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Math Help - partial sum question

  1. #1
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    partial sum question

    hey guys i'm having a little trouble working out a question. the question is to find a formula for the nth partial sum of the series sum of 1/((4n+1)(4n-3)) between n=1 and infinity. any help would be great thanks
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  2. #2
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by newcastle View Post
    hey guys i'm having a little trouble working out a question. the question is to find a formula for the nth partial sum of the series sum of 1/((4n+1)(4n-3)) between n=1 and infinity. any help would be great thanks
    first use partial fractions to write that sum than write few of them (depending how many you need to realize which is formula for nth partial sum)

    show what have you done



     \displaystyle s = \sum _{n=1} ^{N} \frac {N}{4N+1}
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  3. #3
    MHF Contributor chisigma's Avatar
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    Is...

    \displaystyle \frac{1}{(4n+1)\ (4n-3)} = \frac{1}{4}\ (\frac{1}{4n-3} - \frac{1}{4n+1}) (1)

    ... so that...

    \displaystyle \sum_{n=1}^{N} \frac{1}{(4n+1)\ (4n-3)} = \frac{1}{4}\ (1 -\frac{1}{5} + \frac{1}{5} -\frac{1}{9} + \frac{1}{9} -... - \frac{1}{4N+1}) = \frac{1}{4}\ (1 - \frac{1}{4N+1}) (2)

    ... and finally...

    \displaystyle \sum_{n=1}^{\infty} \frac{1}{(4n+1)\ (4n-3)} = \frac{1}{4} (3)

    Kind regards

    \chi \sigma
    Last edited by chisigma; September 9th 2010 at 08:14 AM. Reason: 7 instead 4... error signalled by yeKciM... sorry...
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  4. #4
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by chisigma View Post
    Is...

    \displaystyle \frac{1}{(4n+1)\ (4n-3)} = \frac{1}{7}\ (\frac{1}{4n-3} - \frac{1}{4n+1}) (1)

    ... so that...

    \displaystyle \sum_{n=1}^{N} \frac{1}{(4n+1)\ (4n-3)} = \frac{1}{7}\ (1 -\frac{1}{5} + \frac{1}{5} -\frac{1}{9} + \frac{1}{9} -... - \frac{1}{4N+1}) = \frac{1}{7}\ (1 - \frac{1}{4N+1}) (2)

    ... and finally...

    \displaystyle \sum_{n=1}^{\infty} \frac{1}{(4n+1)\ (4n-3)} = \frac{1}{7} (3)

    Kind regards

    \chi \sigma


    probably just typo it's

    \displaystyle \sum_{n=1}^{\infty} \frac{1}{(4n+1)\ (4n-3)} = \frac{1}{4} (3)





    Edit:


     \displaystyle \frac {1}{(4n+1)(4n-3)} = \frac {A}{4n+1} + \frac {B}{4n-3}

     4A+4B =0 \Rightarrow A=-B

    \displaystyle -3A+B=1 \Rightarrow -4A=1 \Rightarrow A=-\frac {1}{4} \Rightarrow B = \frac {1}{4}

    ...... ........
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