# Limit with binomial coefficients

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. – mhdn Nov 6 '12 at 13:46

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 $k\to\infty$.
$n \to \infty$? – Henry Nov 6 '12 at 14:08
@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