Finding $\sum_{r=1}^{n}{\cot{}^{ - 1}(3r^2-\frac{5}{12})}$

Question

Find the sum of $$\sum_{r=1}^{n}{\cot{}^{ - 1}(3r^2-\frac{5}{12})}$$

I tried to convert all into  $\tan {}^{ - 1}(x) $ using  $\cot {}^{ - 1} (x) = { \tan {}^{ - 1} ( \frac{1}{x} ) }$  and then tried to simplify them using property  $ \tan {}^{ - 1} ( \frac{x - y}{1 - xy} ) = \tan {}^{ - 1} (x) - \tan {}^{- 1} (y) $ so that some terms cancel each other.

But I do not manage to simplify the problem. What to do?

Answer 1

After the fix of sign in front of $\frac{5}{12}$ in the question body, the sum becomes pretty simple.

Let $a_{\pm} = 3(r\pm \frac12)$, notice

$$\frac{a_+ - a_-}{1 + a_+ a_-} = \frac{3}{1 + 9(r^2-\frac14)} = \frac{12}{36r^2 - 5} = \left(3r^2 - \frac{5}{12}\right)^{-1}$$ We have $$\begin{align}\cot^{-1}\left(3r^2 - \frac{5}{12}\right) &= \tan^{-1}\left(\frac{a_+ - a_-}{1 + a_+a_-}\right) = \tan^{-1}a_+ - \tan^{-1}a_-\\ &= \tan^{-1}(3(r + \frac12)) - \tan^{-1}(3(r - \frac12)) \end{align} $$ The new sum is a telescoping one and $$\sum_{r=1}^n \cot^{-1}\left(3r^2 - \frac{5}{12}\right) = \tan^{-1}(3(n + \frac12)) - \tan^{-1}(\frac32)$$

Answer 2

Using CAS help we have:

$$\sum _{r=1}^n \cot ^{-1}\left(3 r^2-\frac{5}{12}\right)=\tan ^{-1}\left(\frac{2}{3}\right)-\tan ^{-1}\left(\frac{2}{6 n+3}\right)$$

Mathematica code:

FullSimplify[Integrate[Sum[D[ArcCot[A*r^2 - 5/12], A], {r, 1, n}] // Normal, 
A] /. A -> 3, n > 0]

(* Luckily, in this case intergration constant C = 0 *)

(*ArcTan[2/3] - ArcTan[2/(3 + 6 n)] *)

By the way, we can still find Sum at n=Infinity

$$\sum _{r=1}^{\infty } \cot ^{-1}\left(3 r^2-\frac{5}{12}\right)=\tan ^{-1}\left(\frac{2}{3}\right)\approx 0.588003$$