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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


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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


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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


    Staff: Mentor

    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


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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


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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


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