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Trying to find interval of convergence for a geometric series

  1. Mar 6, 2012 #1
    1. The problem statement, all variables and given/known data

    here is the series:

    [itex]\sum^{\infty}_{n=0}[/itex]x(-15(x[itex]^{2}[/itex]))[itex]^{n}[/itex]

    2. Relevant equations



    3. The attempt at a solution

    I know that

    -1<-15x[itex]^{2}[/itex]<1 for convergence (because of geometric series properties)

    but I run into a problem here:

    -1/15<x[itex]^{2}[/itex]<1/15

    You can't take the square root of a negative number.. so is the interval of convergence just

    [0,(1/15)[itex]^{1/2}[/itex])?

    Thanks
     
    Last edited: Mar 6, 2012
  2. jcsd
  3. Mar 6, 2012 #2

    LCKurtz

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    No. The left side of that inequality is free so you just have ##x^2<1/15##. When you take the square root of both sides remember ##\sqrt{x^2}=|x|## so you get ##|x|<1/\sqrt{15}## which is ##-1/\sqrt{15}<x<1/\sqrt{15}##.
     
  4. Mar 8, 2012 #3
    OK Thanks!
     
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