Infinite sum of $\cot^{-1} (n^2 + 3/4)$.

Question

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.

Answer 1

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 $xy>1$

Answer 2

Hint. One may use, for $n\ge1$, $$\tan^{-1}\frac{1}{n^2+\frac34}=\tan^{-1}\frac{\left(n+\frac12\right)-\left(n-\frac12\right)}{1+\left(n+\frac12\right)\left(n-\frac12\right)}=\tan^{-1}\left(n+\frac12\right)-\tan^{-1}\left(n-\frac12\right)$$ then use the link between $\cot^{-1}$ and $\tan^{-1}$.