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

 A: 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 $$\lim\limits_{n\to\infty} \frac{a_{n+1}}{a_n} = \frac{3^3}{e^2}.$$
A: GoodDeeds' answer is very nice (and a better answer).  Here is an alternate approach using Stirling's Approximation.  In particular, we need $\ln(n!)=n\ln n-n+O(\log n)$.
We begin by observing that
$$
\left(\frac{(n+1)(n+2)\cdots(3n)}{n^{2n}}\right)^{1/n}=\left(\frac{(3n)!}{n!n^{2n}}\right)^{1/n}.
$$
Now, we continue as above, by applying the logarithm to the limit.  Namely:
\begin{align*}
\lim_{n\rightarrow\infty}\left(\frac{(n+1)(n+2)\cdots(3n)}{n^{2n}}\right)^{1/n}&=\lim_{n\rightarrow\infty}\left(\frac{(3n)!}{n!n^{2n}}\right)^{1/n}\\
&=\operatorname{exp}\left(\ln\left(\lim_{n\rightarrow\infty}\left(\frac{(3n)!}{n!n^{2n}}\right)^{1/n}\right)\right)\\
&=\operatorname{exp}\left(\lim_{n\rightarrow\infty}\left(\ln\left(\frac{(3n)!}{n!n^{2n}}\right)^{1/n}\right)\right)\\&=\operatorname{exp}\left(\lim_{n\rightarrow\infty}\left(\frac{1}{n}\left(\ln((3n!))-\ln(n!)-\ln(n^{2n})\right)\right)\right)
\end{align*}
Next, we apply Stirling's approximation from above to get that this simplifies to 
$$
\operatorname{exp}\left(\lim_{n\rightarrow\infty}\left(\frac{1}{n}\left(3n\ln(3n)-3n+O(\ln(3n))-n\ln(n)+n+O(\ln n)-2n\ln(n)\right)\right)\right)
$$
By using the identity that $\ln(3n)=\ln(3)+\ln(n)$, we can simplify the expression (many terms cancel) above to 
$$
\operatorname{exp}\left(\lim_{n\rightarrow\infty}\left(\frac{1}{n}\left(3n\ln(3)-2n+O(\ln(n))\right)\right)\right)
$$
Since $\lim_{n\rightarrow\infty}\frac{\ln(n)}{n}=0$, the error term vanishes and we are left with
$$
e^{3\ln(3)-2}=\frac{27}{e^2}.
$$
A: For your reference:
Stirling's Formula is useful when one try to evaluate limits involving $n!$.
$$n!\sim\sqrt{2\pi n}\left(\frac ne\right)^n\text{ as } n\to\infty$$  
Alternative solution:
Applying Stirling's Formula:
\begin{align}
L&=\lim_{n\to\infty}\left(\frac{(3n)!}{n!\cdot n^{2n}}\right)^{\frac1n}
\\&=\lim_{n\to\infty}\left(\frac{\sqrt{2\pi (3n)}\left(\frac {3n}e\right)^{3n}}{\sqrt{2\pi n}\left(\frac {n}e\right)^{n}\cdot n^{2n}}\right)^{\frac1n}
\\&=\lim_{n\to\infty}\sqrt3^{\frac1n}\cdot\frac{(\frac{3n}e)^3}{(\frac ne)\cdot n^2}
\\&=\lim_{n\to\infty}\sqrt3^{\frac1n}\cdot\frac{27}{e^2}
\\&=\frac{27}{e^2}
\end{align}
A: $$L=\lim_{n \rightarrow \infty} ((1+\frac1n)(1+\frac2n)\dots(1+\frac{2n}{n}))^{\frac{1}{n}}$$
$$\log L=\log\lim_{n \rightarrow \infty} ((1+\frac1n)(1+\frac2n)\dots(1+\frac{2n}{n}))^{\frac{1}{n}}$$
$$\log L=\lim_{n \rightarrow \infty}\log ((1+\frac1n)(1+\frac2n)\dots(1+\frac{2n}{n}))^{\frac{1}{n}}$$
$$\log L=\lim_{n \rightarrow \infty}\frac1n\log ((1+\frac1n)(1+\frac2n)\dots(1+\frac{2n}{n}))$$
$$\log L=\lim_{n \rightarrow \infty}\frac1n(\log (1+\frac1n)+\log (1+\frac2n)+\dots+\log (1+\frac{2n}{n}))$$
$$\log L=\int_0^2\log(1+x)dx$$
$$\log L=\log(1+x)x-\int\frac{x}{x+1}dx$$
$$\log L=[\log(1+x)x-x+\log|x+1|]_0^2$$
$$\log L=2\log3-2+\log3=\log(3^2)-\log(e^2)+\log3=\log(\frac{27}{e^2})$$
$$L=\frac{27}{e^2}$$
A: This won't guarantee that option 4 is the correct limit, but it does rule out the other three options. We have
$${(n+1)(n+2)\cdots(3n)\over n^{2n}}=\left(1+{1\over n} \right)\cdots\left(1+{n\over n} \right)\left(1+{n+1\over n} \right)\cdots\left(1+{n+n\over n} \right)\ge1\cdots1\cdot2\cdots2=2^n$$
hence
$$\left({(n+1)(n+2)\cdots(3n)\over n^{2n}}\right)^{1/n}\ge2$$
But options 1-3 are all less than $2$.
