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Absolute Converge test for 1/[n*ln(n)]

  1. Dec 13, 2008 #1
    1. The problem statement, all variables and given/known data

    \sum_{n = 2}^{\infty} \frac{1}{n*ln(n)}

    I have to find whether the series absolute converge, conditionally converge or diverge?

    2. The attempt at a solution

    I used the ratio test.

    so, lim(n to infinity) [n*ln(n)]/[(n+1)*ln(n+1)]

    since ln (n+1) will be greater than ln (n) and n+1 will be greater than n, the whole denominator will be greater than the numerator so when i take the limit, the value must be less than 1.

    but i think i have cancel n or ln(n) to show that the whole limit is really less than 1 to converge.


    i m sorry . i dun know how to use the latex code..
  2. jcsd
  3. Dec 13, 2008 #2
    No, the limit is 1 even though each [tex]\frac{n \ln n}{(n + 1) \ln (n + 1)} < 1[/tex], because [tex]\lim_{n \to \infty} \frac{n}{n + 1} = \lim_{n \to \infty} \frac{1}{1 + \frac{1}{n}} = 1[/tex] and [tex]\lim_{n \to \infty} \frac{\ln n}{\ln (n + 1)} = 1[/tex] by l'hopital's rule.

    Since the limit is 1, the ratio test is inconclusive. Do you have any other methods to use?
  4. Dec 13, 2008 #3
    i think i can use "limit comparison test" but i m not sure what i should use for bn.

    if i use either 1/n or 1/ln(n) for bn, the limit goes to zero and since bn diverge, i can't decide on an(original term).

    one more questions abt absolute convergence test. assuming the series is given, not alternative series, if i use one of the test n find out , it is divergent. it is divergent. rt?
    i am confused with when i have to test for conditional convergence. is it only for alternate series.
    all the examples in the text book which turn out to be conditional convergence are alternate series..

    sorry for asking too many questions in one post.
  5. Dec 13, 2008 #4
    Use an integral test:
    [tex]\int 1 / (n*ln|n|)[/tex]
    A certain substitution should eventually lead to an antiderivative of:

    I'll leave the rest to you (remember to look at what lim n->inf: ln(ln(n)) does)
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