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Showing that a series diverges

  1. Nov 13, 2009 #1
    1. The problem statement, all variables and given/known data

    Show that the series [tex]\sum_{0}^{\infty}(2n)!/(n!)^2*(1/4)^n[/tex] diverges

    2. Relevant equations

    I don't know which convergence test to use

    3. The attempt at a solution

    I don't have one, because I don't know which convergence test to use. If someone can tell me what to use, I will be able to figure out this problem.
     
  2. jcsd
  3. Nov 13, 2009 #2

    LCKurtz

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    Why don't you just try the most likely one?
     
  4. Nov 13, 2009 #3
    I know that the basic ones such as ratio, divergence, alternating series, comparison, and integral don't work with this series. I believe it uses a test that I haven't learned about, so I was wondering what that could be.
     
  5. Nov 13, 2009 #4

    LCKurtz

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    I am just leaving for the evening so I don't have time to work on it myself. But I would be very surprised if the ratio test won't settle it. Did you try the root test?

    I will check back later.
     
  6. Nov 13, 2009 #5

    Mark44

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    No, one of the tests you mentioned will work. It's not an alternating series, so you can ignore that test. You wouldn't want to apply the integral test on this series, I don't think, so that eliminates that test.

    Show us what you've done...
     
  7. Nov 14, 2009 #6

    LCKurtz

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    Clarification: Is that [itex](1/4)^n[/itex] in the numerator or the denominator of the fraction? I'm guessing the numerator, making the [itex]4^n[/itex] in the denominator, which makes the problem tougher than I thought at first glance. Is that right? I'm getting the ratio test fails too...
     
  8. Nov 14, 2009 #7
    It's just the first fraction times (1/4)n. I guess there should be two parenthesis around the first fraction so it's just that quantity multiplied by the (1/4)n
     
  9. Nov 14, 2009 #8

    jgens

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    My initial response was deleted by the moderators, but you can prove that the sum diverges by considering the sequence [itex]a_n = 1/n[/itex].
     
    Last edited: Nov 14, 2009
  10. Nov 14, 2009 #9

    Dick

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    You can also learn things about series like this using Stirling's approximation.
     
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