gammas division Can anyone explain me how to calculate this expression?
$$\frac{\Gamma(n/2)}{\Gamma((n-1)/2)}$$
Thank you, I tried with doble factorial, but don´t know really how to continue.
 A: Almost as @J.G. answered, for large values of $n$, using Stirlin approximation twice and continuing with Taylor series
$$\frac{\Gamma \left(\frac{n}{2}\right)}{\Gamma \left(\frac{n-1}{2}\right)}=\sqrt{\frac n 2}\left(1-\frac{3}{4 n}-\frac{7}{32 n^2}-\frac{9}{128 n^3}+\frac{59}{2048
   n^4}+O\left(\frac{1}{n^5}\right)\right)$$ which shows a relative error lower than $0.1$% as soon as $n>2$, lower than $0.01$% as soon as $n>3$, lower than $0.001$% as soon as $n>5$.
If you wish a very good approximation
$$\frac{\Gamma \left(\frac{n}{2}\right)}{\Gamma \left(\frac{n-1}{2}\right)}\sim\sqrt{\frac n 2}\frac{1-\frac{441823}{287784 n}+\frac{119909}{54816 n^2}-\frac{1473029}{877056
   n^3} } {1-\frac{225985}{287784 n}+\frac{697315}{383712 n^2}-\frac{7699031}{18418176
   n^3} }$$
A: It's already in the most helpful form. But its large-$n$ asymptotic behaviour is as $\sqrt{n/2}$, by linearizing $\ln\Gamma(z)$ viz.$$\begin{align}\ln\Gamma(n/2)-\ln\Gamma((n-1)/2)&\approx\frac{\ln\Gamma((n+1)/2)-\ln\Gamma((n-1)/2)}{2}\\&=\frac{\ln((n-1)/2)}{2}\\&=\ln\sqrt{(n-1)/2}.\end{align}$$
A: If your question is connected to your previous one, you can leave the result as shown.
