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?
Best Answer
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 $$\sum_{r=1}^n \cot^{-1}\left(3r^2 - \frac{5}{12}\right) = \tan^{-1}(3(n + \frac12)) - \tan^{-1}(\frac32)$$