How can I calculate a limit involving the Gamma function? I need to find the following limit:
$$
\lim_{n \to \infty} \frac{\Gamma\left(\frac n2\right)}{\Gamma\left(\frac{n+1}2\right)}.
$$
Wolfram Alpha tells me it is 0. Also Wikipedia told me that for $x$ large enough and $\alpha >0$ I have
$$
\Gamma(x + \alpha) \sim \Gamma(x)x^\alpha,
$$
which would yield the answer when I apply this to the denominator for $\alpha = \frac12$. Unfortunately I was not able to proof the identity from Wikipedia (I did not even know how to start...).
I would be glad if someone could suggest a different way to proceed with limit from above or how one can derive the identity from Wikipedia.
 A: Disclaimer - this is not a rigorous proof.
Rewrite the limand as a beta integral:
$$\frac{\Gamma\left(\frac n2\right)}{\Gamma\left(\frac{n+1}2\right)} = \frac1{\sqrt\pi} \operatorname{B}\left(\frac n2, \frac12\right) = \frac1{\sqrt\pi} \int_0^1 t^{\frac n2-1} (1-t)^{-\frac12} = \frac1{\sqrt\pi} \int_0^1 \frac{\sqrt{t^n}}{\sqrt t \sqrt{1-t}} \, dt$$
Substitute:
$$u=\sqrt t \implies t = u^2 \implies dt = 2u\,du$$
$$\frac2{\sqrt\pi} \int_0^1 \frac{u^n}{\sqrt{1-u^2}}\,du$$
Substitute again:
$$u = \sin(v) \implies du = \cos(v) \, dv$$
$$\frac2{\sqrt\pi} \int_0^{\frac\pi2} \sin^n(v) \, dv$$
Recall that $0\le\sin(v)\le1$ for $v\in\left[0,\frac\pi2\right]$. For arbitrarily large $n$, the value of $\sin^n(v)$ will approach $0$ for every point in this interval except $v=\frac\pi2$, so the integral and hence the original limit will also converge to $0$.
A: As @Christophe Leuridan commented, Stirling approximation is very useful
$$f_n=\frac{\Gamma\left(\frac n2\right)}{\Gamma\left(\frac{n+1}2\right)}$$ Take logarithms and use
$$\log(\Gamma(p))=p (\log (p)-1)+\frac{1}{2} (\log (2 \pi )-\log (p))+\frac{1}{12
   p}+O\left(\frac{1}{p^3}\right)$$ Use it twice and continue with Taylor series or long division to obtain
$$\log(f_n)=\frac{1}{2} (\log (2)-\log (n))+\frac{1}{4 n}+O\left(\frac{1}{n^3}\right)$$ that is to say
$$f_n =\sqrt{\frac{2}{n}} \left(1+\frac{1}{4 n}+O\left(\frac{1}{n^2}\right) \right)$$
