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I am interested in the quantity $$ a_{n} = \sqrt{n/2} \frac{\Gamma((n-1)/2)}{\Gamma(n/2)}$$ (this is the geometric bias of the non-central t-distribution with $n$ d.f.) After some plotting, my hunch is that $a_n \approx 1 + \frac{3}{4n} + \mathcal{O}\left(n^{-2}\right)$, as $n\to\infty$. Is there a known asymptotic result of this form? One may assume $n$ is an integer.

A little googling lead me to Feng Qi's excellent survey of inequalities around ratios of $\Gamma$ functions of this form, but I cannot find an asymptotic expansion of this form.

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You could also use the first few terms of the Stirling approximation to $\Gamma$. – Craig Oct 11 '11 at 1:03
@Craig: That's a good idea, and it's probably less mysterious than the DLMF formula I cite in my answer (it would be a bit more computationally intensive, however). – Mike Spivey Oct 11 '11 at 3:02
up vote 5 down vote accepted

You're right that $a_n = 1 + \frac{3}{4n} + O(n^{-2})$. A reference is NIST's Digital Library of Mathematical Functions, Equation 5.11.12, where they have $$\frac{\Gamma(z+a)}{\Gamma(z+b)} \sim z^{a-b} \sum_{k=0}^{\infty} \frac{G_k(a,b)}{z^k},$$ where $G_k(a,b) = \binom{a-b}{k} B^{(a-b+1)}_k(a),$ and $B_n^{(i)}(x)$ is a generalized Bernoulli polynomial.

Your question has $z = n/2$, $a = -1/2$, $b = 0$.

If you take the DLMF formula out one more term, and using $G_0(a,b) =1$, $G_1(a,b) = \frac{1}{2} (a-b)(a+b-1)$, and $G_2(a,b) = \frac{1}{12} \binom{a-b}{2} (3(a+b-1)^2-(a-b+1))$, you get $$a_n = 1 + \frac{3}{4n} + \frac{25}{32n^2} + O(n^{-3}).$$

Added: Believe it or not, Wolfram Alpha can do this for you as well. For example, the link says that $$a_n = 1 + \frac{3}{4n} + \frac{25}{32n^2} + \frac{105}{128n^3} + \frac{1659}{2048 n^4} + \frac{6237}{8192 n^5} + O(n^{-11/2}).$$

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I'm glad I didn't include my hunch for the $n^{-2}$ term, since it was quite a bit off! – shabbychef Oct 11 '11 at 0:02
thanks, again. A little more digging revealed that $a_n^{-1}$ is called '$c_4$' in the statistical process control literature (see e.g. Mathworld has more info under the heading of 'standard deviation distribution' – shabbychef Oct 11 '11 at 18:32

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