Psi and Riemann together?

Note that, $\displaystyle \psi^{(1)}(k) = - \int_{0}^{1} \dfrac{x^{k-1} \ln x }{(1-x)} \mathrm{d}x$

$\displaystyle \implies \sum_{k=1}^{\infty} \dfrac{\psi^{(1)} (k)}{k} = -\sum_{k=1}^{\infty} \int_{0}^{1} \dfrac{x^{k-1} \ln x}{k \ (1-x)} \mathrm{d}x$

$\displaystyle = \int_{0}^{1} \dfrac{\ln x \ln (1-x)}{x(1-x)} \mathrm{d}x$

$\displaystyle = \lim_{a \to 0^{+}} \lim_{b \to 0^{+}} \left(\dfrac{\partial}{\partial a} \left(\dfrac{\partial}{\partial b} \operatorname{B} (a,b)\right) \right)$

$= 2 \zeta (3)$

$\implies A+B = \boxed{5}$

Simply do it like this , Applying Abel's summation , Let a {n} = \frac{1}{n} , b {n} = \ psi (n)
We get , \sum {n=1}^inf \frac{(H {n})}{n^2} = 2 \times \zeta(3)