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Math Help - Cauchy Condensation

  1. #1
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    Cauchy Condensation

    Does 1/ nlognloglogn converge or diverge?

    I am trying to show this with cauchy Condensation using 2^k in the new series. I have

    reduced it to

    (1/log2) E 1/ n(logn+loglog2)

    where E is the sum from n=3 to infinity. I think it diverges, and am trying to show that this expression is larger than

    E 1/ n(logn) which I have seen diverges, but the extra n in the denominator is messing me up. Can someone show me how to bound this, or if I am going about this the wrong way?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Re: Cauchy Condensation

    Quote Originally Posted by veronicak5678 View Post
    Does 1/ nlognloglogn converge or diverge?

    I am trying to show this with cauchy Condensation using 2^k in the new series. I have

    reduced it to

    (1/log2) E 1/ n(logn+loglog2)

    where E is the sum from n=3 to infinity. I think it diverges, and am trying to show that this expression is larger than

    E 1/ n(logn) which I have seen diverges, but the extra n in the denominator is messing me up. Can someone show me how to bound this, or if I am going about this the wrong way?

    But isn't it true (via the limit comparison test for example) that \displaystyle \sum\frac{1}{n\log(n)} converges if and only if \displaystyle \sum\frac{1}{n(\log(n)+\log(\log(2)))} does?
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  3. #3
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    Re: Cauchy Condensation

    We haven't talked about the limit comparison test. How else can you show that?
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Re: Cauchy Condensation

    Quote Originally Posted by veronicak5678 View Post
    We haven't talked about the limit comparison test. How else can you show that?
    Can you explain intuitively why it's true?
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  5. #5
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    Re: Cauchy Condensation

    The first series has terms that are always greater than those of the second, so if the second were to diverge to infinity, the greater series also would.
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