Methods for evaluating $\sum_{n=1}^\infty \frac1{a+(n-1)n}$ I am interested in methods for evaluating the sum $$\sum_{n=1}^\infty \frac1{a+(n-1)n}.$$
Indeed I will give my own answer below using the Residue Theorem.
Please feel free to post other methods for the evaluation, such as Maclaurin series, methods from harmonic/fourier analysis, ...
Related question: Continuity of $f(x) = \sum_{n=0}^{\infty} \frac{1}{n(n+1)+x}$.

Let $s$ and $t$ to be the roots of $n^2-n+a=0$. So
$$\frac{1}{a+n(n-1)}=\frac{1}{(n-s)(n-t)}=\frac{1}{s-t}\left(\frac 1{n-s}-\frac 1{n-t}  \right)$$ Recalling that
$$\sum_{n-1}^p \frac 1{n-x}=\psi(p-x+1)-\psi (1-x)$$ we have that, if
$$S={\sqrt{1-4 a}}\sum_{n-1}^p \frac{1}{a+n(n-1)}$$
$$S=\psi \left(\frac{1}{2}+\frac{1}{2} \sqrt{1-4 a}\right)-\psi
   \left(\frac{1}{2}-\frac{1}{2} \sqrt{1-4 a}\right)+$$ $$\psi \left(p+\frac{1}{2}-\frac{1}{2} \sqrt{1-4 a}\right)-\psi \left(p+\frac{1}{2}+\frac{1}{2} \sqrt{1-4 a}\right)$$ Now, using the reflection formula for the digamma function
$$\psi \left(\frac{1}{2}+\frac{1}{2} \sqrt{1-4 a}\right)-\psi
   \left(\frac{1}{2}-\frac{1}{2} \sqrt{1-4 a}\right)=\pi  \tan \left(\frac{1}{2} \pi  \sqrt{1-4 a}\right)$$ Expanding the remaining terms as series for large values of $p$, then
$$\sum_{n-1}^p \frac{1}{a+n(n-1)}=\frac {\pi  \tan \left(\frac{1}{2} \pi  \sqrt{1-4 a}\right) } {\sqrt{1-4 a}}-\frac{1}{p}+\frac{a}{3 p^3}+O\left(\frac{1}{p^5}\right)$$
