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.
 A: 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.
A: 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
$$\frac{a_{n+1}}{a_n}=\frac{(3n+1)(3n+2)(3n+3)}{(n+1)^3} \cdot (1- \frac{1}{n+1})^{2n} \to \frac{27}{e^2}.$$
A: $(2)^{1/n}\rightarrow 1$, so the answer is not altered by adding an $n$ and dropping $2n$ from the middle of the numerator.
Rearrange the terms in the numerator to produce the following pairing:
$$((n)(3n))((n+1)(3n-1))((n+2)(3n-2))\cdots ((2n-1)(2n+1))$$
This is $\prod\limits_{i=1}^{n} (2n-i)(2n+i) = \prod\limits_{i=1}^{n} (4n^2-i^2)$ 
After dividing by $n^{2n}$, you obtain $\prod\limits_{i=1}^{n} (4-(i/n)^2)$. 
So the limit you are looking for equals to $\exp(\lim\limits_{n\rightarrow \infty} \sum\limits_{i=1}^{n}\frac{1}{n}\log(4-\left(\frac{i}{n}\right)^2))$. 
Clearly, as $n$ tends to infinity, the sum $\sum\limits_{i=1}^{n}\frac{1}{n}\log(4-\left(\frac{i}{n}\right)^2))$ tends to the integral $\int\limits_{0}^{1} \log(4-x^2) \, dx = \int\limits_{0}^{1} \log(2-x) \, dx + \int\limits_{0}^{1} \log(2+x) \, dx$. 
It is easy to compute both definit integrals. 
But your calculation is also good. I that case, you obtain a definite integral on $[0,2]$.
A: Just a comment after Lord Shark the Unknown's answer.
When I see things looking like factorials, my first reaction is to think about Stirling's approximation of them.
$$a_n=\left(\frac{\prod_{i=1}^{2n} (n+i)}{n^{2n}}\right)^{\frac 1n}=\left(\frac{(3n)!}{n^{2n}\,n!}\right)^{\frac 1n}$$
$$\log(a_n)=\frac 1n \left(\log((3n)!)-\log(n!)-2n \log(n)\right)$$ Now, Stirling's approximation
$$\log(p!)=p (\log (p)-1)+\frac{1}{2} \left(\log (2 \pi )+\log
   \left({p}\right)\right)+\frac{1}{12
   p}+O\left(\frac{1}{p^3}\right)$$ Apply it and simplify to get
$$\log(a_n)=\frac 1n \left( (3\log (3)-2)n+\frac{\log (3)}{2}-\frac{1}{18
   n}+O\left(\frac{1}{n^3}\right) \right)$$ $$\log(a_n)=(3\log (3)-2)+\frac{\log (3)}{2n}-\frac{1}{18
   n^2}+O\left(\frac{1}{n^4}\right) $$ Now, using
$$a_n=e^{\log(a_n)}=\frac{27 }{ e^2} \left(1+\frac{\log (3)}{2 n}\right)+O\left(\frac{1}{n^2}\right) $$ which shows the limit and how it is approached.
