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Finding the interval of convergence and the radius of a power series

  1. Jan 24, 2010 #1
    1. The problem statement, all variables and given/known data
    Find the radius and interval of convergence of the power series:
    [tex]\sum_{n=1}^{\infty} \frac{x^{2n-1}}{(n+1)\sqrt{n}}[/tex]


    2. Relevant equations
    ..


    3. The attempt at a solution
    My soltion:
    the ratio test will gives |x^2|=|x|^2
    it converges if |x|^2 < 1
    i.e. if |x|<1
    i.e. if -1<x<1
    But the problem here when i substitute x=1 and x=-1 it will give the same series
    Is this right?
    The resulting series converges by Integral Test.
    Radius = 1
     
  2. jcsd
  3. Jan 24, 2010 #2

    Mark44

    Staff: Mentor

    x = - 1 doesn't give the same series as x = 1. If x = -1, the series is
    [tex]\sum_{n=1}^{\infty} \frac{(-1)^{2n-1}}{(n+1)\sqrt{n}}[/tex]
    Since 2n - 1 is odd for all integers n, the numerator of the expression in the summation is always -1. If x = 1, the numerator is always +1.
     
  4. Jan 24, 2010 #3
    Oh...Thanks
    But will not affect the convergence right?
    since the series which obtained by multiplying a convergent series by a constant is still convergent.
    Right?
     
  5. Jan 24, 2010 #4

    Mark44

    Staff: Mentor

    Right.
     
  6. Jan 24, 2010 #5
    No No Sorry!
    there was a typo!
    its 2n-2 not 2n-1 in the series =)
    Sorry Again.
     
  7. Jan 24, 2010 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Ok, then it is the same series at x=1 and x=(-1). So you only have to check one of them. Is it convergent?
     
  8. Jan 24, 2010 #7
    Thanks you.
    you saved my life !
    It is easy for me.
    It converges by integral test.
     
  9. Jan 24, 2010 #8

    Dick

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    Science Advisor
    Homework Helper

    Sure. Or you could compare it to the convergent p-series 1/n^(3/2).
     
  10. Jan 24, 2010 #9
    Ohhh
    My bad
    How in earth I did not notice this :@
     
  11. Jan 24, 2010 #10

    Dick

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    Science Advisor
    Homework Helper

    I didn't say your way was bad. The integral test works fine.
     
    Last edited: Jan 24, 2010
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