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Radius of convergence of power series

  1. Jan 14, 2007 #1
    1. The problem statement, all variables and given/known data

    Find the radius of convergence of the following series.

    [tex]
    \sum\limits_{k = 1}^\infty {2^k z^{k!} }
    [/tex]


    2. Relevant equations

    The answer is given as R = 1 and the suggested method is to use the Cauchy-Hadamard criterion; [tex]R = \frac{1}{L},L = \lim \sup \left\{ {\left| {a_k } \right|^{\frac{1}{k}} } \right\}[/tex]

    3. The attempt at a solution

    I don't know where to begin. The sequence a_k in the Cauchy-Hadamard criterion is for series of the form [tex]\sum\limits_k^{} {a_k z^k } [/tex] but the series here has z raised to the power of k!, not just k. Substituting something for z (ie. set w = z^2 if the summation was over z^(2k)) doesn't work here. Can someone help me out? Thanks.
     
  2. jcsd
  3. Jan 14, 2007 #2

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    k! is a subsequence of n.

    If it helps that series can be restated as

    [tex]\sum_{n=1}^{\infty} a_nz^n[/tex]

    where a_n is given by:

    if n=k!, then a_n=2^k
    zero otherwise.


    Clearly, terms that are zero have no effect on the limsup, so all you need to work out is

    [tex]\lim\sup (a_n)^{1/n}[/tex]

    well, a_n is either zero, or if n=k!, then (a_n)^1/n = (2^k)^{1/k!}, so the lim sup is easy to work out (and is 1).
     
    Last edited: Jan 14, 2007
  4. Jan 14, 2007 #3
    Thanks matt, I didn't think of looking at the series like that. Perhaps I could have started by considering the sequence of partial sums and the answer might have dropped out. Anyway, thanks again.
     
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