Beauty and beast

[admin]

Find \[\displaystyle\frac{13}{8}+\mathop{\sum}\limits_{n=0}^{\infty}\frac{(-1)^{n+1}(2n+1)!}{(n+2)!\,n!\,4^{2n+3}}\,.\]

[Demetres]

We have \[\sum_{n=0}^{\infty} \frac{(-1)^{n+1}(2n+1)!}{(n+2)!n!4^{2n+3}} = \sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{2(n+2)4^{2n+3}} \binom{2n+2}{n+1} = \frac{1}{8}\sum_{n=0}^{\infty} \frac{1}{n+2} \binom{2n+2}{n+1}\left(-\frac{1}{16} \right)^{n+1}.\]

It is well-known that the generating function of the Catalan numbers is \( \frac{1}{2x}\left(1 - \sqrt{1-4x} \right) \), i.e. \[\frac{1}{2x}\left(1 - \sqrt{1-4x} \right) = \sum_{n=0}^{\infty} \frac{1}{n+1} \binom{2n}{n} x^n\]

where the sum converges for every \(x\) with \(|x| \leqslant 1/4.\) So

\[\frac{13}{8} + \sum_{n=0}^{\infty} \frac{(-1)^{n+1}(2n+1)!}{(n+2)!n!4^{2n+3}} = \frac{13}{8} + \frac{1}{8}\left( -8\left(1 - \sqrt{5/4} \right) -1\right) = \frac{1 + \sqrt{5}}{2}.\]