Explicit closed form $\sum \frac{1}{an^2+bn+c}$ I know that $\sum_{n\geq 1} \frac{1}{n^2}=\pi^2/6$, is there a simple way to get an explicit closed form $\sum \frac{1}{an^2+bn+c}$, where $a,b,c$ are integers, $a\neq 0$?
 A: For all $z\in\mathbb{C}$,
$$
\sum_{n=1}^\infty\left(\frac1n-\frac1{n+z}\right)=\psi(z+1)+\gamma\tag{1}
$$
where $\psi(z)=\frac{\mathrm{d}}{\mathrm{d}z}\log(\Gamma(z))$ is the digamma function.
Partial fractions says
$$
\begin{align}
\frac1{an^2+bn+c}
&=\frac1{\sqrt{b^2-4ac}}\left(\frac1{n+\frac{b-\sqrt{b^2-4ac}}{2a}}-\frac1{n+\frac{b+\sqrt{b^2-4ac}}{2a}}\right)\\
&=\frac1{\sqrt{b^2-4ac}}\left(\frac1n-\frac1{n+\frac{b+\sqrt{b^2-4ac}}{2a}}\right)\\
&-\frac1{\sqrt{b^2-4ac}}\left(\frac1n-\frac1{n+\frac{b-\sqrt{b^2-4ac}}{2a}}\right)\tag{2}
\end{align}
$$
If $b^2\ne4ac$, applying $(1)$ to $(2)$ yields
$$
\begin{align}
\sum_{n=1}^\infty\frac1{an^2+bn+c}
&=\frac1{\sqrt{b^2-4ac}}\psi\left(\frac{b+\sqrt{b^2-4ac}}{2a}+1\right)\\
&-\frac1{\sqrt{b^2-4ac}}\psi\left(\frac{b-\sqrt{b^2-4ac}}{2a}+1\right)\tag{3}
\end{align}
$$

Taking the derivative of $(1)$ gives
$$
\sum_{n=1}^\infty\frac1{(n+z)^2}=\psi^{\,\prime}(z+1)\tag{4}
$$
In the case of $b^2=4ac$, we have
$$
\frac1{an^2+bn+c}=\frac1a\frac1{\left(n+\frac{b}{2a}\right)^2}\tag{5}
$$
Applying $(4)$ to $(5)$ yields
$$
\sum_{n=1}^\infty\frac1{an^2+bn+c}
=\frac1a\psi^{\,\prime}\left(\frac{b}{2a}+1\right)\tag{6}
$$
A: Maple give an  answer 
$$
\sum_{n\geq 1}\frac{1}{an^2+bn+c}=\\
-\Psi \left( 1-1/2\,{\frac {-b+\sqrt {{b}^{2}-4\,ac}}{a}} \right) {
\frac {1}{\sqrt {{b}^{2}-4\,ac}}}+\Psi \left( 1-1/2\,{\frac {-b-\sqrt 
{{b}^{2}-4\,ac}}{a}} \right) {\frac {1}{\sqrt {{b}^{2}-4\,ac}}}
$$
A: Assuming that (b^2 - 4 a c > 0), there is a formula which is  
(H[(b + d)/(2 a)] - H[(b - d)/(2 a)] ) / d 
where d = Sqrt[b^2 - 4 a c] and H is the harmonic number.  
In the case where d=0, this reduces to PolyGamma[1, 1 + b/(2 a)] / a, as shown also by Leox. 
