Asymptotic approximation of the binomial coefficient Asymptotic approximation of the binomial coefficient
\begin{equation}
\binom{-1/2}{m}
\end{equation}
What is it?
Thanks in advance.

$${-1/2 \choose n}=\frac{1}{n!}\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)...\left(-\frac{1}{2}-n+1\right)=\frac{(-1)^n}{n!}\frac{1\cdot3\cdot ... \cdot (2n-1)}{2^n}=\frac{(-1)^n (2n)!}{n!^2 \cdot 4^n}=\frac{(-1)^n}{4^n} {2n \choose n}$$
Now use either Stirling's formula or Wallis' integrals to prove that
$${-1/2 \choose n} \sim \frac{(-1)^n}{\sqrt{n \pi}}$$
