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Comparison Test

  1. Apr 18, 2009 #1
    http://img205.imageshack.us/img205/5117/summation.jpg [Broken]

    I think next step is: http://img205.imageshack.us/img205/7044/summation2.jpg [Broken]

    but the questions are: how & why?

    why is that the next step?
    how do I solve such problems?\how do I choose the second part?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Apr 18, 2009 #2
    For sure we are comparing so there is a "less than" sign..
    I`m just concerned on how to choose the terms on the left?
     
  4. Apr 18, 2009 #3

    Mark44

    Staff: Mentor

    For the series
    [tex]\sum_{n = 1}^\infty \frac{1}{2^n - 1}[/tex]

    the obvious choice would be a comparison with the convergent series
    [tex]\sum_{n = 1}^\infty \frac{1}{2^n}[/tex]
    Unfortunately, all of the terms in the first series are larger than those of the convergent series, so that comparison won't do any good.

    I have no idea what you're trying to do with this series:
    [tex]\sum_{n = 1}^\infty \frac{1}{2^n - 1^{n - 1}}[/tex]

    Do you know the limit comparison test?
     
  5. Apr 18, 2009 #4
    The teacher have just talked about it today,
    and he talked about the comparison test first..

    This problem, he solved it before talking about LCT..
    So, its solved by CT..
    Now why is it solved like that, or specifically, why is that second part chosen to be like that, the teacher himself said that he doesn't know!
    Well, he wasn't going to say it.. but someone asked (and I was going to ask) about that step.. then he said "well, its just that you choose it like that, this is how the problem is solved.."

    So I`m not here seeking for a final answer.. the question is: how do I choose the 2nd parts.. and why aren`t you, nor my teacher, able to know how to choose the 2nd term?
    (what are even the steps that I should take to put/know the second term?)
     
  6. Apr 18, 2009 #5

    Mark44

    Staff: Mentor

    OK, here's a convergent series you can use for comparison:
    [tex]\sum_{n = 1}^\infty \frac{1}{(3/2)^n}[/tex]

    2n - 1 > (3/2)n for n > 1
    so 1/(2n - 1) < 1/(3/2)n for n > 1

    Can you finish it?
     
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