Find the value of : $\lim\limits_{n\to \infty} \sqrt [n]{\frac{(3n)!}{n!(2n+1)!}} $ 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 :)
 A: 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
$$\lim_{n\to\infty} \frac{a_{n+1}}{a_n} =\frac{(3n+3)(3n+2)(3n+1)}{(n+1)(2n+3)(2n+2)} = \frac{27}4.$$
A: Using Stirling's formula, $n!\approx n^ne^{-n}\sqrt{2\pi n}$ (I'll leave out the strict explanation of what $\approx$ means in this context), we have
$$ \frac{(3n)!}{n!(2n)!}\approx \frac{(3n)^{3n}e^{-3n}\sqrt{6\pi n}}{n^ne^{-n}\sqrt{2\pi n}(2n)^{2n}e^{-2n}\sqrt{4\pi n}}=\frac{27^{n}}{4^{n}}\cdot \sqrt{\frac{3}{4n}}.$$
From this you should find that the limit is $\frac{27}{4}$.
A: Without Stirling: 
consider $e^{\frac{\log a_n}{n}}$, where $a_n  = \frac{(3n)!}{n!(2n+1)!}$ and consider that $\log n!  = \sum_{k=1}^{n} \log k \sim \int_{1}^{\infty} \log x dx = n \log n - (n-1)$. A lot of terms will cancel out, many will divide $n$ and the rest will be of the form $e^{\frac{1}{n}} \to_n 1$.   
