Sum of arctangents: sum [n=1 to infty] tan^(-1)(2/(n^2+2n+6)

[mma]

Hello you all.

I am not able to evaluate the following sum:

. . . . .\(\displaystyle \displaystyle \sum_{n=1}^{\infty}\, \tan^{-1}\left(\dfrac{2}{n^2\, +\, 2n\,+\, 6}\right)\)

I'm trying to find p and q so that the sum telescopes, but it is not easy for me

 

[Deleted member 4993]

I am not able to evaluate the following sum:

. . . . .\(\displaystyle \displaystyle \sum_{n=1}^{\infty}\, \tan^{-1}\left(\dfrac{2}{n^2\, +\, 2n\,+\, 6}\right)\)

I'm trying to find p and q so that the sum telescopes but it is not easy for me

What are "p" and "q" - how are those related to the given expression or the series?

 

[mma]

Thanks Khan for your interest.
I mean that the arctangent telescopes when, for instance, it is possible to find p=p(n) and q=p(n+k) with k some integer so that:

. . . . .\(\displaystyle \tan^{-1}(p)\, -\, \tan^{-1}(q)\, =\, \tan^{-1}\left(\dfrac{p\, -\, q}{1\, +\, pq}\right)\)

In other words, I can evaluate the sum if I can find p and q such that:

. . . . .\(\displaystyle \dfrac{p\, -\, q}{1\, +\, pq}\, =\, \dfrac{2}{n^2\, +\, 2n\, +\, 6}\)