Another infinite series involving Gamma function It's not hard to see that Mathematica expresses this series in terms of hypergeometric function,
but how about finding a way of expressing the result in terms of elementary functions only?
Is that possible? How would you recommend me to proceed?
$$\sum_{n=1}^{\infty} (-1)^{n+1} \frac{n\Gamma(2n-1)}{2^{2n}\Gamma\left(2n+\frac{1}{2}\right)}$$
 A: $$\begin{eqnarray*}S=\sum_{n=1}^{+\infty}(-1)^{n+1}\frac{n\,\Gamma(2n-1)}{4^n\,\Gamma(2n+1/2)}&=&\sum_{n=1}^{+\infty}\frac{(-1)^{n+1}n}{4^n\Gamma(3/2)}B(2n-1,3/2)\\&=&\frac{2}{\sqrt{\pi}}\sum_{n=1}^{+\infty}\frac{(-1)^{n+1}n}{4^n}\int_{0}^{1}x^{2n-2}(1-x)^{1/2}\,dx\\&=&\frac{1}{2\sqrt{\pi}}\int_{0}^{1}\sum_{n=0}^{+\infty}\frac{(-1)^n(n+1)}{4^n}x^{2n}(1-x)^{1/2}\,dx\\&=&\frac{8}{\sqrt{\pi}}\int_{0}^{1}\frac{(1-x)^{1/2}}{(4+x^2)^2}\,dx\\&=&\frac{8}{\sqrt{\pi}}\int_{0}^{1}\frac{x^{1/2}}{(5-2x+x^2)^2}\,dx\\&=&\frac{16}{\sqrt{\pi}}\int_{0}^{1}\frac{x^2\,dx}{(5-2x^2+x^4)^2}.\end{eqnarray*}$$
Now the last integral can be evaluated with standard techniques.
We get the terrifying identity:
$$ S =\frac{\phantom{}_3 F_2\left(\frac{1}{2},1,2;\frac{5}{4},\frac{7}{4};-\frac{1}{4}\right)}{3\sqrt{\pi}}=\\=\scriptstyle{\frac{1}{4\sqrt{5\pi }\, }\sqrt{2+ \sqrt{5}} \left(\pi -\arctan\left(2 \sqrt{2+\sqrt{5}}\right)+2\sqrt{5}\log 2-2 \sqrt{5} \log\left(1+\sqrt{5}+\sqrt{2 \left(1+\sqrt{5}\right)}\right)+2 \log\left(2+\sqrt{5}+2 \sqrt{2+\sqrt{5}}\right)\right)}\\=0.16981854\ldots.$$
