Fairly Impossible#3(last one ,phew!)

You would be grossly disappointed about this explanation if you hadn't seen the discussion "on the iterated properties of Beta Function".Since, $\frac{\Gamma(x+y)\Gamma(y+1)}{\Gamma(x)\Gamma(y)}\sum_{n=1}^{\infty}\frac{\Gamma(x+n-1)}{\Gamma(x+y+n)}(\psi(x+n-1)-\psi(x+y+n))= \psi(x)-\psi(x+y)$ .Then simply letting $x=1/2,y=1/2$ we have,

$\frac{\Gamma(1)\Gamma(3/2)}{\Gamma(1/2)\Gamma(1/2)}\sum_{n=1}^{\infty}\frac{\Gamma(n-1/2)}{\Gamma(n+1)}(\psi(n-1/2)-\psi(1+n))=\psi(1/2)-\psi(1)$ Since $\psi(1/2)=-2\ln 2-\gamma,\Gamma(1/2)=\sqrt\pi and\psi(1)=-\gamma$ .Then multiplying both sides by $\frac{\pi}{\Gamma(3/2)}$ and simplifying we have

$\sum_{n=1}^{\infty}\frac{\Gamma(n-1/2)}{\Gamma(n+1)}(\psi(n-1/2)-\psi(1+n))=-4\sqrt\pi\ln 2$ .Hence a=4 and b=2 ,a+b=6 .Sorry for complexity.