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Math Help - Calculus II - Comparison Test(s)

  1. #1
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    Calculus II - Comparison Test(s)

    I have two similar questions:

    1.) Sigma from n=1 to infinity (1 / (sqrt(n) * (ln n)^4))

    2.) Sigma from n=1 to infinity ((ln n)^3 / n^2)

    I must use the direct comparison/limit comparison tests for these, but I'm not sure how to go about choosing a bn.

    My first intuition for #1 was to choose bn to be 1/n, but 1/n is divergent and greater than (1 / (sqrt(n) * (ln n)^4)).

    Any help would be appreciated because I'm very lost.
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  2. #2
    MHF Contributor matheagle's Avatar
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    since log x is a slowly varying function, you can more or less ignore it in this case

    By that I mean the first series will diverge and the second will converge.
    NOW, you can either do the direct comp or limit comp tests.

    Since the essential p=.5 in the first series you can use any p in (.5,1] say p=2/3.
    Compare that series to

    \sum {1\over n^{2/3}}=\infty or better yet \sum {1\over n}=\infty

    Next show that {{1 \over \sqrt{n} (\ln n)^4}\over {1\over n}}\to\infty and you're done

    Showing this limit is infinite means your terms are larger hence your series is larger
    and we all know that the harmonic series diverges.
    Last edited by matheagle; August 11th 2009 at 12:07 AM.
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  3. #3
    MHF Contributor matheagle's Avatar
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    Same idea with \sum {(\ln n)^3 \over n^2}

    This converges, BUT it's slightly bigger than \sum {1\over n^2}

    SO here, pick any p in (1,2) and you win, say p=3/2...

    \sum {1\over n^{3/2}}<\infty

    Finally take either ratio and show that your terms are much smaller than {1\over n^{3/2}}

    It doesn't matter which ratio you use.
    One will go to zero, the other to infinity.
    The point is, you found a larger sum that's finite.
    Since all the terms in both series are positive your series will converge too.
    A direct comp test works just as well.
    The only problem is that we may need to ignore the first few terms.
    Last edited by matheagle; August 11th 2009 at 12:09 AM.
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