Correct way to solve a limit. I have the following limit,
$$\lim_{n\rightarrow \infty}\left \{ \frac{\left ( n+1 \right )\left ( n+2 \right )...3n}{n^{2n}} \right \}^{\frac{1}{n}}$$
My procedure of solving (which is wrong).
Step 1: I break up the expression in the following manner,
$$\lim_{n\rightarrow \infty }\left  \{ \left ( 1+\frac{1}{n} \right )\left ( 1+\frac{2}{n} \right )...\left ( \frac{3n}{n} \right ) \right  \}^{\frac{1}{n}}$$
Step 2: I apply the limits,
$$\left \{ \left ( 1+\frac{1}{\infty } \right ) \left ( 1+\frac{2}{\infty } \right )...\left ( 2+\frac{1}{\infty } \right )...3\right \}^{\frac{1}{\infty }}$$
which makes it, $$\left \{ 1^{\infty } *2^{\infty }*3\right \}^{0}$$
I am stuck here and could not proceed. Can somebody please help me out?
 A: A simpler approach is to use the following theorem on sequences:

Theorem: If $a_{n}$ is a sequence of positive terms such that $\lim\limits_{n \to \infty}\dfrac{a_{n + 1}}{a_{n}} = L$ then $\lim\limits_{n \to \infty}\sqrt[n]{a_{n}} = L$.

Let $$a_{n} = \frac{(n + 1)(n + 2)\cdots(n + 2n)}{n^{2n}}$$ and the sequence in question is $b_{n} = \sqrt[n]{a_{n}}$. We have
\begin{align}
\frac{a_{n + 1}}{a_{n}} &= \frac{(n + 2)(n + 3)\cdots(3n + 3)}{(n + 1)^{2n + 2}}\cdot\frac{n^{2n}}{(n + 1)(n + 2)\cdots(n + 2n)}\notag\\
&= \frac{3(3n + 1)(3n + 2)}{(n + 1)^{2}}\left(\frac{n}{n + 1}\right)^{2n}\notag\\
&= 27\cdot\dfrac{\left(1 + \dfrac{1}{3n}\right)\left(1 + \dfrac{2}{3n}\right)}{\left(1 + \dfrac{1}{n}\right)^{2}}\cdot\left\{\left(1 + \dfrac{1}{n}\right)^{n}\right\}^{-2}\notag\\
&\to \frac{27}{e^{2}}\text{ as }n \to \infty\notag
\end{align}
and therefore the sequence in question $b_{n} = a_{n}^{1/n} \to 27/e^{2}$ as $n \to \infty$.
A: Note that we can write
$$\frac{n(n+1)\cdots 3n}{n^{2n}}=\prod_{k=1}^{2n}\left(1+\frac kn\right)$$
Proceeding we find 
$$\lim_{n\to\infty}\left(\prod_{k=1}^{2n}\left(1+\frac kn\right)\right)^{1/n}=\lim_{n\to\infty}e^{\frac 1n \sum_{k=1}^{2n}\log\left(1+\frac kn\right)}=e^{\int_0^2\log(1+x)\,dx}$$
Can you finish now?
A: $$ \text{ let }y = \left \{ \frac{\left ( n+1 \right )\left ( n+2 \right )...3n}{n^{2n}} \right \}^{\frac{1}{n}}.$$ then 
$$\begin{align} \ln y &= \frac 1n\left(\ln\left(1+\frac1n\right)+ \ln\left(1+\frac2n\right) +\cdots+ \ln\left(1+\frac {2n}n\right)\right)\\
&= \int_1^3ln(x) \ dx+\cdots\\
&= \left(x\ln x - x\right)_1^3+\cdots\\
&= 3\ln(3) - 2 = \ln\left(\frac{27}{e^2}\right)+\cdots\end{align} $$ therefore $$\lim_{n\to \infty}y =  \frac{27}{e^2}.$$
A: Hint: Apply $\ln $ to the expression to obtain a Riemann sum whose limit is a recognizable integral.
A: Try this: 
$$\lim_{n\rightarrow \infty} \left(\dfrac{n+1}{n}\cdot\dfrac{n+2}{n}\cdot\cdot\cdot\dfrac{3n}{n}\right)^{\dfrac{1}{n}}=\lim_{n\rightarrow \infty} \left(\left(1+\dfrac{1}{n}\right)\cdot\left(1+\dfrac{2}{n}\right)\cdot\cdot\cdot3\right)^{\dfrac{1}{n}}=\lim_{n\rightarrow \infty}\left(1+\dfrac{1}{n}\right)^{\dfrac{1}{n}}\cdot\lim_{n\rightarrow \infty}\left(1+\dfrac{2}{n}\right)^{\dfrac{1}{n}}\cdot\cdot\cdot\lim_{n\rightarrow \infty}3^{\dfrac{1}{n}}=1\cdot 1\cdot1\cdot\cdot\cdot1=1$$
