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Math Help - conegrens of sine mmn11 2a

  1. #1
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    conegrens of sine mmn11 2a

    is it converges absolutly ,or conditionally convergent
    or is it diverges?


    \sum_{k=1}^{infinity}(-1)^{k}sin\frac{1}{k}
    the absolute value series is |sin(1/k)|<=|1/k|
    which is diverges

    but it gives me nothing
    because if it where convergent then the smaller one would converge
    but here the bigger one diverges(harmonic series)
    ??
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  2. #2
    Ted
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    Try to use the limit comparison test with \displaystyle b_k=\dfrac{1}{k}
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  3. #3
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by transgalactic View Post
    is it converges absolutly ,or conditionally convergent
    or is it diverges?


    \sum_{k=1}^{infinity}(-1)^{k}sin\frac{1}{k}
    the absolute value series is |sin(1/k)|<=|1/k|
    which is diverges

    but it gives me nothing
    because if it where convergent then the smaller one would converge
    but here the bigger one diverges(harmonic series)
    ??
    Pauls Online Notes : Calculus II - Alternating Series Test

    Kind regards

    \chi \sigma
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  4. #4
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    i did compared it to 1/k
    but 1/k is diverging

    so it doesnt say anything about it
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  5. #5
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by transgalactic View Post
    i did compared it to 1/k
    but 1/k is diverging

    so it doesnt say anything about it
    Of course \displaystyle \sum_{k=1}^{\infty} \frac{1}{k} diverges... but also is \displaystyle \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k} = \ln 2...
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  6. #6
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    first we need to compare the absolute value
    the absolute value series is |sin(1/k)|<=|1/k|
    which is diverges

    but it gives me nothing
    because if it where convergent then the smaller one would converge
    but here the bigger one diverges(harmonic series)
    ??
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