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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?

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    $\begingroup$ Did you try to use the residue theorem? en.wikipedia.org/wiki/Residue_theorem $\endgroup$
    – Nicolas
    Jan 10, 2021 at 18:41
  • $\begingroup$ 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. $\endgroup$
    – V.G
    Jan 10, 2021 at 18:42
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    $\begingroup$ Basically this here $\endgroup$ Jan 10, 2021 at 18:58
  • $\begingroup$ @Buraian: How is that helpful here? $\endgroup$
    – V.G
    Jan 10, 2021 at 19:00
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    $\begingroup$ $$\sum _{r=1}^{\infty } \text{arccot}\left(3 r^2-r-\frac{1}{3}\right)=\frac{\pi}{4}$$ $\endgroup$
    – Raffaele
    Jan 10, 2021 at 20:34

1 Answer 1

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

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    $\begingroup$ I think this is wonderful. $\endgroup$
    – Matthew H.
    Jan 11, 2021 at 3:44
  • $\begingroup$ Why did you assume the form of $f(r)$ has a linear fraction? $\endgroup$ Jan 11, 2021 at 7:08
  • $\begingroup$ @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. $\endgroup$
    – bonsoon
    Jan 11, 2021 at 7:34
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    $\begingroup$ 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 $\frac{A}{D} r + \frac{B}{D}$ $\endgroup$ Jan 11, 2021 at 12:41
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    $\begingroup$ So you can solve it this way only if $a^2+4ac = b^2 +4$ $\endgroup$ Jan 11, 2021 at 13:46

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