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
Best Answer
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.