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Homework Help: Convergent Series

  1. May 4, 2010 #1
    1. The problem statement, all variables and given/known data
    ∑ ln((n)/(n+1)) I was assuming this would be [tex]\infty[/tex]/[tex]\infty[/tex]
    and if I divide through by n it gives me 1/1 or 1 so would this just be divergent?

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 4, 2010 #2
    So is this a specific test you're trying to use?

    The limit goes to 0. Factor out an n from n/(n+1) to get [n(1)]/[n(1 + 1/n)].

    Cancel the n to get 1/[1+ 1/n]. When you apply the limit, you should get ln(1/1) which just equals 0.

    The only problem with this is I'm not sure if you have to change the n's to x's to make it a continuous function in order to cancel the n's, but either way that's how you get the answer.
     
  4. May 4, 2010 #3

    Mark44

    Staff: Mentor

    [itex]\infty/\infty[/itex] is meaningless as a final answer to anything (it is called indeterminate), and you're ignoring the fact that the general term in your series is ln(n/(n + 1)). As n gets large, n/(n + 1) --> 1, so your general term --> 0. This tells you precisely nothing about your series, so you're going to need to do something else. What other tests do you know?
     
  5. May 4, 2010 #4
    Hey Mark, do you have to make it a continuous function to cancel the n's after factoring them out, the way I did it?
     
  6. May 4, 2010 #5
    OK so I'm trying to use the Integral test and I've gotten to lim[tex]\int[/tex]ln(x/x+1)dx but now I'm not sure how to proceed. I was wondering if I could divide this into two separate integrals. I'm a little confused.
     
  7. May 4, 2010 #6

    Mark44

    Staff: Mentor

    I'm not sure I understand your question, but I'll take a stab at it. lim [ln(f(x))] = ln[lim (f(x))] as long as f is continuous. In this case f(x) = x/(x + 1), which is continuous for x > - 1. For the series in this problem, it's not shown, but I suspect that n ranges from 1 to infinity.
     
  8. May 4, 2010 #7
    Apply one of the logarithmic laws, and you will face a telescoping series
     
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