What would be the value of $\sum\limits_{n=0}^\infty \frac{1}{an^2+bn+c}$ I would like to evaluate the sum
$$\sum_{n=0}^\infty \frac{1}{an^2+bn+c}$$
Here is my attempt:

Letting  
$$f(z)=\frac{1}{az^2+bz+c}$$
The poles of $f(z)$ are located at 
$$z_0 = \frac{-b+\sqrt{b^2-4ac}}{2a}$$
and
$$z_1 = \frac{-b-\sqrt{b^2-4ac}}{2a}$$
Then
$$
b_0=\operatorname*{Res}_{z=z_0}\,\pi \cot (\pi z)f(z)=
\lim_{z \to z_0} \frac{(z-z_0)\pi\cot (\pi z)}{az^2+bz+c}=
\lim_{z \to z_0} \frac{\pi\cot (\pi z)+(z_0-z)\pi^2\csc^2 (\pi z)}{2az+b}
$$
Using L'Hopital's rule. Continuing, we have the limit is
$$
\lim_{z \to z_0} \frac{\pi\cot (\pi z)+(z_0-z)\pi^2\csc^2 (\pi z)}{2az+b}=
\frac{\pi\cot (\pi z_0)}{2az_0+b}
$$
For $z_0 \ne 0$
Similarly, we find
$$b_1=\operatorname*{Res}_{z=z_1}\,\pi \cot (\pi z)f(z)=\frac{\pi\cot (\pi z_1)}{2az_1+b}$$
Then 
$$\sum_{n=-\infty}^\infty \frac{1}{an^2+bn+c} = -(b_0+b_1)=\\
-\pi\left( \frac{\cot (\pi z_0)}{2az_0+b} + \frac{\cot (\pi z_1)}{2az_1+b}\right)= 
-\pi\left( \frac{\cot (\pi z_0)}{\sqrt{b^2-4ac}} + \frac{\cot (\pi z_1)}{-\sqrt{b^2-4ac}}\right)=
\frac{-\pi(\cot (\pi z_0)-\cot (\pi z_1))}{\sqrt{b^2-4ac}}=
\frac{\pi(\cot (\pi z_1)-\cot (\pi z_0))}{\sqrt{b^2-4ac}}
$$
Then we have
$$\sum_{n=0}^\infty \frac{1}{an^2+bn+c} = \frac{\pi(\cot (\pi z_1)-\cot (\pi z_0))}{2\sqrt{b^2-4ac}}$$

Is this correct?  I feel like I made a mistake somewhere.  Could someone correct me?  Is there an easier way to evaluate this sum?

Take $a=1,\ b=3, \ c=2$, then $z_0=-2, \ z_1=-1$, and so you have to compute $\cot(-\pi)$ and $\cot(-2\pi)$ which make no sense. However
$$
\sum_{n=0}^\infty\frac{1}{n^2+3n+2}=\sum_{n=0}^\infty(\frac{1}{n+1}-\frac{1}{n+2})
=\lim_{m\to \infty}(1-\frac{1}{m+2})=1.
$$
