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
 A: In this case I'd just write it out recursively. Denote by $a_n$ the $n^{th}$ term of the sequence, so that
$$ a_n = \frac{n!(2n)!}{(3n)!} = \frac{n(2n)(2n-1)}{(3n)(3n-1)(3n-2)}a_{n-1}$$
The basic idea behind understanding sequences is finding something simpler to compare it to. So let's take apart that factor:
$$ \frac{n(2n)(2n-1)}{(3n)(3n-1)(3n-2)} = \frac{1}{3}\frac{2n-1}{3n-1}\frac{2n-1}{3n-2} < \frac{1}{3} $$
since each of those latter two fractions is less than $1$.
That means every new term in the sequence is less than a third the previous term: $a_n < a_{n-1}/3$. Your sequence starts with $1$, so $a_n < (1/3)^n$. 
So now, ask a simpler question: what does $(1/3)^n$ do? Then use the answer to answer your original question.
A: 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 $n\to +\infty$.
A: Hint:
$$\frac{n! (2n)!}{(3n)!} = \prod_{i=1}^{n}\frac{i}{2n+i},$$ and
$$\frac{i}{2n+i} \le \frac{1}{3}.$$
A: Let $a_n = \dfrac{n!(2n)!}{(3n)!}$. Then
$$
\frac{a_{n+1}}{a_n}
= \frac{(n+1)!(2n+2)!}{(3n+3)!}\frac{(3n)!}{n!(2n)!}
= \frac{(n+1)(2n+2)(2n+1)}{(3n+3)(3n+2)(3n+1)}
= \frac{4n^3+\cdots}{27n^3+\cdots}
\to \frac{4}{27} < 1
$$
Therefore, $a_n \to 0$.
A: Hint: You don't need to use l'Hospital's rule here at all; simply write out what the factorials mean... for example, $$(3n)!=(3n)\times (3n-1)\times (3n-2)\times\cdots\times (3)\times (2)\times(1).$$ Now write out in similar fashion the $(2n)!$ and you'll see a trend in the $(2n)!$ that also occurs in the $(3n)!$. All that is left is considering the $n!$, and if you write it out in the same way as the other two, you can make some more (though less obvious) connections to the $(3n)!$.
A: Correct me if wrong :
$a_n=\dfrac{(n!)(2n!)}{(3n!)}=$
$\dfrac{n(n-1)(n-2)....1}{(2n+1)(2n+2)........(2n+n)}$
$\lt \dfrac{n^n}{(2n)^n}= \dfrac{1}{2^n}.$
Hence $\lim_{n \rightarrow \infty} a_n =0$.
A: As  Lord Shark the Unknown commented, every time I see products/quotients of factorials, my first reaction is to use Stirling approximation.
$$a_n=\frac{n!\times(2n)!}{(3n)!}\implies \log(a_n)=\log(n!)+\log((2n)!)-\log((3n)!)$$
Now, Stirling approximation
$$\log(p!)=p (\log (p)-1)+\frac{1}{2} \left(\log (2 \pi )+\log
   \left({p}\right)\right)+O\left(\frac{1}{p}\right)$$ making
$$\log(a_n)=\log \left(2 \sqrt{\frac{\pi
   }{3}}\right)+n (2 \log (2)-3 \log (3))+\frac{1}{2} \log \left({n}\right)+O\left(\frac{1}{n^3}\right)$$ and $(2 \log (2)-3 \log (3))<0$ from which you can conclude.
The advantage is that you can easily compute the value of $a_n$; for example, the above formula would give
$$\log(a_{10})=-17.2279\implies a_{10}=3.296 \times 10^{-8}$$ while the exact value is 
$$a_{10}=\frac{1}{30045015}\approx 3.328 \times 10^{-8}$$
