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Math Help - conditional convergence v.s. converges absolutely

  1. #1
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    [SOLVED]conditional convergence v.s. converges absolutely

    1.After I use Alternating Series Test and result is converges
    Which one do I say? converges or converges absolutely
    Is it only way to prove series is converges absolutely by Ratio Test for Absolute Convergence???

    2.I know how to prove the series is converges absolutely by Ratio Test for Absolute Convergence

    How to prove a series is conditional convergence? Which method do I need to use?


    Thank you!
    Last edited by soleilion; April 15th 2008 at 07:09 PM.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Ok

    Quote Originally Posted by soleilion View Post
    1.After I use Alternating Series Test and result is converges
    Which one do I say? converges or converges absolutely
    Is it only way to prove series is converges absolutely by Ratio Test for Absolute Convergence???

    2.I know how to prove the series is converges absolutely by Ratio Test for Absolute Convergence

    How to prove a series is conditional convergence? Which method do I need to use?


    Thank you!
    If a series |a_n| converges, then it conditionally converges...but if |a_n| and a_n converges it is absolutely convergent
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  3. #3
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    Quote Originally Posted by Mathstud28 View Post
    If a series |a_n| converges, then it conditionally converges...but if |a_n| and a_n converges it is absolutely convergent
    ok
    How to say a series is conditional converges
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Ok

    Quote Originally Posted by soleilion View Post
    ok
    How to say a series is conditional converges
    You would say a series is...o wait I am sorry I mistyped it is actually absolutely convergent if |a_n| converges...for example since \sum_{n=0}^{\infty}\frac{(-1)^{n+1}}{n} converges but \sum_{n=0}^{\infty}\bigg|\frac{(-1)^{n+1}}{n}\bigg|=\frac{1}{n} diverges it is only conditionally convergent
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  5. #5
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    Quote Originally Posted by Mathstud28 View Post
    You would say a series is...o wait I am sorry I mistyped it is actually absolutely convergent if |a_n| converges...for example since \sum_{n=0}^{\infty}\frac{(-1)^{n+1}}{n} converges but \sum_{n=0}^{\infty}\bigg|\frac{(-1)^{n+1}}{n}\bigg|=\frac{1}{n} diverges it is only conditionally convergent
    that means if I can prove uk is converges and ak is diverges
    then, I can say the series is conditional converges
    right?
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by soleilion View Post
    that means if I can prove uk is converges and ak is diverges
    then, I can say the series is conditional converges
    right?
    I have no idea what u_k and a_k is but if you could prove that a series converges but the absolute value of that series diverges it is conditionally convergent
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  7. #7
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    Quote Originally Posted by Mathstud28 View Post
    I have no idea what u_k and a_k is but if you could prove that a series converges but the absolute value of that series diverges it is conditionally convergent
    ok
    I understood
    Thank you very much
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by soleilion View Post
    ok
    I understood
    Thank you very much
    No problem...on the ones I didnt mistype
    Last edited by ThePerfectHacker; April 15th 2008 at 07:22 PM.
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