1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Difficult Infinite Series
  1. Infinite series (Replies: 3)

  2. Infinite series (Replies: 4)

  3. Infinite Series (Replies: 2)

  4. Infinite series. (Replies: 4)

  5. Infinite series. (Replies: 1)