What is the limit of $n!\cdot (2n)! / (3n)!$ $$\lim_{n\to\infty}{n!\cdot (2n)!\over (3n)!}$$
Unsure as to whether to try and divide each term by $(3n)!$ Or where to start really
 A: Let 
$$ a_n = \frac{n!(2n)!}{(3n)!} = \frac{1}{\binom{3n}{n}}. $$
We have, for every $n\geq 1$:
$$ 0\leq \frac{a_{n+1}}{a_n}=\frac{1}{3}\cdot\frac{(2n+2)(2n+1)}{(3n+2)(3n+1)}\leq\frac{1}{3}\cdot\frac{3}{5}=\frac{1}{5}$$
hence our limit is just zero.
A: Start with Stirling approximation
$$n!\sim \sqrt{2\pi n}\left({n\over e}\right)^n$$
Replace $n$ by $2n$ et $3n$ we get
$${n!(2n)!\over (3n)!}\sim 2\sqrt{\pi}\sqrt{{n\over 3}}\left({4\over 27}\right)^n$$
So the limit we are looking for is $0$
A: After cancellation, we have
$$\prod_{k=0}^{n-1}\frac{n-k}{3n-k} \le \prod_{k=0}^{n-1}\frac{n-k}{3n-3k} = \prod_{k=0}^{n-1}\frac{1}{3} = \frac{1}{3^n} \to 0.$$
A: In a similar fashion to this answer, we can bound
$$
\frac{\left(\frac{27}4\right)^n}{\sqrt{\frac{4\pi}3\left(n+\frac{16}{65}\right)}}\le\binom{3n}{n}\le\frac{\left(\frac{27}4\right)^n}{\sqrt{\frac{4\pi}3\left(n+\frac7{36}\right)}}
$$
for all $n\ge0$.
The left hand inequality tells us that
$$
\frac{n!\,(2n)!}{(3n)!}=\frac1{\binom{3n}{n}}\le\frac{\sqrt{\frac{4\pi}3\left(n+\frac{16}{65}\right)}}{\left(\frac{27}4\right)^n}
$$
Which vanishes rapidly as $n\to\infty$.
