What's gamma doing inside?

Note that the series can be written as

$\displaystyle \sum_{k=1}^{\infty} \dfrac{(-1)^k}{k \ \Gamma(k+1) \Gamma \left( \frac{1}{2} - k\right)} \qquad (\because \Gamma(x+1) = x!)$

By the infinite binomial expansion of $(1-x)^{n}$ , we have,

$\displaystyle (1-x)^{n} = 1 + \sum_{k=1}^{\infty} (-1)^k \prod_{r=1}^{k} \left(\dfrac{n-r+1}{r}\right) x^k$

Now,

$\displaystyle \prod_{r=1}^{k} \left(\dfrac{n-r+1}{r}\right) = \dfrac{1 \cdot 2 \cdot 3\cdot \ldots \cdot (n-k+1)}{k!}$

$\displaystyle = \dfrac{\Gamma (n-k+1)}{\Gamma (n-k+1)} \times \dfrac{1 \cdot 2 \cdot 3\cdot \ldots \cdot (n-k+1)}{\Gamma (k+1)}$

$\displaystyle = \dfrac{\Gamma(n+1)}{\Gamma(n-k+1) \Gamma(k+1)} \qquad (\because x \Gamma (x) = \Gamma(x+1))$

Thus,

$\displaystyle (1-x)^{n} = 1 + \sum_{k=1}^{\infty} (-1)^k \dfrac{\Gamma(n+1)}{\Gamma(n-k+1) \Gamma(k+1)} x^k$

Putting $\displaystyle n = -\dfrac{1}{2}$ and dividing by $x$ , we have,

$\displaystyle \sum_{k=1}^{\infty} (-1)^k \dfrac{\Gamma \left(\frac{1}{2}\right)}{\Gamma \left(\frac{1}{2}-k \right) \Gamma \left(k+1 \right)} x^{k-1} = \dfrac{1}{x\sqrt{1-x}} - \dfrac{1}{x}$

$\displaystyle \implies \sum_{k=1}^{\infty} (-1)^k \dfrac{\Gamma \left(\frac{1}{2} \right)}{k \ \Gamma \left(\frac{1}{2}-k \right) \Gamma(k+1)} = \int_{0}^{1} \left(\dfrac{1}{x\sqrt{1-x}} - \dfrac{1}{x} \right) \mathrm{d}x$

$\displaystyle \implies \sum_{k=1}^{\infty} \dfrac{ (-1)^k }{k \ \Gamma \left(\frac{1}{2}-k \right) \Gamma(k+1)} = \dfrac{1}{\Gamma\left(\frac{1}{2}\right)} \left[ -2 \ln (\sqrt{1-x} + 1) \right]_{0}^{1}$

$\displaystyle \implies \sum_{k=1}^{\infty} \dfrac{ (-1)^k }{k \ \Gamma \left(\frac{1}{2}-k \right) \Gamma(k+1)} = \dfrac{2\ln 2}{\sqrt{\pi}} \qquad \left(\because \Gamma\left(\frac{1}{2}\right) = \sqrt{\pi} \right)$

$\implies A+B+C+D = \boxed{7}$