Factorial series

Prove that

$\sum_{n=0}^{\infty} \frac{(4n)!}{(4n+4)!} = \frac{\ln 2}{4} - \frac{\pi}{24}$

Solution

We have successively:

$\begin{align*} \sum_{n =0}^{\infty} \frac{(4n)!}{(4n+4)!} & =\sum_{n=0}^{\infty} \frac{\Gamma(4n+1)}{\Gamma(4n+5)} \\ &= \frac{1}{6} \sum_{n=0}^{\infty} \mathrm{B}(4n+1,4) \\ & = \frac{1}{6}\sum_{n=0}^{\infty} \int_{0}^{1}x^{4n}(1-x)^3\, \mathrm{d}x \\ & = \frac{1}{6}\int_{0}^{1} \frac{(1-x)^2}{(1+x^2)(1+x)} \, \mathrm{d}x \\ &= \frac{1}{6} \int_{0}^{1} \left ( \frac{2}{x+1} - \frac{x+1}{x^2+1} \right )\, \mathrm{d}x \\ &= \frac{1}{6} \int_{0}^{1} \left ( \frac{2}{x+1} - \frac{x}{x^2+1} - \frac{1}{x^2+1} \right )\, \mathrm{d}x \\ & = \frac{\ln 2}{4}-\frac{\pi}{24} \end{align*}$