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Difficult Infinite Series

  1. Mar 8, 2010 #1
    1. The problem statement, all variables and given/known data
    Determine whether the following series converges absolutely, converges conditionally, or diverges.
    [tex]\sum_{n=1}^{\infty} \frac{(-1)^n}{n^3 - ln(n)}[/tex]

    2. Relevant equations
    The assortment of different tests.

    3. The attempt at a solution
    Okay, first of all, I tried using Alternating Series Test. This worked and the series satisfied all 3 conditions (decreasing, alternating, and limit as n approaches infinity = 0). This means the series must converge either conditionally or absolutely since I haven't eliminated the possibility of it converging absolutely yet.

    So I take the absolute value of the series and if it converges, it's absolutely convergent.
    If it diverges, then it's not absolutely convergent. Therefore, that means it's conditionally convergent because I already proved that it must converge in some manner.
    So this is
    [tex]\sum_{n=1}^{\infty} |\frac{(-1)^n}{n^3 - ln(n)}|[/tex]
    [tex]=\sum_{n=1}^{\infty} |\frac{1}{n^3 - ln(n)}|[/tex]

    Now how do I find that this series converges or diverges? I tried every test I'm aware of and each was inconclusive. I tried WolframAlpha and it said that the tests were inconclusive, but it gave a number.
    Does this mean it absolutely converges? If so, how would I show my work?

    I tried using the Direct Comparison test and compared the series with the absolute values to 1000/n^3. However, I'm not entirely sure if 1000/n^3 is greater than the absolute value series for all n terms.
    Last edited: Mar 8, 2010
  2. jcsd
  3. Mar 8, 2010 #2


    Staff: Mentor

    It should be fairly easy to show that n^3 - ln(n) > n^2 for all n >= some N, using induction or some other means.

    Once you get that, then 1/(n^3 - ln(n)) < 1/(n^2), and it's known that sum(1/(n^2)) is a convergent p-series.
  4. Mar 8, 2010 #3
    Okay, cool. Thanks for the example. Do you think that 1000/n^3 works? I tried checking to see if it it's greater for all n using the calculator, and it sure seemed like it, but I can't truly be sure.

    In other words, how can I prove that n^3/1000 < or = n^3 - ln(n) for all n or that it isn't? Solving the inequality, I get
    [tex]n^3 \ge \frac{1000}{999}ln(n)[/tex]
    And here I'm stuck. I'm gonna be mad if I have to use the Lambert W function or whatever. :\

    Could I argue that they never intersect and because n^3 is greater than the other function at one point, it must therefore be greater at all n terms? Then again, can I even fool around with the equation by solving for the inequality if I'm not even sure if it's true in the first place?
    Last edited: Mar 8, 2010
  5. Mar 9, 2010 #4


    Staff: Mentor

    Yeah, I think that will work. You can show that n^3 >= (1000/999)ln(n) by showing that n^3 - (1000/999)ln(n) >= 0 for all n larger than some fixed N. One way to do that is to call define f(x) = x^3 - (1000/999)ln(x), find some x0 for which f(x) > 0, and then show that f'(x) > 0 for all x > x0.
  6. Mar 10, 2010 #5
    Thanks!! I absolutely love you guys.

    And sorry for bumping this thread up.
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