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Convergence of a sequence

  1. Mar 10, 2010 #1
    I am having difficulty determining whether or not the following sequence can be classified as convergent or divergent:

    [tex]^{\infty}_{k=1}{\sum}[/tex][tex]\frac{1}{k^{ln(k)}}[/tex]

    This can be simplified to:

    [tex]^{\infty}_{k=1}{\sum}[/tex][tex]\frac{1}{e^{{ln(k)}^{2}}}[/tex]

    Both the ratio test and root test are inconclusive (giving values of 1), while attempting the integral test doesn't work as I am unable to integrate this as a function.

    Any suggestions?
     
    Last edited: Mar 10, 2010
  2. jcsd
  3. Mar 10, 2010 #2
    If by k, you mean n, then consider the following.
    [tex]
    y = e^{(ln n)^2} \rightarrow
    y' = y \frac{2 \ln n}{n} > 0.
    [/tex]

    Therefore, the terms are strictly nonincreasing.

    Consider the following:

    http://en.wikipedia.org/wiki/Cauchy_condensation_test

    I'm sure you can do the rest.
     
  4. Mar 10, 2010 #3
    Yes, by n I meant k.

    I have actually never encountered the Cauchy condensation test until now.

    I was able to finish it. Thank you very much.
     
  5. Mar 11, 2010 #4

    Gib Z

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    Homework Helper

    A comparison to a p-series would have also worked.
     
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