# 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.

• have you tried Wolfram Alpha? – Dr. Sonnhard Graubner Sep 22 '17 at 13:33
• Actually I am not interested in the value but approach. Can writing the terms in telescopic form and cancelling out terms be applied? – jnyan Sep 22 '17 at 13:35
• i think no that we can use this method here – Dr. Sonnhard Graubner Sep 22 '17 at 13:36
• @jnyan There is a nice closed form, so I do wonder. – George Coote Sep 22 '17 at 13:39
• i have a formula for the finite sum – Dr. Sonnhard Graubner Sep 22 '17 at 13:40

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 $\cot^{-1}\left(n^2+\dfrac34\right)=\cot^{-1}\left(\dfrac{1+\dfrac{2n+1}2\cdot\dfrac{2n-1}2}{\dfrac{2n+1}2-\dfrac{2n-1}2}\right)$

$=\cot^{-1}\left(\dfrac{2n-1}2\right)-\cot^{-1}\left(\dfrac{2n+1}2\right)$

• It was plain stupid of me. Thank you. – jnyan Sep 22 '17 at 13:47
• – lab bhattacharjee Sep 22 '17 at 18:05

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 $\tan^{-1}$.