Dear friends, for $\displaystyle \alpha \in\mathbb{R}$, it appears that $\displaystyle \displaystyle \sum_{k \ge 0}\frac{1}{k^2+k-\alpha} = \frac{\pi\tan\left(\frac{1}{2}\pi\sqrt{4\alpha+1}\ right)}{\sqrt{4\alpha+1}}$.

But I'm unable to show it! Can you show it? I tried to use the identity:

$\displaystyle \displaystyle \frac{1}{a^2+a-b} = \frac{2}{\sqrt{4b+1}\left(2a-\sqrt{4b+1}+1\right)}-\frac{2}{\sqrt{4b+1}\left(2a+\sqrt{4b+1}+1\right)}$,

and turn it into an integral, but I faced divergence issues and the whole thing evaluated to $\displaystyle 1$.