<s>[INST] Infinite sum of $\cot^{-1} (n^2 + 3/4)$. I am trying to find the infinite sum 
$$\sum_{n=1}^\infty \cot^{-1} (n^2 + ( \frac{3}{4})),$$
 I tried to get a telescopic series but I couldn't find one. 
 [/INST] As $\cot(A-B)=\dfrac{\cot A\cot B+1}{\cot B-\cot A}$
$$\dfrac{4n^2+3}4=1+\dfrac{4n^2-1}4=1+\dfrac{2n+1}2\cdot\dfrac{2n-1}2$$
Again,$\dfrac{2n+1}2-\dfrac{2n-1}2=1$
So, we can write $\cot^{-1}\left(n^2+\dfrac34\right)=\cot^{-1}\left(\dfrac{1+\dfrac{2n+1}2\cdot\dfrac{2n-1}2}{\dfrac{2n+1}2-\dfrac{2n-1}2}\right)$
$=\cot^{-1}\left(\dfrac{2n-1}2\right)-\cot^{-1}\left(\dfrac{2n+1}2\right)$
See also: Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$
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