Prove the validity of gamma function equation $\Gamma(n)\Gamma(n+1/2) = 2^{1-2n}\sqrt{\pi}\;\Gamma(2n)$

Question

How to prove this identity for natural $n$? $$\Gamma(n)\Gamma(n+1/2) = 2^{1-2n}\sqrt{\pi}\;\Gamma(2n)$$

Firstly, I set $n=1$ and looked at general gamma equation. How to simplify or... ?

Answer 1

Assuming $n$ to be a natural, and using $\Gamma(x) = (x-1) \Gamma(x-1)$, and $\Gamma(1/2) = \sqrt{\pi}$,

$\Gamma(n+\frac{1}{2} )= \underbrace{\frac{2n -1}{2} \frac{2n-3}{2} \dots \frac{1}{2}}_{\text{$n$ terms}} \sqrt{\pi}$

and

$\Gamma(n) = \underbrace{\frac{2n-2}{2} \frac{2n-4}{2} \dots \frac{2}{2}}_{\text{$n-1$ terms}} $

It then follows that

$$\Gamma(n)\Gamma(n+\frac{1}{2}) = \frac{(2n - 1)!}{2^{2n -1}}\sqrt{\pi}$$

Lastly, $\Gamma(2n) = (2n-1)!$ finishes the argument.

Answer 2

Split up $~\Gamma(2n)~=~(2n-1)!~$ into two sub-products, the former consisting only of odd numbers, and the latter having only even factors; then, in both, factor $2$ outside each term, and divide then multiply the entire expression by $\Gamma\bigg(\dfrac12\bigg)=\sqrt\pi$.