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I'm practising solving some limits and, currently, I'm trying to solve $\lim\limits_{x\to\infty}{\left({{(x!)^2}\over{(2x)!}}\right)}$.

What I have done:

  • I have attempted to simplify the fraction until I've reached an easier one to solve, however, I'm currently stuck at the following:

$$ \lim_{x→\infty}{\left({{(x!)^2}\over{(2x)!}}\right)}= \lim_{x→\infty}{\left({{(\prod_{i=1}^{x}i)^2}\over{\prod_{i=1}^{2x}i}}\right)}= \lim_{x→\infty}{\left({ { {\prod_{i=1}^{x}i}\cdot{\prod_{i=1}^{x}i} }\over{ { {\prod_{i=1}^{x}}i}\cdot{\prod_{i=x+1}^{2x}i} } }\right)}= \lim_{x→\infty}{\left({ {\prod_{i=1}^{x}i}\over{ {\prod_{i=x+1}^{2x}i}} }\right)}. $$

  • Instinctively, I can see that the limit is equal to $0$, since the numerator is always less than the denominator, thus approaching infinity slower as $x→\infty$.

Question:

  • How can I continue solving the above limit w/o resorting to instinct to determine it equals $0$ ?
  • If the above solution can't go any further, is there a better way to approach this problem?
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  • $\begingroup$ You use the variable $x$ for integers, is that correct? (If so, that's slightly non-standard -- people usually go for $n$, $k$, or $m$ for integer variables) $\endgroup$ – Clement C. Feb 27 '18 at 21:13
  • $\begingroup$ @ClementC. I think that's not a problem at all. $\endgroup$ – Botond Feb 27 '18 at 21:14
  • $\begingroup$ Also, two questions: (i) are you familiar with/allowed to use Stirling's approximation? (ii) if not, are you familiar with properties of the Binomial distribution? $\endgroup$ – Clement C. Feb 27 '18 at 21:14
  • $\begingroup$ @Botond It is not a problem, and I never wrote it was. It'd be equally correct to write $\clubsuit$ for the variable, for that matter. But it is somewhat off-putting at first glance, that's all. $\endgroup$ – Clement C. Feb 27 '18 at 21:15
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Continuing from what you have mentioned, $$0 \le \lim_{x\to\infty}{\left({ {\prod_{i=1}^{x}i}\over{ {\prod_{i=x+1}^{2x}i}} }\right)} = \lim_{x\to\infty}\prod_{i=1}^{x}\frac{i}{i+x} \le \lim_{x\to\infty}\prod_{i=1}^{x}\frac{x}{x+x} = \lim_{x\to\infty}\frac{1}{2^x}=0.$$

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Let $a_n={{(n!)^2}\over{(2n)!}}$ and note that by ratio test

$$\frac{a_{n+1}}{a_n}={{((n+1)!)^2}\over{(2n+2)!}}{{(2n)!}\over{(n!)^2}}=\frac{(n+1)^2}{(2n+2)(2n+1)}\to \frac14$$

then $$a_n\to 0$$

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  • $\begingroup$ @ReneSchipperus Are you sure? It was proved to me during my course in calculus and analysis also for sequence, see the link. $\endgroup$ – user Feb 27 '18 at 21:32
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    $\begingroup$ @ReneSchipperus if the corresponding series converges (which is here proved by the ratio test) then its limit is $0$ as $n$ approaches infinity. +1 from me! $\endgroup$ – user372003 Feb 27 '18 at 21:32
  • $\begingroup$ @user372003 it is also a well known result for sequecences, as root test and ratio-root criteria, Thanks $\endgroup$ – user Feb 27 '18 at 21:34
  • $\begingroup$ I believe you should interchange numerator and denominator in the definition of $a_n$. $\endgroup$ – user Feb 27 '18 at 21:53
  • $\begingroup$ ops...thanks just a typo of course! $\endgroup$ – user Feb 27 '18 at 21:54
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Hint: ${n\choose k}={n-1\choose k-1}+{n-1\choose k}$

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Define $a_n=\frac{(n!)^2}{(2n)!}$. Then it is easy to show that $a_{n+1}<a_n$ and $a_n\in (0,1)$. So $\lim_{n\to\infty}a_n$ exists. Then notice that $$a_{n+1}=a_n\cdot \frac{n}{(2n+1)(2n+2)}.$$ Take limit $n\to\infty$ on both sides, we get $\lim_{n\to\infty}a_n=0$.

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Look into Stirling's Approximation and use asymptotics.

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    $\begingroup$ ah, darn, changed it. $\endgroup$ – Samuel Feb 27 '18 at 21:21
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Use stirling's approximation, namely that $$ n!\sim\sqrt{2\pi n}\left(\frac{n}{e}\right)^n. $$ Hence $$ \lim_{n\to\infty}\frac{(n!)^2}{(2n)!}=\lim_{n\to\infty}\frac{2\pi n\left(\frac{n}{e}\right)^{2n}.} {\sqrt{4\pi n}\times2^{2n}\left(\frac{n}{e}\right)^{2n}} =\lim_{n\to\infty}\frac{\sqrt{\pi n}}{2^{2n}}=0. $$

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Stirling formula may be difficult to remember, but the simpler one below is extremely useful and allows you to solve most asymptotic results with factorials:

Factorial Inequality problem $\left(\frac n2\right)^n > n! > \left(\frac n3\right)^n$

You get $\quad\dfrac{n^{2n}}{9^n}< (n!)^2 < \dfrac{n^{2n}}{4^n}$

And also $\quad\dfrac{4^nn^{2n}}{9^n}< (2n)! < \dfrac{4^nn^{2n}}{4^n}$

So by dividing positive quantities $0<\dfrac{(n!)^2}{(2n)!}<\dfrac{n^{2n}9^n}{4^n4^nn^{2n}}=\left(\dfrac 9{16}\right)^n\to 0$


You can also notice that in this case $\dfrac{(2n)!}{(n!)^2}=C_{2n}^n\ge C_{2n}^1\ge 2n\to\infty$ so its inverse is going to zero.

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