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Another Test for Convergence Question

  1. Sep 10, 2013 #1
    1. The problem statement, all variables and given/known data
    Test the following series for convergence or divergence.
    $$\sum_{n=1}^{\infty} \frac{1}{3^{\ln n}}$$

    3. The attempt at a solution
    I've tried to compare this to geometric series ##3^n## but obviously the target term is larger overall than its geometric counterpart. Comparing with the reciprocal of ##\ln n## also brings no avail because the term is smaller than the former. Ratio test gives ##ρ=1## which is inconclusive. It certainly passes preliminary test because the limit goes to 0 once n goes to infinity. I haven't done integral test because it's quite ugly and I don't know how to integrate the corresponding function.

    Can you guys give me suggestion to tame this series down nicely?

    Thank You
  2. jcsd
  3. Sep 11, 2013 #2


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    Use that 3=e^(ln(3)) and use the laws of exponentials to rearrange it a bit. It becomes a power series.
  4. Sep 11, 2013 #3
    Edit: Dick beat me to it. Quick.
    Last edited: Sep 11, 2013
  5. Sep 11, 2013 #4


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    Try using 3 =eln(3) to rewrite your summand
  6. Sep 11, 2013 #5
    Strange things happened when I arrange the summation.

    ##3=e^{\ln 3}##
    $$3^{\ln n}=(e^{\ln 3})^{\ln n}$$
    $$3^{\ln n}=e^{\ln 3 ° \ln n}$$
    $$3^{\ln n}=n^{\ln 3}$$

    If that's the case then the series must be convergent because of p-test, with p being larger than 1.
  7. Sep 11, 2013 #6


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    I don't know if I'd call that strange, but it is the correct conclusion.
  8. Sep 11, 2013 #7
    Well maybe I'm not quite used to manipulating exponents yet. Thanks for your help guys!
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