Limit $\lim_{n \to \infty}\frac{(n^2)!}{(2n)!}$ Limit as n goes to infinity for  $$\lim_{n \to \infty}\frac{(n^2)!}{(2n)!}$$.
In my approach, I break down the numerator as 
$${(n^2)(n^2-1)...(2n+1)(2n)!}$$ 
and therefore the value of the original fraction would be infinity. But when I check with wolframalfa, i get the answer as zero. Where am I going wrong here?
Edit: Some comments have suggested that the parentheses in the numerator might not be correct. This is indeed the case.
-Thank you
 A: You may notice that for any $n>2$
$$ \frac{(n^2)!}{(2n)!} = n^2\cdot(n^2-1)\cdot\ldots\cdot(2n+1) \geq n^2$$
holds, hence the limit is obviously $+\infty$. I suspect the original problem was about finding $\lim_{n\to +\infty}\frac{n!^2}{(2n)!}$, instead. In such a case, we may notice that by the Cauchy-Schwarz inequality
$$ \frac{(2n)!}{n!^2} = \binom{2n}{n} = \sum_{k=0}^{n}\binom{n}{k}^2 \stackrel{\text{CS}}{>}\frac{4^n}{n+1} $$
hence $\lim_{n\to +\infty}\frac{n!^2}{(2n)!}=0$.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
{\pars{n^{2}}! \over \pars{2n}!} & \,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,
{\root{2\pi}\pars{n^{2}}^{n^{2} + 1/2}\expo{-n^{2}} \over
\root{2\pi}\pars{2n}^{2n + 1/2}\expo{-2n}} =
{n^{2n^{2} - 2n + 1/2}\expo{-n^{2} + 2n} \over 2^{2n + 1/2}}
\,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\to}\,\,\,\bbx{\ds{\infty}}
\end{align}
A: Since $n^2 > 2n$ for $n > 2,$ we have for these $n$ that
$$\frac{(n^2)!}{(2n)!} = n^2(n^2-1)\cdots (2n+1)\frac{(2n)!}{(2n)!} \ge n^2.$$
Since $n^2\to \infty,$ the limit is $\infty.$
