Calculation of $\Gamma(n+\frac12)$

Question

I want to show that $$\Gamma\left(n+\dfrac12\right)=\dfrac{1\cdot 3\cdot 5\cdots(2n-1)}{2^n}\sqrt{\pi}=2^{1-2n}\dfrac{\Gamma(2n)}{\Gamma(n)}\sqrt{\pi}.$$ I know that $\Gamma(\frac12)=\sqrt{\pi}$. The definitions of $\Gamma$ are such as $$\Gamma(z)=\lim_{n\rightarrow\infty}\dfrac{n^zn!}{z(z+1)\ldots(z+n)}$$ and

$$\Gamma(z) = \lim_{n\to \infty} e^{-\gamma z}\frac1z\prod_{k=1}^n \left(1+\frac{z}{k}\right)^{-1}e^{z/k}$$

How can those help?

Answer 1

The properties $$\Gamma(1/2)=\sqrt{\pi} $$ and $$\Gamma(x+1)=x \Gamma(x) $$ are sufficient.

EDIT: In order to prove the second one, observe that $$x \Gamma(x)= \lim_{n \to \infty} \frac{n^x n!}{(x+1) \dots(x+n)}=\lim_{n \to \infty} \frac{n^{x+1} (n-1)!}{(x+1) \dots (x+n)}= \\\lim_{n \to \infty}\frac{(n-1)^{x+1} (n-1)!}{(x+1) \dots (x+n)} \frac{n^{x+1}}{(n-1)^{x+1} }, $$ and use limits arithmetics.

Answer 2

I will prove it using mathematical induction for $n\in \mathbb{N}$

for the case $n=1$ we have

$$\Gamma\left(\dfrac32\right)=\frac{1}{2}\Gamma \left( \frac{1}{2}\right) =\frac{\sqrt{\pi}}{2}$$

Now assume that

$$\tag{1}\Gamma\left(n+\dfrac12\right)=\dfrac{1\cdot 3\cdot 5\cdots(2n-1)}{2^n}\sqrt{\pi}$$

For the case

$$\Gamma\left(n+\dfrac32\right)=\dfrac{1\cdot 3\cdot 5\cdots(2n+1)}{2^{n+1}}\sqrt{\pi}$$

We start by the LHS

$$\Gamma\left(n+\dfrac32\right) = \left( n+\frac{1}{2}\right)\Gamma\left(n+\frac{1}{2} \right)$$

Now use (1)

$$\Gamma\left(n+\dfrac32\right)=\left( n+\frac{1}{2}\right)\dfrac{1\cdot 3\cdot 5\cdots(2n-1)}{2^{n}}\sqrt{\pi}=\dfrac{1\cdot 3\cdot 5\cdots(2n+1)}{2^{n+1}}\sqrt{\pi}$$

For the second equality use that

$$\Gamma(n) = (n-1)! \,\,\, \to \,\,\, \Gamma(2n)= (2n-1)!$$

$$\frac{(2n-1)! }{(n-1)!}= \frac{(2n-1)\cdot (2n-2) \cdots 3\cdot 2\cdot 1}{(n-1)\cdot (n-2) \cdots 3\cdot 2\cdot 1}=2^{n-1}\left( 1\cdot 3\cdot 5\cdots(2n-1)\right)$$

Which follows from the first proof.