Product under Summation (Part 3)

Let $\displaystyle f(a) = \dfrac{1}{1+a} + \sum_{r=1}^{\infty} \left[ \dfrac{1}{(2r+a+1)}\prod_{k=1}^{r}\left(\dfrac{2k-1}{2k}\right) \right]$

$\displaystyle = \int_{0}^{1} x^{a} \mathrm{d}x + \sum_{r=1}^{\infty} \left[ \prod_{k=1}^{r}\left(\dfrac{2k-1}{2k}\right) \int_{0}^{1} x^{2r+a} \mathrm{d}x \right]$

$= \displaystyle \int_{0}^{1} x^{a} \sum_{r=1}^{\infty} \left[ x^{2r} \prod_{k=1}^{r}\left(\dfrac{2k-1}{2k}\right) \mathrm{d}x \right] + \int_{0}^{1} x^{a} \mathrm{d}x$

Using infinite binomial series, we have,

$f(a) = \displaystyle \int_{0}^{1} x^a \left(\dfrac{1}{\sqrt{1-x^2}} - 1 \right) \mathrm{d}x + \int_{0}^{1} x^a \mathrm{d}x$

$= \displaystyle \int_{0}^{1} \dfrac{x^a}{\sqrt{1-x^2}} \mathrm{d}x$

Let $x = \sin \theta$

$\displaystyle \implies f(a) = \int_{0}^{\frac{\pi}{2}} \sin^a \theta \ \mathrm{d}\theta$

$\displaystyle = \dfrac{1}{2} \operatorname{B} \left(\dfrac{p+1}{2}, \dfrac{1}{2}\right)$

$= \displaystyle \dfrac{1}{2} \dfrac{\Gamma \left(\dfrac{a+1}{2}\right) \Gamma \left(\dfrac{1}{2}\right)}{\Gamma\left(\dfrac{a+2}{2}\right)}$

Using Gamma duplication formula, we have,

$f(a) = \displaystyle \dfrac{\pi}{2^{a+1}} \dfrac{\Gamma (a+1)}{\Gamma^2\left(\dfrac{a+2}{2}\right)} \quad ; \quad a > -1$

Required series is $-f'(0)$ . Evaluating it, we have,

$\text{S} = \boxed{\dfrac{\pi \ln 2}{2}}$