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I Convergence of (ln n)^x

  1. Jun 12, 2016 #1
    Find the values of ##x## for which the following series is convergent.
    Screen Shot 2016-06-13 at 1.25.44 am.png

    I compared the series with the harmonic series and deduced it is always divergent. I used ##y^p<e^y## for large ##y##. I used a different method from the answer given, which I don't understand.

    Screen Shot 2016-06-13 at 1.25.21 am.png

    When ##k=1##, ##M_1=e=2.72## and ##M_0=1##. Since ##M_0+1\leq n\leq M_1##, we have ##n=2##, giving us the term ##\frac{1}{1^X}##.

    When ##k=2##, ##M_2=e^2=7.39## and ##M_1=e=2.72##. Since ##M_1+1\leq n\leq M_2##, we have ##n=4, 5, 6, 7##, giving us the terms ##\frac{1}{2^X}+\frac{1}{2^X}+\frac{1}{2^X}+\frac{1}{2^X}##.

    Finding the possible values of ##n## is troublesome.

    Is there a typo in the given answer?
     
    Last edited: Jun 12, 2016
  2. jcsd
  3. Jun 12, 2016 #2

    mfb

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    You get of the order of e terms of 1/1X, e2 terms of 1/2X, e3 terms of 1/3X and so on (up to the constant prefactor). Following the same logic as your approximation, ##\displaystyle \frac{e^n}{n^X}## grows to infinity instead of going to zero.
     
  4. Jun 12, 2016 #3
    Why is the series in the question and the series in the answer equivalent?

    Their first terms are clearly different. The former's is ##\frac{1}{(\ln 2)^X}## while the latter's is ##\frac{1}{1^X}##.

    If they are not equivalent, how do we show that the former is always bigger than the latter for every term?
     
  5. Jun 12, 2016 #4

    mfb

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    The first fraction is always larger than the second by construction of the Mk. So your series is larger than a divergent series.
     
  6. Jun 12, 2016 #5
    Could you explain how? I don't see it.
     
  7. Jun 12, 2016 #6

    mfb

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    ln(2) < 1
    ln(3) < 2, ln(4) < 2, ln(5) < 2, ln(6) < 2, ln(7) < 2

    Therefore 1/ln(2)X > 1/1X and so on for positive X.
     
  8. Jun 12, 2016 #7
    The given answer is so complicated vs if we just compare the series with the harmonic series.
     
  9. Jun 12, 2016 #8

    mfb

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    How do you do that for x=-3?
     
  10. Jun 13, 2016 #9
    Let ##y=\ln n##. For large ##n, y^3<e^y##. So ##(\ln n)^3<n##. Then ##\frac{1}{(\ln n)^3}>\frac{1}{n}##. The series is bigger than the harmonic series for every term. Hence it is also divergent.
     
  11. Jun 13, 2016 #10

    mfb

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    Ah right, that works as well, and it is easier.
     
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