Value of $\lim_{n \to \infty} \left({\frac{(n+1)(n+2)(n+3)...(3n)}{n{^{2n}}}}\right)^{1/n}$ I was asked to evaluate the following expression:
$\lim_{n \to \infty} \left({\frac{(n+1)(n+2)(n+3)...(3n)}{n{^{2n}}}}\right)^{1/n}$
My first step was to assume that the limit existed, and set that value to $y$. 
$ y = \lim_{n \to \infty} \left({\frac{(n+1)(n+2)(n+3)...(3n)}{n{^{2n}}}}\right)^{1/n}$
And then, I took the natural logarithm of both sides of the equation. I obtained the expression:
$ \ln y = \lim_{n \to \infty} \frac{1}{n} \cdot \left(\ln(1+\frac{1}{n}) + \ln(1+\frac{2}{n}) + ... + \ln(1+\frac{2n}{n})\right) $
This simplified to: 
$ \ln y = \lim_{n \to \infty} \frac{1}{n} \cdot \sum_{k = 1}^{\color{Red}{2n}} \ln(1+\frac{k}{n}) $
I realize that this is similar to the form of a Riemann sum, which can then be manipulated to give the expression in the form of a definite integral. However, the part bolded in red, which is $ 2n$, throws me off. I have only seen Riemann sums be evaluated when the upper limit is $ n - k $, where $k$ is a constant.
Therefore, how would I go about evaluating this expression?
Thank you for all help in advance.

We use the following result: if $(a_n)$ is a sequence such that $a_n >0$ for all $n$ and $(\frac{a_{n+1}}{a_n})$ is convergent, then $(a_n^{1/n})$ is convergent and
$$\lim_{n \to \infty}\frac{a_{n+1}}{a_n}=\lim_{n \to \infty}a_n^{1/n}.$$
Now let $a_n:= \frac{(n+1)(n+2)(n+3)...(3n)}{n{^{2n}}}$.
Some easy computations give
$$\frac{a_{n+1}}{a_n}=\frac{(3n+1)(3n+2)(3n+3)}{(n+1)^3} \cdot (1- \frac{1}{n+1})^{2n} \to \frac{27}{e^2}.$$
