# Arctangent Series

1. Nov 8, 2005

### amcavoy

I am trying to compute:

$$\sum_{k=0}^{n}\arctan{\left(\frac{1}{k^{2}+k+1}\right)}.$$

I have used some trig identities and reduced it, but before I can do anything more I am stuck on:

$$\sum_{k=0}^{n}\arctan{\left(k\right)}.$$

Is there a formula for this sum?

Thanks for the help.

2. Nov 8, 2005

### amcavoy

I have two ways to do this (probably more, but two that I can think of):

$$\sum_{k=0}^{n}\arctan{\left(\frac{1}{k^{2}+k+1}\right)}=\frac{\pi n}{2}-\sum_{k=0}^{n}\arctan{\left(k^{2}+k+1\right)},$$

and

$$\sum_{k=0}^{n}\arctan{\left(\frac{1}{k^{2}+k+1}\right)}=\sum_{k=0}^{n}\arctan{\left(k+1\right)}-\sum_{k=0}^{n}\arctan{\left(k\right)}.$$

Which one is better to work with? Thanks for the help.