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Arctangent Series

  1. Nov 8, 2005 #1
    I am trying to compute:

    [tex]\sum_{k=0}^{n}\arctan{\left(\frac{1}{k^{2}+k+1}\right)}.[/tex]

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

    [tex]\sum_{k=0}^{n}\arctan{\left(k\right)}.[/tex]

    Is there a formula for this sum?

    Thanks for the help.
     
  2. jcsd
  3. Nov 8, 2005 #2
    I have two ways to do this (probably more, but two that I can think of):

    [tex]\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)},[/tex]

    and

    [tex]\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)}.[/tex]

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