$\lim_{n \to \infty} (\frac{(n+1)(n+2)\dots(3n)}{n^{2n}})^{\frac{1}{n}}$ is equal to : 
$\lim_{n \to \infty} (\frac{(n+1)(n+2)\dots(3n)}{n^{2n}})^{\frac{1}{n}}$  is equal
to :

*

*$\frac{9}{e^2}$

*$3 \log3−2$

*$\frac{18}{e^4}$

*$\frac{27}{e^2}$


My attempt :
$\lim_{n \to \infty} (\frac{(n+1)(n+2)\dots(3n)}{n^{2n}})^{\frac{1}{n}}$
$=\lim_{n \to \infty} (\frac{(n+1)(n+2)\dots(n+2n)}{n^{2n}})^{\frac{1}{n}}$
$=\lim_{n \to \infty} (\frac{{n^{2n}}\{(1+1/n)(1+2/n)\dots(1+2n/n)\}}{n^{2n}})^{\frac{1}{n}}$
$=\lim_{n \to \infty} (\frac{{n^{2n}}\{(1+1/n)(1+2/n)\dots(1+2)\}}{n^{2n}})^{\frac{1}{n}}$
$=\lim_{n \to \infty} (\frac{{n^{2n}}\{(1+1/n)(1+2/n)\dots(3)\}}{n^{2n}})^{\frac{1}{n}}$
$=\lim_{n \to \infty} (\{(1+1/n)(1+2/n)\dots(3)\})^{\frac{1}{n}}$
I'm stuck here.

Can you please explain?


Similarly as in this question we can use the fact that:

If $a_n$ is a sequence of positive real numbers and the limit $\lim\limits_{n\to\infty} \frac{a_{n+1}}{a_n}$ exists, then $\lim\limits_{n\to\infty} \sqrt[n]{a_n}$ also exists and
  $$\lim\limits_{n\to\infty} \sqrt[n]{a_n} = \lim\limits_{n\to\infty} \frac{a_{n+1}}{a_n}.$$

For the proof see, for example Limit of ${a_n}^{1/n}$ is equal to $\lim_{n\to\infty} a_{n+1}/a_n$ or How to show that $\lim_{n \to \infty} a_n^{1/n} = l$?

Now if we apply the above to the sequence 
$$a_n=\frac{(n+1)(n+2)\dots(3n)}{n^{2n}}$$ 
we get
\begin{align*}
\frac{a_{n+1}}{a_n} 
&= \frac{(3n+1)(3n+2)(3n+3)}{n+1} \cdot \frac{n^{2n}}{(n+1)^{2(n+1)}}\\
&= \frac{(3n+1)(3n+2)(3n+3)}{(n+1)^3} \cdot \frac{n^{2n}}{(n+1)^{2n}}\\
&= \frac{(3n+1)(3n+2)(3n+3)}{(n+1)^3} \cdot \frac{1}{\left(1+\frac1n\right)^{2n}}.
\end{align*}
We now see that $$\lim\limits_{n\to\infty} \frac{a_{n+1}}{a_n} = \frac{3^3}{e^2}.$$
