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

By Stirling's inequality the answer is clearly $\left(\frac{3}{5e}\right)^3$. To prove it, you may notice that by setting $$ a_n = \frac{(3n)!}{(5n)^{3n}} $$ you have: $$ \frac{a_{n+1}}{a_n} = \frac{(3n+3)(3n+2)(3n+1)(5n)^{3n}}{(5n+5)^{3n+3}} = \frac{\frac{3n+3}{5n+5}\cdot\frac{3n+2}{5n+5}\cdot\frac{3n+1}{5n+5}}{\left(1+\frac{1}{n}\right)^{3n}}\to\frac{\left(\frac{3}{5}\right)^3}{e^3}$$ as $n\to +\infty$.

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.

$$ (3(n+1))! \neq 3 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 4 \cdots 3 \cdot (n+1) $$ 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.