Evaluating $\sum_{r=1}^{\infty} \cot^{-1}(ar^2+br+c)$ 
Evaluate the series $$S=\sum_{r=1}^{\infty} \cot^{-1}(ar^2+br+c)$$

I have tried many values of $(a,b,c)$ and plugged into Wolframalpha, it always converges. I know that for particular values of $a,b,c$, we solve it by forming a telescoping series by using the fact that $\displaystyle \arctan x-\arctan y=\arctan\left(\dfrac{x-y}{1+xy}\right)$ and converting it into a form $f(r+1)-f(r)$.
But I think that we cannot convert all types into this form. Even if this was possible, what is the way for us to know what $f$ to use? In particuar, I was evaluating $\displaystyle \sum_{r=1}^{\infty} \cot^{-1}\left(3r^2-r-\frac13\right)$, but couldn't convert it into telescoping series. So, how then, do we solve this? and for what values of $(a,b,c)$ is the sum convergent?
 A: Given quadratic $ar^2 +br+c$, suppose that we can find some function $f(r)$ such that
$$
(\dagger)\quad\quad\frac{1+f(r+1)f(r)}{f(r+1)-f(r)} = ar^2 + br + c. 
$$
Then as $$\cot^{-1}(ar^2+br+c)=\arctan\frac{1}{ar^2+br+c}\\
=\arctan \frac{f(r+1)-f(r)}{1+f(r+1)f(r)}=\arctan(f(r+1))-\arctan(f(r)),
$$
we can proceed to find $\displaystyle \sum_{r=1}^\infty \cot^{-1}(ar^2 +br+c)$ telescopically as mentioned. Indeed we would have $$\sum_{r=1}^\infty \cot^{-1}(ar^2 +br+c)=-\arctan(f(1))+\lim_{r\to\infty}\arctan(f(r)),$$
provide convergence.
Now to find $f(r)$, we propose to seek it in the form $$f(r)=\frac{Ar+B}{Cr+D},$$ a linear fraction. If so, then the LHS of the $(\dagger)$ equation will be a quadratic in $r$. Indeed, substituting this linear fractional into the LHS yields
$$\frac{A^2+C^2}{AD-BC}r^2 + \frac{A^2+C^2+2AB+2CD}{AD-BC} r + \frac{B^2+D^2+AB+CD}{AD-BC}.$$
So the task becomes to find $A,B,C,D$ such that
$$ \begin{eqnarray}a &=& \frac{A^2+C^2}{AD-BC} \\
b &=& \frac{A^2+C^2+2AB+2CD}{AD-BC}\\
c &=& \frac{B^2+D^2+AB+CD}{AD-BC}\end{eqnarray}$$
Since here we overparametrize $a,b,c$ with $A,B,C,D$, we can in principle (see remark below) find these values. For instance you could set $A=0$ to simplify your search. Also note that as $\displaystyle \lim_{r\to\infty}f(r) = \frac AC$, we will have $ \displaystyle \lim_{r\to\infty}\arctan(f(r)) = \arctan\left(\frac AC\right)$.

An example, find $\displaystyle \sum_{r=1}^\infty \cot^{-1}\left(3r^2-r-\frac 13\right)$. We seek $A,B,C,D$ as above that works. Take $A=0$, we have by wolfram alpha $B = 1,C = -3 ,D = 2$ (among many other possible solutions). So $f(r) = \dfrac{1}{-3r+2}$ and
$$
\sum_{r=1}^\infty \cot^{-1}\left(3r^2-r-\frac13\right)\\
=-\arctan(f(1))+\lim_{r\to\infty}\arctan(f(r))\\
=-\arctan(-1)= \arctan(1) = \frac{\pi}4.
$$

Remark. There is a limitation to this, for instance say $a\neq 0 $ for this $f(r)$ to take this form. Indeed, if $a =0$, then we see that $A=C=0$, which will give a contradiction if $b\neq 0 $. So if $f(r) = \dfrac{Ar + B}{Cr+D}$, then $(a,b,c)$ needs to be in the range of the function
$$(A,B,C,D)\mapsto \left(\frac{A^2+C^2}{AD-BC},\frac{A^2+C^2+2AB+2CD}{AD-BC},\frac{B^2+D^2+AB+CD}{AD-BC}\right).$$
Remark 2. Despite this limitation, you can use this the other way: Pick your favorite four numbers $A,B,C,D$ and write down $f(r) = \dfrac{Ar + B}{Cr+D}$. This generates a quadratic $ar^2 + br + c$, and with this you will have the value of $\displaystyle \sum_{r=1}^\infty \cot^{-1}(ar^2 +br+c) = -\arctan\left(\dfrac{A + B}{C+D}\right)+\arctan\left(\dfrac{A}{C}\right)$.
