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Math Help - Summation to infinity

  1. #1
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    Summation to infinity

    Determine \sum_{n = 0}^\infty \frac{1}{4r^2-8r+3} given that \frac{2}{4r^2-8r+3}=\frac{1}{2r-3}-\frac{1}{2r-1}
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    Quote Originally Posted by Punch View Post
    Determine \sum_{n = 0}^\infty \frac{1}{4r^2-8r+3} given that \frac{2}{4r^2-8r+3}=\frac{1}{2r-3}-\frac{1}{2r-1}
    With regard to your previous thread you know that, \sum_{r=1}^{n}(\frac{1}{4r^2-8r+3})=\frac{1}{2}\left(-1-\frac{1}{2n-1}\right) So take the limit as n tends to infinity.
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    Quote Originally Posted by Sudharaka View Post
    With regard to your previous thread you know that, \sum_{r=1}^{n}(\frac{1}{4r^2-8r+3})=\frac{1}{2}\left(-1-\frac{1}{2n-1}\right) So take the limit as n tends to infinity.
    as n tends to infinity, \frac{1}{2n-1} tends to 0.

    \sum_{n = 0}^\infty \frac{1}{4r^2-8r+3}=-\frac{1}{2}

    but ans is -\frac{1}{6}
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    Quote Originally Posted by Punch View Post
    as n tends to infinity, \frac{1}{2n-1} tends to 0.

    \sum_{n = 0}^\infty \frac{1}{4r^2-8r+3}=-\frac{1}{2}

    but ans is -\frac{1}{6}
    In your previous thread we have found that, \sum_{r=1}^{n}(\frac{1}{4r^2-8r+3})=\frac{1}{2}\left(-1-\frac{1}{2n-1}\right)

    Here we have to find, \sum_{r = 0}^{n}\frac{1}{4r^2-8r+3}=\frac{1}{3}+\sum_{r = 1}^{n}\frac{1}{4r^2-8r+3}

    Therefore taking the limit as n tend to infinity,

    \sum_{r = 0}^{\infty}\frac{1}{4r^2-8r+3}=\frac{1}{3}+\sum_{r = 1}^{\infty}\frac{1}{4r^2-8r+3}=\frac{1}{3}-\frac{1}{2}=-\frac{1}{6}

    Hope I did not confuse you with my last post, there I did not notice that the summation must be taken from r=0.
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  5. #5
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    Quote Originally Posted by Sudharaka View Post
    In your previous thread we have found that, \sum_{r=1}^{n}(\frac{1}{4r^2-8r+3})=\frac{1}{2}\left(-1-\frac{1}{2n-1}\right)

    Here we have to find, \sum_{r = 0}^{n}\frac{1}{4r^2-8r+3}=\frac{1}{3}+\sum_{r = 1}^{n}\frac{1}{4r^2-8r+3}

    Therefore taking the limit as n tend to infinity,

    \sum_{r = 0}^{\infty}\frac{1}{4r^2-8r+3}=\frac{1}{3}+\sum_{r = 1}^{\infty}\frac{1}{4r^2-8r+3}=\frac{1}{3}-\frac{1}{2}=-\frac{1}{6}

    Hope I did not confuse you with my last post, there I did not notice that the summation must be taken from r=0.
    thanks i didnt notice that it starts from zero too
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