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Trying to find the radius of convergence of this complicated infinite series

  1. Mar 9, 2012 #1
    1. The problem statement, all variables and given/known data

    k is a positive integer.

    [itex]\sum^{\infty}_{n=0}[/itex] [itex]\frac{(n!)^{k+2}*x^{n}}{((k+2)n)!}[/itex]

    2. Relevant equations



    3. The attempt at a solution

    I have no idea.. this is too confusing. I tried the ratio test (which is the only way I know how to deal with factorials) but I get stuck at the following

    lim n->[itex]\infty[/itex] of | x(n+1)[itex]^{k+2}[/itex][itex]\frac{[(k+2)n]!}{[(k+2)(n+1)]!}[/itex] |

    I can't seem to find a way to cancel out the factorials in the fractional portion of that
     
  2. jcsd
  3. Mar 9, 2012 #2
    Here's a thought. k is just some positive integer, right? The issue of convergence of a series arises in the "tail" of the series - that is, if you want to look at the convergence of [itex]\sum_{i=0}^{\infty}\ a_n x^n[/itex], you could just as easily look at the convergence of [itex]\sum_{i=N}^{\infty}\ a_n x^n[/itex]. With that in mind, try splitting the series into n < k and n > k to simplify things slightly. Also you can drop the absolute values around those factorials, since they're positive.
     
  4. Mar 9, 2012 #3

    Dick

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    Try and focus on the problem by doing the special case k=0. Then try k=1. Can you generalize?
     
  5. Mar 9, 2012 #4
    When I cancel factorials, I usually expand the factorial into maybe three or more factors, e.g. [tex]n!=n(n-1)(n-2)...[/tex] and [itex](n-1)!=(n-1)(n-2)...[/itex]

    Then it is clear that [itex]\frac{n!}{(n-1)!}=n[/itex]
     
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