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Math Help - Summation

  1. #1
    MHF Contributor Mathstud28's Avatar
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    Summation

    I was doing something and I got \sum_{n=1}^{m}arctan\bigg(\frac{1}{n^2+n+1}\bigg)=  arctan(m+1)-\frac{\pi}{4}...does anyone concur?
    Last edited by Mathstud28; April 12th 2008 at 07:17 PM.
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  2. #2
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    Quote Originally Posted by Mathstud28 View Post
    I was doing something and I got \sum_{n=1}^{m}arctan\bigg(\frac{1}{n^2+n+1}\bigg)=  arctan(m+1)-\frac{\pi}{5}...does anyone concur?
    It's not true for m = 1 or m = 2 .... Did you do such a check?
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  3. #3
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    care to post your method or the complete question ?
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Ok

    Quote Originally Posted by Mathstud28 View Post
    I was doing something and I got \sum_{n=1}^{m}arctan\bigg(\frac{1}{n^2+n+1}\bigg)=  arctan(m+1)-\frac{\pi}{4}...does anyone concur?
    Now that I fixed my typo is this correct?
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  5. #5
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    Quote Originally Posted by Mathstud28 View Post
    Now that I fixed my typo is this correct?
    An improvement.
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    I am sorry

    Quote Originally Posted by mr fantastic View Post
    An improvement.
    Does that mean I am correct?
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  7. #7
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    Quote Originally Posted by Mathstud28 View Post
    Does that mean I am correct?
    Well, it means it's correct for m = 1 and m = 2 .....

    I haven't done a detailed analysis - perhaps you could post all your working for possible critique.
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Ehh

    Quote Originally Posted by mr fantastic View Post
    Well, it means it's correct for m = 1 and m = 2 .....

    I haven't done a detailed analysis - perhaps you could post all your working for possible critique.
    I think I will tomorrow its a lot of work..I think I am right =D....at least up to m=10
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