Proof Of A Gamma Function - Double Factorial Identity How does one prove the following identity?
$$\sqrt{(-1)^n\frac{\Gamma(n+1/2)}{\Gamma(1/2-n)}} = \frac{(2n-1)!!}{2^n}$$
I attempted to prove this using the definition of the double factorial, however I couldn't continue and feel like there is a better method. I would appreciate it if somebody could show me a proof. Thank you
 A: We use the identity
\begin{align*}
\Gamma(1-z)\Gamma(z)=\frac{\pi}{\sin (\pi z)}\qquad\qquad z\notin \mathbb{Z}
\end{align*}
evaluated at $z=n-\frac{1}{2}$ and obtain
\begin{align*}
\Gamma\left(1-\left(n+1/2\right)\right)\Gamma\left(n+1/2\right)&=\frac{\pi}{\sin\left(\pi\,\frac{2n+1}{2}\right)}\\
\color{blue}{\Gamma(1/2-n)\Gamma(n+1/2)}&\color{blue}{=(-1)^n\,\pi}\tag{1}
\end{align*}
We obtain with (1) and application of the identities
\begin{align*}
\Gamma(z+1)&=z\Gamma(z)\qquad\qquad\qquad z\in\mathbb{C}\setminus\{0,-1,-2,\ldots\}\tag{2}\\
\Gamma(1/2)&=\sqrt{\pi}\tag{3}
\end{align*}
\begin{align*}
\color{blue}{\sqrt{(-1)\frac{\Gamma(n+1/2)}{\Gamma(1/2-n)}}}
&=\frac{1}{\sqrt{\pi}}\Gamma(n+1/2)\tag{$\to$ (1)}\\
&=\frac{1}{\sqrt{\pi}}\left(n-\frac{1}{2}\right)\Gamma(n-1/2)\tag{$\to$ (2)}\\
&=\frac{1}{\sqrt{\pi}}\left(n-\frac{1}{2}\right)\left(n-\frac{3}{2}\right)\Gamma(n-3/2)\tag{$\to$ (2)}\\
&=\cdots\\
&=\frac{1}{\sqrt{\pi}}\left(n-\frac{1}{2}\right)\left(n-\frac{3}{2}\right)\cdots\left(\frac{1}{2}\right)\Gamma(1/2)\tag{$\to$ (2)}\\
&=\left(n-\frac{1}{2}\right)\left(n-\frac{3}{2}\right)\cdots\left(\frac{1}{2}\right)\tag{$\to$ (3)}\\
&=\frac{1}{2^n}(2n-1)(2n-3)\cdots 1\\
&\,\,\color{blue}{=\frac{1}{2^n}(2n-1)!!}
\end{align*}
and the claim follows.
