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Trying to find the interval of convergence of this series: run into a problem

  1. Mar 8, 2012 #1
    1. The problem statement, all variables and given/known data

    The series:

    [itex]\sum^{n=\infty}_{n=0}[/itex] [itex]\frac{(-1)^{n+2}(x^{3}+8)^{n+1}}{n+1}[/itex]

    2. Relevant equations



    3. The attempt at a solution

    using the ratio test, I get the following:

    |x[itex]^{3}+8|[/itex]<1, but I know that the radius of convergence must be in the form:
    |x-a|<b, where the degree of x is 1. I can't find a way to get the degree of x to 1! Can someone help me? Thanks
     
  2. jcsd
  3. Mar 8, 2012 #2

    LCKurtz

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    That form is for a power series in ##x##. You have a power series in ##x^3+8##. Just figure out what ##x## give ##|x^3+8|<1##, whether or not you get an interval.
     
  4. Mar 8, 2012 #3
    When I solve for x I get that the interval of convergence is (-3,(-7)[itex]^{1/3}[/itex]). So in order to solve |x[itex]^{3}[/itex]+8|<1, x must fall in that interval. But what I don't get is how to get the radius of convergence from this..
     
  5. Mar 8, 2012 #4

    LCKurtz

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    ##(-9)^{-\frac 1 3}## isn't -3. If ##a## is the center of your interval you should be able to express it in the form ##|x-a|<b##. But I'm not sure whether or not I would call ##b## a radius of convergence in this type of problem. You might not even get an interval. For example a similar problem with squares instead of cubes might come out something like ##4 < x^2 < 9## giving ##-3<x<-2## and ##2 < x < 3##. Then you don't have an "interval of convergence".
     
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