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Homework Help: Does the series Converge or Diverge ?

  1. Mar 8, 2012 #1
    1. The problem statement, all variables and given/known data

    Ʃ ln(n!)/(n^3)*ln(n)

    2. Relevant equations

    Ratio test, which states:
    Lim as n tends to infinity of |(An+1)/An|=ρ would lead to convergent series if the limit is less than 1
    it diverges if ρ is greater than 1
    test is inconclusive if ρ=1

    3. The attempt at a solution
    I thought of applying Ratio test but didn't know how to cancel out terms
    Last edited by a moderator: Mar 8, 2012
  2. jcsd
  3. Mar 8, 2012 #2


    Staff: Mentor

    n! = 2 * 3 * 4 * ... * (n - 1) * n

    ln(a*b) = ln(a) + ln(b)

    Is ln(n) in the numerator or denominator? The way you write the series, most would read as
    $$ \sum_{n = 2}^{\infty} \frac{ln(n!) \cdot ln(n)}{n^3}$$
  4. Mar 8, 2012 #3
    ln(n) is in the deonminator
  5. Mar 8, 2012 #4


    User Avatar
    Science Advisor
    Homework Helper

    The ratio test seems to work. Care to show us how far you got along in trying to simplify?
  6. Mar 9, 2012 #5
    [ln((n+1)!)]/[ln(n+1)*(n+1)^3] * [ln(n)*n^3]/[ln(n!)]

    now at infinity n^3/(n+1)^3 will give 1 & ln(n)/ln(n+1) will give 1 I guess
    but what about ln((n+1)!)/ln(n!) what is this going to give & how to prove...

    Can this be done by Direct Comparison Test or Limit Comparison Test??
  7. Mar 9, 2012 #6


    Staff: Mentor

    Then you need to write the expression with more grouping symbols, like so:
  8. Mar 9, 2012 #7
    The separation of ln(n!) will help me how ???
  9. Mar 9, 2012 #8


    Staff: Mentor

    It might or might not be helpful in the Ratio Test.
  10. Mar 10, 2012 #9
    I guess I found the solution

    fact is
    ##X! < X^x## so we say that by Direct Comparison Test $$\frac{ln(n!)}{n^3 *ln(n)} < \frac {ln(n^n)}{n^3 *ln(n)} = \frac {n*ln(n)}{n^3 *ln(n)} = \frac {1}{n^2 } $$

    since $$\frac {1}{n^2 } $$ is a convergent series of p=2, the $$Ʃ\frac{ln(n!)}{n^3 *ln(n)}$$ converges by Direct comparison test
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