3

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.

Larry
  • 5,090
Larry Freeman
  • 10,433
  • 4
    It should suffice to show that $\Gamma(x-1)/\Gamma(x/2)^2$ is increasing on $(2,\infty)$, which means showing the derivative is positive. Might look into that. – anon Apr 15 '13 at 02:15
  • 2
    If that turns out, please answer your own question to close the matter. – vonbrand Apr 15 '13 at 02:55
  • Thanks, @vonbrand. I will attempt a proof and post it as an answer. Cheers. – Larry Freeman Apr 15 '13 at 03:06

1 Answers1

6

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.

Larry Freeman
  • 10,433
  • 1
    Hey Larry -- I checked and your answer is correct, so you should accept it. Also, another way to see $\Psi(x-1) > \Psi(x/2)$ for $x\geq 1$ is because $\Psi$ itself is an increasing function. – Douglas B. Staple Apr 16 '13 at 20:44