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Proof that $(n!)^2/(2n)!$ converges to $0$. I take following steps:

  1. $(n!)^2/(2n)(2n-1)\cdots(n!) = (n!)/(2n)(2n-1)\cdots(n-1)$. I assume (do I need to prove?) that $n!$ divides $(2n)(2n-1)\cdots(n-1)$.
  2. So I have at the end $1/K$ ($K$ is the remainder after division of the denominator by $n!$).
  3. $1/K$ as increases with increasing $n$ converges to $0$, is a null sequence.

Thanks for any advice.

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$K$ isn't a remainder, it is the quotient. You haven't shown that $K$ increases as $n$ increases, much less that $K$ has infinite limit, so you haven't proven anything yet. – Thomas Andrews Apr 23 '12 at 20:37
Where you have $n-1$, you actually get $n+1$. – Thomas Andrews Apr 23 '12 at 20:40
I would say that you do need to prove that $n!$ divides $(2n)!/(n!)$. You also need to at least provide some sort of lower-bound of $K$ as a function of $n$. – trutheality Apr 23 '12 at 20:43
K increases.Consider n+1. (2(n+1))!/(n+1)!^2. Gives us (2n+2)*(2n+1)*(2n)....(n-1) divided by (n+1)! (n+1)! already was divised by stopping at n-1. – Ignace Apr 24 '12 at 21:20
consider n+1 then (2(n+1))!/(n+1)!^2. We get (2n+2)*(2n+1)/(n+1) when we divide twice by (n+1)!. This increases K surely. 1/K converges to 0. – Ignace Apr 24 '12 at 21:30

I think your first step leads to an easy inequality :

$$\frac{(n!)^2}{(2n)!} = \frac{1}{n+1} \underbrace{\frac{2}{n+2}}_{\le 1}\ldots \underbrace{\frac{n}{2n}}_{\le 1} \le \frac{1}{n+1} \underset{n \to \infty}{\longrightarrow} 0$$

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I don't get the equality. n!in the nominator, denominator ???. I see my assumption n!/2n*(2n-1)...*(n-1). What am i missing here. – Ignace Apr 24 '12 at 21:38
We have $$\frac{(n!)^2}{(2n)!} = \frac{(n!)^2}{n! \times (n+1) \times \ldots \times (2n-1) \times (2n)} = \frac{(n!)}{(n+1) \times \ldots \times (2n-1) \times (2n)} = \frac{1}{n+1} \ldots \frac{n-1}{2n-1} \frac{n}{2n}$$ – Joel Cohen Apr 25 '12 at 12:32
Yes, this confirms my assumption where n! is divided by the serie 2n...n+1. And now ? all terms in n! are dividers of 2n...n+1 (proof). So we end up with 1/K. And to be proved that K increases as n en 2n increase. I suppose i can do this by in induction. – Ignace Apr 26 '12 at 12:57

A combinatorial argument shows that $\frac{(2n)!}{n!^2}=\sum_{j=0}^n\binom nj^2$ (take a set $S$ with $2n$ elements, $S_1$ a set with $n$ elements and $S_2$ its complement; to choose $n$ elements, is to take $k$ elements in $S_1$ and $n-k$ in $S_2$), so $\frac{(2n)!}{n!^2}\geq \binom n1^2=n^2$ and $\frac{n!^2}{(2n)!}\leq \frac 1{n^2}$.

In fact, $\frac{n!^2}{(2n)!}$ behaves like $C\sqrt n4^{-n}$. You can find the constant $C$ thanks to Stirling's formula.

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I think it is better to remark $\frac{(2n)!}{n!^2} = {2n \choose n}$. Also the asymptotics are really $\frac{\sqrt{\pi n}}{4^n}$, not $C4^{-n}$. – sdcvvc Apr 23 '12 at 23:00
@sdcvvc In fact I didn't make the computation, so you result is probably the good one. – Davide Giraudo Apr 24 '12 at 9:18

In Joel's form of the formula, each fraction in the product is less than or equal to 1/2, and since the number of fractions in the product is increasing...

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