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Math Help - Telescoping series sum

  1. #1
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    Telescoping series sum

    so i have a telescoping series that converges and I need to find the sum. How do I do this?

    Thanks
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  2. #2
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    series from zero to infinite of 4/((x squared)-1)

    sorry about the way i wrote it out i will learn with time
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by dogface View Post
    series from zero to infinite of 4/((x squared)-1)

    sorry about the way i wrote it out i will learn with time
    Ok you have \sum_{n=0}^{\infty}\frac{4}{x^2-1}\Rightarrow\sum_{n=0}^{\infty}\frac{4}{(x-1)(x+1)}

    then by using partial fractions decomposition we get


    \sum_{n=0}^{\infty}\bigg[\frac{2}{x-1}-\frac{2}{x+1}\bigg]...now write out the first few terms of the sequence...do you notice a pattern of cancellations?
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  4. #4
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    got it thank you
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by dogface View Post
    got it thank you
    No problem! See we are not so bad here!
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  6. #6
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    Quote Originally Posted by dogface View Post
    series from zero to infinite of 4/((x squared)-1)

    sorry about the way i wrote it out i will learn with time
    There is a problem: What happens when x = 1 ......? So assuming you instead want


    = \sum_{x = 2}^{\infty} \left( \frac{2}{x-1} - \frac{2}{x+1} \right) = 2 \sum_{x = 2}^{\infty} \left( \frac{1}{x-1} - \frac{1}{x+1} \right)


    using partial fractions


    = 2 \left( \left[ 1 - \frac{1}{3}\right] + \left[ \frac{1}{2} - \frac{1}{4}\right] + \left[ \frac{1}{3} - \frac{1}{5}\right] + \left[ \frac{1}{4} - \frac{1}{6}\right] + .....\right)


    and yu should be able to see that after cancelling you're left with 2 \left(1 + \frac{1}{2} \right) = 3.
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  7. #7
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    Quote Originally Posted by Mathstud28 View Post
    Ok you have \sum_{n=0}^{\infty}\frac{4}{x^2-1}\Rightarrow\sum_{n=0}^{\infty}\frac{4}{(x-1)(x+1)}
    This is abuse of notation. When something implies another thing, a previous equality symbol should be there, so instead an \implies that's actually an =.

    As for dogface's question, I think you should find problems about your issue and post it (by makin' new threads of course) to see where you're getting stuck.
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