Simplifying fraction with factorials: $\frac{(3(n+1))!}{(3n)!}$ 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!

$$ (3(n+1))! \neq 3 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 4 \cdots 3 \cdot (n+1) $$  To make it clear what the problem is, let's write the right-hand side with brackets: $$ (3 \cdot 2) \cdot (3 \cdot 3) \cdot (3 \cdot 4) \cdots (3 \cdot (n+1)) $$  That's just multiplying all the positive multiples of 3 less than $ 3(n+1) $ together; the factorial is defined as multiplying together all positive integers less than $ 3(n+1) $.  So the correct expansions are \begin{align*}
(3(n+1))! &= 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdots (3n-2) \cdot (3n-1) \cdot 3n \cdot (3n+1) \cdot (3n+2) \cdot 3(n+1) \\
(3n)! &= 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdots (3n-2) \cdot (3n-1) \cdot 3n
\end{align*}
which clearly have the quotient Mathematica gave you.  
