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Radius of Convergence Power Series

  1. Apr 2, 2013 #1
    1. The problem statement, all variables and given/known data

    Determine the radius of convergence and the interval of convergence og the folling power series:

    n=0 to infinity
    Ʃ=[itex]\frac{(2x-3)^{n}}{ln(2n+3)}[/itex]

    2. Relevant equations

    Ratio Test

    3. The attempt at a solution
    Well I started with the ratio test but I have no clue where to go with it

    [itex]\frac{(2x-3)^{n+1}}{ln(2(n+1)+3)}[/itex] * [itex]\frac{ln(2n+3)}{(2x-3)^{n}}[/itex]

    I know that I will be left with the 2x-3 and I can pull that out in front of the limit to use when doing the interval of convergence, but I am lost with the natural log. should I distribute the two and make it

    [itex]\frac{(2x-3)^{n+1}}{ln(2n+5)}[/itex] * [itex]\frac{ln(2n+3)}{(2x-3)^{n}}[/itex]

    2x-3 Lim as n approaches infinity of : ln(2n+3)-ln(2n+5)??

    Any help would be appreciated
     
  2. jcsd
  3. Apr 2, 2013 #2

    Zondrina

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    Homework Helper

    The limit of ##\frac{ln(2n+3)}{ln(2n+5)}## as n goes to infinity is just 1.

    Can you continue?

    EDIT : It might be informative to wrap your limit in an absolute value.
     
    Last edited: Apr 2, 2013
  4. Apr 2, 2013 #3
    Thanks,

    let me solve and see what I get
     
  5. Apr 2, 2013 #4
    So I can now set up the interval of convergence to be

    -1 < 2x-3 < 1
    1 < x < 2

    so any number between 1 and two will cause the series to converge

    now i have to check endpoints seperately

    x:1 Ʃ=[itex]\frac{(2x-3)^{n}}{ln(2n+3)}[/itex]
    [itex]\frac{(-1)^{n}}{ln(2n+3)}[/itex] which is an alternating series

    the limit is 0 and it is also decreasing

    X:2 Ʃ=[itex]\frac{(2x-3)^{n}}{ln(2n+3)}[/itex]
    [itex]\frac{(1)^{n}}{ln(2n+3)}[/itex]

    this limit is also 0

    so the interval is [1,2]?

    Radius of convergence would be [itex]\frac{1}{2}[/itex]?

    EDIT: when x=2 I feel like I have to do something different for solving for the limit? Or is it correct?
     
  6. Apr 2, 2013 #5

    Zondrina

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    Your radius of convergence is what you get from the calculation of ##lim_{n→∞} |\frac{a_{n+1}}{a_n}| = L##.

    If L is finite, then your radius of convergence is 1/L.
    If L is 0, then your radius of convergence is ∞.
    If L = ∞, then your radius of convergence is 0.

    Since you got 1 as the limit, the radius is R = 1/1 = 1. So your series will converge for |2x-3| < 1, aka 1 < x < 2. So you're doing good.

    So now to get a concrete interval of convergence, you need to check the endpoints.

    At x = 1, it looks like your series is conditionally convergent.

    At x = 2, it appears your series diverges. Could you tell me why?

    What's your interval now?
     
    Last edited: Apr 2, 2013
  7. Apr 2, 2013 #6
    At x = 2, I don't know why it is divergent. I thought it was convergent to zero. If it isnt convergent the interval would become [1,2)
     
  8. Apr 2, 2013 #7

    Zondrina

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    Okay good about the interval. In the case of x=2, can you compare your series to anything? Think about minimizing the denominator.
     
  9. Apr 2, 2013 #8
    Basically we have

    [itex]\frac{1}{ln(2n+3)}[/itex]

    The limit of ln(2n+3) is ∞. [itex]\frac{1}{∞}[/itex] = 0

    is there something I am missing?
     
  10. Apr 2, 2013 #9

    Zondrina

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    You're missing the whole picture.

    Why are you taking limits? All you have to do is check the convergence of your series at the endpoints.

    So when you did x = 1. You saw ##\sum |a_n|## diverged, but ##\sum (-1)^na_n## converges by the alternating series test, so it was conditionally convergent.

    Now when x = 2. You don't need the A.S.T. It's quite literally a simple comparison.
     
    Last edited: Apr 2, 2013
  11. Apr 2, 2013 #10
    How about:

    If I use the comparison test for
    [itex]\frac{1}{ln(2n+3)}[/itex] and compare with [itex]\frac{1}{n}[/itex]

    [itex]\frac{1}{n}[/itex] is a divergent p series
    using the comparison test, [itex]\frac{1}{ln(2n+3)}[/itex] is also divergent???
     
  12. Apr 2, 2013 #11

    Zondrina

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    Yes that's the idea.
     
  13. Apr 2, 2013 #12
    Thank you Zondrina!
     
  14. Apr 7, 2013 #13
    My professor gave me a 4 out of 10 possible points....he passed out the solutions to the problems and oddly enough I have the same answer.
     
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