}{(3n)!} = \sum_{n\geq 0}(3n+1) B(n+1,2n+1)=\int_{0}^{1}\sum_{n\geq 0}(3n+1)(1-x)^{n}x^{2n}\,dx $$
equals $\phantom{}_3 F_2\left(\frac{1}{2},1,1;\frac{1}{3},\frac{2}{3};\frac{4}{27}\right)$ or $\int_{0}^{1}\frac{1+2x^2-2x^3}{(1-x^2+x^3)^2}\,dx$, which is a convergent integral (and pretty close to $\sqrt{2}$), the main term of the original series goes to zero as $n\to +\infty$.
