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!
 A: 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 $n\to +\infty$.
A: 
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.

A: $$ (3(n+1))! \neq 3 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 4 \cdots 3 \cdot (n+1) $$  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.  
A: My favorite elementary inequality
for $n!$:
$
\def\lfrac#1#2{{\large\frac{#1}{#2}}}
$
$(n/e)^n
< n!
< (n/e)^{n+1}
$.
Taking n-th roots,
$n/e
< (n!)^{1/n}
< (n/e)^{1+1/n}
$.
Therefore
$\sqrt[n]{\lfrac{(3n)!}{(5n)^{3n}}}
=\lfrac{((3n)!)^{1/n}}{(5n)^{3}}
>\lfrac{((3n/e)^{3n})^{1/n}}{(5n)^{3}}
=\lfrac{(3n/e)^{3}}{(5n)^{3}}
=\lfrac{27n^3}{125e^3n^{3}}
=\lfrac{27}{125e^3}
$.
Arguing the other way,
$\sqrt[n]{\lfrac{(3n)!}{(5n)^{3n}}}
=\lfrac{((3n)!)^{1/n}}{(5n)^{3}}
<\lfrac{((3n/e)^{3n+1})^{1/n}}{(5n)^{3}}
=\lfrac{(3n/e)^{3}(3n/e)^{1/n}}{(5n)^{3}}
=\lfrac{27n^3(3n/e)^{1/n}}{125e^3n^{3}}
=\lfrac{27}{125e^3}(3n/e)^{1/n}
$.
Since,
as $n \to \infty$,
$n^{1/n}
\to 1
$
and
$a^{1/n}
\to 1
$
for any $a > 0$,
$(3n/e)^{1/n}
\to 1
$ so the limit is
$\lfrac{27}{125e^3}
$.

A more general result:
$\sqrt[n]{\lfrac{(an)!}{(bn)^{an}}}
=\lfrac{((an)!)^{1/n}}{(bn)^a}
\gt \lfrac{((an/e)^{an})^{1/n}}{(bn)^a}
=\lfrac{(an/e)^a}{(bn)^a}
=\lfrac{a^a n^a}{b^ae^a}
=\lfrac{a^a}{b^ae^a}
=\left(\lfrac{a}{be}\right)^a
$.
The other way goes exactly as above,
so that
$\lim_{n \to \infty} \sqrt[n]{\lfrac{(an)!}{(bn)^{an}}}
=\left(\lfrac{a}{be}\right)^a
$.
