Is this correct $\Gamma(n+\frac{1}{2}) = \frac{\sqrt{\pi} }{2^n}\cdot\prod_{k=1}^{n} (2k - 1)$?

Question

This WolframAlpha Gamma Function displays

$$ \frac{\sqrt{\pi}}{2},\frac{3\sqrt{\pi}}{4},\frac{15\sqrt{\pi}}{8},\frac{105\sqrt{\pi}}{16}, \frac{945\sqrt{\pi}}{32}, \frac{10395\sqrt{\pi}}{64} $$

Notice denominators are $2^n$ and numerators are $1,3,15,105,954,10395$. Searching OEIS found A001147.

So, the general pattern is

$$\Gamma\left(n+\frac{1}{2}\right) = \frac{\sqrt{\pi} }{2^n}\cdot\prod_{k=1}^{n} (2k - 1)$$

I'm very rusty with integration, so I wasn't sure how to verify that pattern.

Answer 1

Use the property

$$\Gamma(z+1)=z\Gamma(z)~~~\text{and}~~~\Gamma(1/2)=\sqrt\pi$$

Let $z=n-\frac{1}2$, so we have

$$\Gamma(n+\frac{1}2)=(n-\frac{1}2)\Gamma(n-\frac{1}2)$$

Similarly,

$$\Gamma(n-\frac{1}2)=(n-\frac{3}2)\Gamma(n-\frac{3}2)$$

So we get a chain,

$$\Gamma(n+\frac{1}2)=(n-\frac{1}2)\cdot(n-\frac{3}2)\dots\frac{3}2\cdot \frac{1}2\cdot\Gamma(\frac{1}2)=\frac{(2n-1)\cdot (2n-3)\dots 2\cdot 1}{2^n}\sqrt\pi$$

Therefore,

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