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.

Consider
$$\int_0^2f(x)\,dx$$
where
$$f(x)=\ln(1+x).$$
Splitting $[1,2]$ into $2n$ intervals of length $1/n$ gives a Riemann
sum
$$\frac1n\sum_{k=1}^{2n}f(k/n)=\frac1n\sum_{k=1}^{2n}\ln\left(
1+\frac kn\right)$$
which is exactly yours.
Alternatively you could use Stirling's formula.
