How to compare two ratios of gamma functions?

Question

How would I show that for $x \ge 2$:

$$\frac{\Gamma(x-1)}{[\Gamma(\frac{x}{2})]^2} \le \frac{\Gamma(x)}{[\Gamma(\frac{x+1}{2})]^2}$$

Any hints or suggestions are greatly appreciated.

Answer 1

The answer is to show that $\frac{d}{dx}(\frac{\Gamma(x-1)}{\Gamma([\frac{x}{2}]^2)}) > 0$

This will be established if I show that $\frac{d}{dx}[\ln{\Gamma(x-1)}- 2\ln\Gamma(\frac{x}{2})] > 0$.

I will use this series ψ:

$$\frac{d}{dx}(\ln\Gamma(x)) = \frac{\psi(x)}{dx} = -\gamma + \sum_{k=0}^\infty(\frac{1}{k+1} - \frac{1}{k + x})$$

Applying this gives:

$$\frac{d}{dx}[\ln{\Gamma(x-1)}- 2\ln\Gamma(\frac{x}{2})] = \psi(x-1) - \frac{2\psi(\frac{x}{2})}{2} = \psi(x-1) - \psi(\frac{x}{2})$$

$$\psi(x-1) - \psi(\frac{x}{2}) = \sum_{k=0}^{\infty}(\frac{1}{k+\frac{x}{2}} - \frac{1}{k+x-1}) $$

We can see that for $x \ge 2$, $\frac{x}{2} < x-1$ so it follows that $\frac{1}{k + \frac{x}{2}} > \frac{1}{k+x-1}$ and it is therefore increasing.