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For which values of a does this series converge?

  1. May 17, 2010 #1
    [/PHP]1. The problem statement, all variables and given/known data

    For which values of a does this series converge?

    [tex]\sum[/tex] (n!)^2/(an)!

    3. The attempt at a solution

    I know a cannot be a negative integer because you cannot have a negative factorial.

    If a is 0, then it's limit is infinity. ie. lim (n!)^2/ 0 = infinity

    If a is +1, then lim cn+1/cn = ((n+1)!)^2/(n+1)! x n!/(n!)^2

    = lim (n+1)^2/(n+1) = lim (n+1)/1 = infinity

    For the series to converge it's limit from n to infinity must be equal to 0 right? So is there any value of a where it converges, or is my math wrong?
  2. jcsd
  3. May 17, 2010 #2
    If a = 0, 0! = 1
  4. May 17, 2010 #3
    thx. The limit is still infinity though.
  5. May 17, 2010 #4


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    Science Advisor

    Try using Stirling's formula. I think you'll see then that the series does converge as long as a is greater than some number.
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