How to show $\Gamma(2s)/\Gamma(s+1/2)$ is log convex?

Question

I read in a book (Mathematics by Experiment, 2nd Edition, Example 5.13), that $$f=s \to \log(\Gamma(2s)/\Gamma(s+1/2))$$ is convex function. But the book does not explain it and I cannot see why this is true.

Answer 1

As an alternative to Lord Shark's perfectly fine answer, you may consider that by Legendre's duplication formula $$ \frac{\Gamma(2s)}{\Gamma\left(s+\frac{1}{2}\right)} = \frac{4^s}{2\sqrt{\pi}}\,\Gamma(s) \tag{1} $$ where $\frac{4^s}{2\sqrt{\pi}}$ is blatantly log-convex and $\Gamma(s)$ is log-convex by Bohr-Mollerup's theorem, or as a consequence of Holder's inequality applied to $\Gamma(s)=\int_{0}^{+\infty} x^s\frac{dx}{x e^x}$. I am assuming $s>0$, of course.

Answer 2

You need to show its second derivative which is $$4\psi'(2s)-\psi'(s+1/2)$$ is positive. Here $\psi$ is the digamma function, and $$\psi'(s)=\sum_{n=0}^\infty\frac{1}{(s+n)^2}.$$ Therefore $$4\psi'(2s)-\psi'(s+1/2) =\sum_{n=0}^\infty\frac{1}{(s+n/2)^2} -\sum_{m=0}^\infty\frac{1}{(s+(2m+1)/2)^2}.$$ The second sum is the odd $n$ terms of the first sum...