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Improper integrals and Infinite Series

  1. Jun 6, 2010 #1
    Supose [tex] a_n=f(n) [/tex]
    What is the relationship between convergence or divergence of:

    [tex] \sum_{n=1}^\infty a_n[/tex] and [tex] \int_{1}^{\infty}f(x)dx[/tex] ?

    Besides the integral test (which only works for special cases).
     
    Last edited: Jun 6, 2010
  2. jcsd
  3. Jun 6, 2010 #2

    LCKurtz

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    There is no relation between the two, even for nonnegative an if you don't require f(x) to be decreasing. That is, you can have the series converge and the integral diverge or the series diverge and the integral converge.
     
  4. Jun 6, 2010 #3
    But why? What is the theory/intuition behind this?
     
  5. Jun 6, 2010 #4

    LCKurtz

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    I'm not sure what you mean by "why?". That's just the way it is.

    Have you tried to construct an example where the series converges but the integral doesn't. Or the other way around?
     
  6. Jun 6, 2010 #5
    [tex] \sum_{n=1}^\infty \frac {2-sin(n)}{n} [/tex] is convergent while

    [tex] \int_{n=1}^\infty \frac {2-sin(n)}{n}dn[/tex] is divergent.

    In vice versa I can only think about alternating function

    [tex] \int_{n=1}^\infty sin(n^2)dn [/tex] is convergent while

    [tex] \sum_{n=1}^\infty sin(n^2) [/tex] is divergent.

    But I can't think of positive function.
     
  7. Jun 6, 2010 #6

    LCKurtz

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    Man, you are making this much more difficult than it is.

    Is that obvious?

    Is that obvious? Is it even true?

    How about something easier like this -- suppose an= 0 for all n so the series obviously converges. Now can you build an f(x) such that f(n) = 0 for integers n but whose area under the curve goes to infinity? Once you do that then start thinking about the other way.
     
  8. Jun 7, 2010 #7

    LCKurtz

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    Hey Estro, I keep getting email notification that you have posted a new reply in this thread and when I click on the link to view it, it isn't there. I suppose you are deleting the posts later after you post them. Personally, I find that very annoying and it makes it very difficult to assist you or even to want to.
     
  9. Jun 7, 2010 #8
    I'm sorry, but I think I've posted in the wrong topic.
    I reformulated my question in a new topic.
    https://www.physicsforums.com/showthread.php?t=408533
     
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