Show $\frac{\Gamma((n-1)/2)}{\Gamma(n/2)} \approx \frac{\sqrt{2}}{\sqrt{n-2}}$


Using the facts:

  • $(1 + \alpha/m)^m = e^\alpha ( 1+ r_m)$, where $\lim_{m \to \infty} \sqrt{m}r_m = 0$
  • $\Gamma(n+1) = n^{n + 1/2} e^{-n} \sqrt{2 \pi} (1 + r_n)$, where $\lim_{n \to \infty} \sqrt{n}r_n = 0$

Note :

$$ \begin{aligned} \frac{\Gamma((n-1)/2)}{\Gamma(n/2)} &= \frac{\left(\frac{n-3}{2}\right)^{(n-2)/2} e^{-(n-3)/2}(\sqrt{2\pi}(1 + r_{1n})}{\left(\frac{n-2}{2}\right)^{(n-1)/2} e^{-(n-2)/2}(\sqrt{2\pi}(1 + r_{2n})} \\ &=\left( \left(\frac{n-3}{n-2}\right)^{(n-2)/2} \sqrt{e} \frac{1+r_{1n}}{1 + r_{2n}} \right) \times \frac{\sqrt{2}}{\sqrt{n-2}} \end{aligned} $$

But I'm stuck at how I should proceed to eliminate the $\left( \left(\frac{n-3}{n-2}\right)^{(n-2)/2} \sqrt{e} \frac{1+r_{1n}}{1 + r_{2n}} \right)$ term.

  • 2
    $\begingroup$ What is your definition of $\Gamma$? According to it, proving Gautschi's inequality might be more or less trivial. $\endgroup$ – Jack D'Aurizio Nov 21 '18 at 2:21
  • 2
    $\begingroup$ what is your definition of $\approx$?? $\endgroup$ – Masacroso Nov 21 '18 at 2:39

A possible approach:

$$ \frac{\Gamma\left(\tfrac{n-1}{2}\right)}{\Gamma\left(\tfrac{n}{2}\right)}=\tfrac{1}{\sqrt{\pi}}\,B\left(\tfrac{n-1}{2},\tfrac{1}{2}\right)=\frac{1}{\sqrt{\pi}}\int_{0}^{1}x^{\frac{n-3}{2}}(1-x)^{-1/2}\,dx=\frac{2}{\sqrt{\pi}}\int_{0}^{1}\frac{x^{n-2}}{\sqrt{1-x^2}}\,dx $$ gives $$ \frac{\Gamma\left(\tfrac{n-1}{2}\right)}{\Gamma\left(\tfrac{n}{2}\right)}= \frac{2}{\sqrt{\pi}}\int_{0}^{\pi/2}\left(\cos\theta\right)^{n-2}\,d\theta\sim\frac{2}{\sqrt{\pi}}\int_{0}^{+\infty}\exp\left[-(n-2)\frac{\theta^2}{2}\right]d\theta=\sqrt{\frac{2}{n-2}}. $$ Notice that the integral representation (as a moment) instantly gives that the LHS is a log-convex function. The asymptotic equivalence $\sim$ can be seen as an instance of Laplace/Hayman's method.


Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.