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Radius of convergence

  1. May 30, 2009 #1
    Hey ,
    I was wondering if anyone could help me out with this question regarding calculating the radius of convergence of the infinity series of (1/n!)x^(n!)

    This is my work

    First we consider when abs(x) < 1
    then we have 0 <= abs(x^n!) <= abs(x^n)
    so we know that the series converges whenever abs(x) < 1 , so by the comparison test, we conclude that the series converges for abs(x) < 1

    Now consider abs(x) > 1
    we have 0<= abs (x) <= abs(x^n!)
    so the series by comparison test diverges for abs(x) > 1

    After that , I consider what happen at x=1 and x=-1 , it turned out that they converges as well .
    So my radius of convergence is [-1,1] and the series converges for abs(x) <= 1

    is this approach correct ? because i thought that for question like this , ratio test is not a good approach.
  2. jcsd
  3. May 30, 2009 #2


    Staff: Mentor

    I don't see anything wrong, except that your radius of convergence, R, is 1. The interval of convergence is [-1, 1], and the radius of this interval is 1.
  4. May 31, 2009 #3
    Alright thanks a lot =D .
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