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I am trying to compute the following limit (k is a fixed constant): $$ \lim_{n\to\infty} \frac{ {n/2 - 1\choose(k-1)/2} {n/2 \choose (k-1)/2} }{n-1 \choose k-1} $$

I expanded the binomial coefficient but I got stuck and couldn't get anywhere from there. In theory, if my approach is correct, this should converge to a constant relative to k.

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What is your definition of, say $\binom{3/2}{1/2}$? –  Dennis Gulko Nov 6 '12 at 13:23
    
@Dennis: if you look at math.stackexchange.com/questions/231212 you will see $k$ is probably intended to be odd. So perhaps $n$ is supposed to be even. –  Henry Nov 6 '12 at 13:46
    
I forgot to mention that $n$ is even and $k$ is odd, so the binomial operator still holds. –  Mohamad Nov 6 '12 at 13:46

1 Answer 1

up vote 3 down vote accepted

Keeping only the leading terms in $n$ yields (with $\ell=k-1$)

$$ \frac{(n/2)^{\ell/2}(n/2)^{\ell/2}}{n^\ell}\cdot\frac{\ell!}{(\ell/2)!^2}=2^{-\ell}\binom\ell{\ell/2}\approx\frac1{\sqrt{\pi\ell/2}}\;, $$

where the estimate on the right is asymptotic for $k\to\infty$.

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$n \to \infty$? –  Henry Nov 6 '12 at 14:08
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@Henry: How do you mean? This gives the limit as $n\to\infty$, as desired. I thought that was clear from "keeping only the leading terms in $n$". The numerator and the denominator are polynomials in $n$ of the same degree, so we only need to look at the leading terms. –  joriki Nov 6 '12 at 14:09
    
@joriki Makes sense.. Thanks a lot –  Mohamad Nov 6 '12 at 14:40

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