TITLE: Find the value of : $\lim\limits_{n\to \infty} \sqrt [n]{\frac{(3n)!}{n!(2n+1)!}} $
QUESTION [7 upvotes]: First of all, sorry if something similar to this has been posted before (it's my first time in this web). I need to calculate the limit as $n\rightarrow \infty$ for this:
$$\lim\limits_{n\to \infty} \sqrt [n]{\dfrac{(3n)!}{n!(2n+1)!}} $$
But I don't know which steps I need to follow in order to do it.
Thank you everyone in advance :)

REPLY [4 votes]: You can use the following result:

Let $(a_n)$ be a sequence of positive real numbers. If $\lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}=L$, then $\sqrt[n]{a_n}$ converges too and $\lim\limits_{n\to\infty}\sqrt[n]{a_n}=L$. 

See, for example, the following posts (and other posts shown there among linked questions): 

Convergence of Ratio Test implies Convergence of the Root Test
Finding the limit of $\frac {n}{\sqrt[n]{n!}}$
Inequality involving $\limsup$ and $\liminf$: $ \liminf(a_{n+1}/a_n) \le \liminf((a_n)^{(1/n)}) \le \limsup((a_n)^{(1/n)}) \le \limsup(a_{n+1}/a_n)$ 


If you apply the above result to $a_n=\frac{(3n)!}{n!(2n+1)!}$ you get that the limit is
$$\lim_{n\to\infty} \frac{a_{n+1}}{a_n} =\frac{(3n+3)(3n+2)(3n+1)}{(n+1)(2n+3)(2n+2)} = \frac{27}4.$$