<s>[INST] 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?
 [/INST] 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
$$
\frac{1}{a}\sum_{n=0}^\infty \frac{1}{(n-z_0)(n-z_1)}=
\frac{1}{az_0z_1}\sum_{n=0}^\infty \frac{\Gamma(n-z_0)\Gamma(n-z_1)n!}{\Gamma(n-z_0+1)\Gamma(n-z_1+1)n!}=
\frac{1}{z_0z_1a}{_3}F_2(-z_0, -z_1, 1;1-z_0, 1-z_1;1)
$$
</s>