Question about the limit of a sequence. I am stuck on the following problem to find the limit of the sequence $a_n=\frac{(n!)^2 \cdot (2n)!}{(4n)!}$. I show $\lim_{n \rightarrow \infty} |\frac{a_{n+1}}{a_n}| < 1$, thus $a_n \rightarrow 0$. However, I would like to figure out a more algebraic way of doing this, although I am stuck on how to simplify this expression. Could someone offer a hint on how to get started, not a full solution please.
Thank you.
 A: At the request of the OP, we present herein a purely algebraic way forward.
HINT:
Let $a_n=\frac{(n!)^2\,(2n)!}{(4n)!}$.  Then, we can write
$$\begin{align}\log(a_n)&=2\log(n!)+\log((2n)!)-\log((4n)!)\\\\&=2\sum_{k=1}^n\log(k) +\sum_{k=1}^{2n}\log(k)-\sum_{k=1}^{4n}\log(k)\\\\&=2\sum_{k=1}^n\log(k)-\sum_{2n+1}^{4n}\log(k)\end{align}$$
SPOLIER ALERT:  Scroll Over the Highlighted Area to Reveal the Full Solution

\begin{align}\log(a_n)&=2\log(n!)+\log((2n)!)-\log((4n)!)\\\\&=2\sum_{k=1}^n\log(k) +\sum_{k=1}^{2n}\log(k)-\sum_{k=1}^{4n}\log(k)\\\\&=2\sum_{k=1}^n\log(k)-\sum_{2n+1}^{4n}\log(k)\\\\&=2\sum_{k=1}^n\log(k)-\sum_{k=1}^{2n}\log(k+2n)\\\\&\le 2n\log(n)-2n\log(2n+1)\\\\&=-2n\log\left(2+\frac1n\right)\\\\&<-2n\end{align}Inasmuch as $\lim_{n\to \infty}e^{-2n}=0$, we find that $$\lim_{n\to \infty}a_n=0$$

A: Start with this and simplify
$$
\frac{a_{n+1}}{a_n}
=
\frac{((n+1)!)^2 \cdot (2(n+1))!}{(4(n+1))!}
\cdot
\frac{(4n)!}{(n!)^2 \cdot (2n)!}\\
=
\frac{((n+1)n!)^2 \cdot (2n+2)(2n+1)(2n)!}{(4n+4)(4n+3)(4n+2)(4n+1)(4n)!}
\cdot
\frac{(4n)!}{(n!)^2 \cdot (2n)!}
$$
