Factorial quotient with same number of factors

Question

I have this limit:

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

How can I evaluate this? I know that the answer is $+\infty$

Answer 1

Note that $$ \frac{(3n)!}{(2n)!} = \frac{(3n)(3n-1)(3n-2)...}{(2n)(2n-1)(2n-2)...} = (3n)(3n-1)(3n-2)...(2n+2)(2n+1) $$ and similarly $$ \frac{(2n)!}{n!} = \frac{(2n)(2n-1)(2n-2)...}{(n)(n-1)(n-2)...} = (2n)(2n-1)(2n-2)...(n+2)(n+1). $$ So $$ \frac{n!(3n)!}{(2n)!(2n)!} = \frac{(3n)(3n-1)(3n-2)...(2n+2)(2n+1) }{(2n)(2n-1)(2n-2)...(n+2)(n+1)}\ge \left(\frac{3}{2}\right)^n, $$ which is unbounded.

Answer 2

You can use Stirling's approximation $n! \approx n^ne^{-n}\sqrt{2 \pi n}$. Plug that in and see what happens to the numeric factors. All the $n$s will disappear, as will the $e$s

Answer 3

HINT

One way is to cancel the terms: $$ \frac{n! (3n)!}{(2n)!^2} = \frac{\prod_{k=1}^n k \times \prod_{k=1}^{3n} k} {\prod_{k=1}^{2n} k^2} = \frac{(2n+1) \times \ldots \times 3n} {(n+1) \times \ldots \times 2n} = \prod_{k=1}^n \frac{2n+k}{n+k} = \prod_{k=1}^n \left(1 + \frac{n}{n+k}\right) $$ so each product term is actually above one, and later ones are closer to $1.5$, and the number of such terms increases without bound as $n \to \infty$.

Answer 4

As an inductive step, for $n\ge1$, $$ \begin{align} \frac{n!(3n)!}{(2n)!^2} &=\frac{n\,3n(3n-1)(3n-2)}{2n(2n-1)\,2n(2n-1)}\frac{(n-1)!(3n-3)!}{(2n-2)!^2}\\ &=\frac{27}{16}\frac{\left(1-\frac1{3n}\right)\left(1-\frac2{3n}\right)}{\left(1-\frac1{2n}\right)\left(1-\frac1{2n}\right)}\frac{(n-1)!(3n-3)!}{(2n-2)!^2}\\ &=\frac{27}{16}\frac{1-\frac1n+\frac2{9n^2}}{1-\frac1n+\frac1{4n^2}}\frac{(n-1)!(3n-3)!}{(2n-2)!^2}\\ &\ge\frac32\frac{(n-1)!(3n-3)!}{(2n-2)!^2} \end{align} $$

Answer 5

Interesting is the case of $$A_n=\frac{(a n)!\, (b n)!}{\left(\left(\frac{a+b}{2} n \right)!\right)^2}$$ Using three times Stirling approximation $$\log(A_n)= \left(a \log (a)+b \log (b)-(a+b) \log \left(\frac{a+b}{2}\right)\right) n+\log \left(\frac{2 \sqrt{a b}}{a+b}\right)+O\left(\frac{1}{n}\right)$$