TITLE: Simplifying fraction with factorials: $\frac{(3(n+1))!}{(3n)!}$
QUESTION [6 upvotes]: I was trying to solve the limit:
$$\lim_{n \to \infty} \sqrt[n]{\frac{(3n)!}{(5n)^{3n}}}$$
By using the root's criterion for limits (which is valid in this case, since $b_n$ increases monotonically):
$$L= \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$
Now I realise using Sterling's formula would make everything easier, but my first approach was simplifying the factorial after applying the criterion I mentioned before. So, after a few failed attempts I looked it up on Mathematica and it said that $\frac{(3(n+1))!}{(3n)!}$ (which is one of the fractions you have to simplify) equals $3(n+1)(3n+1)(3n+2)$. Since I can't get there myself I want to know how you would do it.
Just so you can correct me, my reasoning was:
$$\frac{(3(n+1))!}{(3n)!} = \frac{3\cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 4 \cdot (...) \cdot 3 \cdot (n+1)}{3 \cdot 1 \cdot 3 \cdot 2 \cdot 3 \cdot 3 \cdot (...) \cdot 3 \cdot n } = $$
$$= \frac{3^n(n+1)!}{3^{n}n!} = \frac{(n+1)!}{n!} = n+1$$
Which apparently isn't correct. I must have failed at something very silly. Thanks in advance!

REPLY [3 votes]: I thought it might be useful to present an approach that relies on elementary tools only.  To that end, we proceed.

First, we write 
$$\begin{align}
\frac{1}{n}\log((3n!))&=\frac1n\sum_{k=1}^{3n}\log(k)\\\\
&=\left(\frac1n\sum_{k=1}^{3n}\log(k/n)\right)+3\log(n) \tag 1
\end{align}$$
Now, note that the parenthetical term on the right-hand side of $(1)$ is the Riemann sum for $\int_0^3 \log(x)\,dx=3\log(3)-3$.  
Using $(1)$, we have
$$\begin{align}
\lim_{n\to \infty}\sqrt[n]{\frac{(3n)!}{(5n)^{3n}}}&=\lim_{n\to \infty}e^{\frac1n \log((3n)!)-3\log(5n)}\\\\
&=\lim_{n\to \infty}e^{\left(\frac1n\sum_{k=1}^{3n}\log(k/n)\right)+3\log(n)-3\log(5n)}\\\\
&=\lim_{n\to \infty}e^{\left(\frac1n\sum_{k=1}^{3n}\log(k/n)\right)-3\log(5)}\\\\
&= e^{3\log(3)-3-3\log(5)}\\\\
&=\left(\frac{3}{5e}\right)^3
\end{align}$$
as expected!

Tools used:  Riemann sums and the continuity of the exponential function.