in a probability computation problem on research, I am facing a conditional probability, which can be modeled with the following simplified formulation:
$\sum_{x=0}^{\infty} \frac{1}{(x-a)(x-b)}$, $a>0$, $b>0$
I have investigated this problem for a while and no conclusive result found. I would appreciate if you would please let me know a solution clue.
Assume $a, b \not\in \mathbb{N}$ and $a \ne b$, then $$\begin{align} \sum_{x=0}^\infty \frac{1}{(x-a)(x-b)} &= \frac{1}{a-b}\sum_{x=0}^\infty\left(\frac{1}{x-a} - \frac{1}{x-b}\right)\\ &= \frac{1}{b-a}\sum_{x=0}^\infty\left[ \left(\frac{1}{x+1} - \frac{1}{x-a}\right) - \left(\frac{1}{x+1} - \frac{1}{x-b}\right) \right]\\ &= \frac{1}{b-a}\left[(\gamma+\psi(-a)) - (\gamma+\psi(-b))\right]\\ &= \frac{\psi(-a)-\psi(-b)}{b-a} \end{align} $$ where $\psi(z)$ is the digamma function.