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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

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    You likely put it into WolframAlpha wrong. Check your parenthesis. – 2017-02-28
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    Wolframalpha is wrong then, or there was an error in its input. – 2017-02-28
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    Alpha [has it right](http://www.wolframalpha.com/input/?i=Limit%5B((n%5E2)!)%2F((2n)!),n-%3Einf%5D). – 2017-02-28
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    Try stirling approximation for both the numerator and the denominator – 2017-02-28

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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 $\lim_{n\to +\infty}\frac{n!^2}{(2n)!}=0$.

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    (OP appears to have $(n^2)!$, not $(n!)^2$...) – 2017-02-28
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    @StevenStadnicki: my bad, reworking. – 2017-02-28
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    I suspecct the original problem probably was supposed to be $\frac {(n!)^2}{2n!}$. $\frac {n^2!}{2n!}$ seems absurdly obvious. – 2017-02-28
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    @fleablood: well, I guess I will cover both cases, then. – 2017-02-28
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    I'm just speculating. I have no idea where the question came from. But for $n> 2$, $n^2 > 2n$ and for $m > n$, $\frac {m!}{n!} \ge m$ and so $\lim \frac {(n^2!)}{2n!} \ge \lim n^2 = \infty$... well...That *could* be a legitimate question but it's kind of ... silly. – 2017-02-28
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    No the problem is noted correctly. The original form of the question is correct – 2017-03-01
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

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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 $\infty.$