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Determine the values of x for series convergence

  1. Nov 5, 2013 #1
    [itex][/itex]1. The problem statement, all variables and given/known data

    Determine the values of x for which the following series converges. Remember to test the end points of the interval of convergence.
    [itex]^{∞}_{n=0}\sum\frac{(1-)^{n+1}(x+4)^{n}}{n}[/itex]

    2. Relevant equations

    I worked it down to
    |x+4|<1
    ∴-5<x<-3

    3. The attempt at a solution
    When I come to test the end point when x=-3

    Ʃ=[itex]\frac{(-1)^{n+1}(-1)^{n}}{n}[/itex]
    ALTERNATING SERIES
    Test 1. See if [itex]lim_{n→∞}a_{n}[/itex]=0

    [itex]lim_{n→∞}[/itex][itex]\frac{(1)^{n}}{n}[/itex]
    =[itex]lim_{n→∞}[/itex][itex]\frac{(1)}{n}[/itex] =0

    test 2. See if [itex]a_{n+1}<a_{n}[/itex]

    [itex]a_{n+1}= \frac{1}{n+1}<\frac{1}{n}[/itex]
    ∴[itex]a_{n+1}<a_{n}[/itex]

    hence, series converges when x=-3


    When x=-5
    Ʃ=[itex]\frac{(-1)^{n+1}(-1)^{n}}{n}[/itex]
    Test 1 see if [itex]lim_{n→∞}a_{n}[/itex]=0\

    [itex]lim_{n→∞}[/itex][itex]\frac{(-1)^{n}}{n}[/itex]
    =[itex]lim_{n→∞}[/itex][itex]\frac{\frac{(-1)^{n}}{n}}{1}[/itex] =0


    Test 2. [itex]a_{n+1}<a_{n}[/itex]
    [itex]a_{n+1}[/itex]=[itex]\frac{-(-1)^{n}}{n+1}<\frac{(-1)^{n}}{n+1}[/itex] (since negative of the other?

    Hence, series converges when x=-5

    ∴ series converges when -5≤x≤-3

    Not sure if I have the right answer. but I don't know what to do when I get [itex]a_{n}[/itex] has the [itex](-1)^{n}[/itex]
     
    Last edited: Nov 5, 2013
  2. jcsd
  3. Nov 5, 2013 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    (-3+ 4)= 1, not -1. You should not have the [itex](-1)^n[/itex] term.
    However, that means the series is an alternating series so what you have is correct. See below for the x= -5 case.

    [itex](-1)^{n+1}(-1)^n= (-1)^{2n+1}= (-1)^{2n}(-1)= -1[/itex]
    The sum is just [itex]-\sum\frac{1}{n}[/itex] which does NOT converge

     
  4. Nov 5, 2013 #3
    mate, you're a legend!
     
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