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?

Here is my answer in terms of the digamma function (cf. Abramowitz and Stegun).  Using the decomposed version of the sum given by Eric Naslund♦, we find:
$$\sum_{n=0}^\infty \frac{1}{an^2+bn+c}=
\frac{1}{\sqrt{b^{2}-4ac}}\sum_{n=0}^{\infty}\left(\frac{1}{n-z_{0}}-\frac{1}{n-z_{1}}\right)=
\frac{\psi (-z_0)-\psi(-z_1)}{\sqrt{b^{2}-4ac}}
$$
I note that this answer is in fact quite similar to the sum's closed form from $-\infty$ to $\infty$.

Here is an alternate solution:
$$\sum_{n=0}^\infty \frac{1}{an^2+bn+c}=
\frac{1}{a}\sum_{n=0}^\infty \frac{1}{(n-z_0)(n-z_1)}
$$
If $c_n$ is the $n^{th}$ term of the second version, then
$$\frac{c_{n+1}}{c_n}=\frac{(n-z_0)(n-z_1)}{(n-z_0+1)(n-z_1+1)}$$
which shows that this sum can be written as a hypergeometric function
$$
\frac{1}{a}\sum_{n=0}^\infty \frac{1}{(n-z_0)(n-z_1)}=
\frac{1}{az_0z_1}\sum_{n=0}^\infty \frac{\Gamma(n-z_0)\Gamma(n-z_1)n!}{\Gamma(n-z_0+1)\Gamma(n-z_1+1)n!}=
\frac{1}{z_0z_1a}{_3}F_2(-z_0, -z_1, 1;1-z_0, 1-z_1;1)
$$
