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Homework Help: Series converge or diverge

  1. Mar 23, 2010 #1
    1. The problem statement, all variables and given/known data

    Determine whether the series converges or diverges.

    [tex] \sum \frac{lnk}{k^3} [/tex]


    2. Relevant equations



    3. The attempt at a solution

    Since lnk always less than 0, so [tex] \frac{lnk}{k^3} \leq \frac{1}{k^3}[/tex] and [tex]\frac{1}{k^3} [/tex]diverges so[tex] \frac{lnk}{k^3}[/tex] diverges.
     
  2. jcsd
  3. Mar 23, 2010 #2

    Mark44

    Staff: Mentor

    I count three mistakes here.
    1. Why do you think that ln(k) is always < 0? I'm assuming that k = 1, 2, 3, ...
    2. The series whose general term is 1/k3 converges.
    3. If you want to show that a given series diverges, its terms must be larger than those of a divergent series. If the terms of the given series are smaller than those of a divergent series, you can't conclude anything.
     
  4. Mar 23, 2010 #3
    Can I argue that [tex]ln(k) < k [/tex] then [tex] \frac{ln(k)}{k^3} < \frac{k}{k^3} [/tex]then it converge because [tex]\frac{1}{k^2} [/tex] converges?
     
  5. Mar 23, 2010 #4

    Mark44

    Staff: Mentor

    I can buy that. However, your should be able to convince yourself that ln(x) < x.
     
  6. Mar 23, 2010 #5
    Yeah since ln(x) is the inverse of e^x and they reflect on the y =x.
     
  7. Mar 23, 2010 #6

    Mark44

    Staff: Mentor

    The fact that y = ln(x) and y = ex are reflections across the line y = x doesn't prove that ln(x) < x. If y = ex happens to cross this line a few times means that ln(x) will do so, also.
     
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