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Homework Help: Power Series

  1. Apr 6, 2010 #1
    Im given the following power series:
    [tex]\sum[/tex] (x-3)^n/ n

    I determined that the radius of convergence is R=1 and the interval of convergence is [2, 4)

    They ask what values of x for which series converges absolutely?
    and values of x for which series converges conditionally?

    From what i read, my series should converge absolutely at |x-3|<1 (am i right?)

    Im not sure about values of x where the series converges conditionally.
     
  2. jcsd
  3. Apr 6, 2010 #2

    LCKurtz

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    Check the end points, x = 2 and 4.
     
  4. Apr 6, 2010 #3
    yeah i did.
    it converged on x=2
    but diverged on x=4

    so is that what they mean by absolute and conditionally?
     
  5. Apr 6, 2010 #4

    LCKurtz

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    A series is conditionally convergent if is convergent but not absolutely convergent. So at x = 2, since it converges, you have to also check whether it converges absolutely. If it doesn't, then it is conditionally convergent.
     
  6. Apr 6, 2010 #5
    I did the algebra and it came out that x=2 does converge abs.

    so there are no other values that converge absolutely? what about in between the interval? do i have to check those too?

    all other values outside the interval would diverge the series correct?
     
  7. Apr 6, 2010 #6

    LCKurtz

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    We don't say that "x=2 does converge". We say "the series converges for x=2". But you better check again the "absolutely" part for x = 2.
    I thought you had already understood that you have absolute convergence inside the radius of convergence. And, of course, divergence beyond it.
     
  8. Apr 6, 2010 #7
    well what i did to check the absolute convergence for x= 2 is I used the ratio test for:
    E (x-3)^n/ n

    And what I did was I just plugged in 2 for the x value and it came out to be -1 when I took the limit. And according to Ratio Test, <1 means abs conv.
     
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