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Math Help - Does this series converge or diverge?

  1. #1
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    Does this series converge or diverge?

    Does this series converge or diverge?

    Summation of 2/[k(lnk)^2]
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by WartonMorton View Post
    Does this series converge or diverge?

    Summation of 2/[k(lnk)^2]
    That's a typical stuation where you want to apply Cauchy's Condensation Test (Cauchy condensation test - Wikipedia, the free encyclopedia)

    This theorem says that if the terms a_k\geq 0 are decreasing, the series \sum_k^\infty a_k converges or diverges if and only if the series \sum_n^\infty 2^n a_{2^n} converges or diverges, respectively (i.e. these two series behave the same way as regards convergence/divergence).

    So let's take a closer look, therefore, at the series

    \sum_n^\infty 2^n \frac{1}{2^n \ln^2(2^n)}=\sum_n^\infty \frac{1}{n\ln^2(2)}=\frac{1}{\ln^2(2)}\sum_n^\inft  y \frac{1}{n}

    which turns out to be essentially the same as the well known divergent series \sum_n^\infty \frac{1}{n}

    Thus, the series \sum_k^\infty \frac{1}{k\ln^2(k)} diverges.

    Edited:
    Plato is right, of course (he is almost always right): I made a silly mistake, since n comes out squared, one gets a series that converges:

    \sum_n^\infty 2^n \frac{1}{2^n \ln^2(2^n)}=\sum_n^\infty \frac{1}{n^2\ln^2(2)}=\frac{1}{\ln^2(2)}\sum_n^\in  fty \frac{1}{n^2}

    So, since \sum_n^\infty \frac{1}{n^2} converges, the original series converges.
    Last edited by Failure; May 6th 2010 at 11:28 AM.
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  3. #3
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    Quote Originally Posted by Failure View Post
    Thus, the series \sum_k^\infty \frac{1}{k\ln^2(k)} diverges.
    That statement is incorrect.
    This is the classic example for the integral test.
    \sum_k^\infty \frac{1}{k\ln^2(k)} converges.
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