# 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?

• Did you try to use the residue theorem? en.wikipedia.org/wiki/Residue_theorem Jan 10, 2021 at 18:41
• Unfortunately, I haven't learnt it yet, sir. However, a solution by that is also welcomed, since it would provide the final closed form which would be helpful for me.
– V.G
Jan 10, 2021 at 18:42
• Basically this here Jan 10, 2021 at 18:58
• @Buraian: How is that helpful here?
– V.G
Jan 10, 2021 at 19:00
• $$\sum _{r=1}^{\infty } \text{arccot}\left(3 r^2-r-\frac{1}{3}\right)=\frac{\pi}{4}$$ Jan 10, 2021 at 20:34

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 $$\displaystyle \sum_{r=1}^\infty \cot^{-1}(ar^2 +br+c) = -\arctan\left(\dfrac{A + B}{C+D}\right)+\arctan\left(\dfrac{A}{C}\right)$$.

• I think this is wonderful. Jan 11, 2021 at 3:44
• Why did you assume the form of $f(r)$ has a linear fraction? Jan 11, 2021 at 7:08
• @Buraian I should clarify that $f$ need not be a linear fraction -- What we really want is some $f$ that $(\dagger)$ holds, namely that combination giving a quadratic, and I am also looking for $f$ such that $\lim f(r)$ exists. So a linear fraction comes to mind. Note not all quadratic can be realized this way, and in that case you will need to find a suitable $f$ that will work. For the example quadratic $3r^2 -r -1/3$, this linear fraction form of $f$ will work, but possibly not for other quadratic. Jan 11, 2021 at 7:34
• The fraction approach does not provide any solutions that could not have been solved with $f(r) =Ar+B.$ Proof: If $(A_0,B_0,C_0,D_0)$ was a solution represented by the vectors $(A_0,C_0)^T$ and $(B_0,D_0)^T$, then we can get another solution $(A_1,B_1,C_1,D_1)$ which is represented by the vectors $(A_1,C_1)^T$ and $(B_1,D_1)^T$ simply by rotating $(A_0,C_0)^T$ and $(B_0,D_0)^T$ by the same angle. We always can rotate $(A_0,C_0)^T$ such that $C_1=0$, which means that we could as well have started with a linear function $\frac{A}{D} r + \frac{B}{D}$ Jan 11, 2021 at 12:41
• So you can solve it this way only if $a^2+4ac = b^2 +4$ Jan 11, 2021 at 13:46