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Math Help - Power Series

  1. #1
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    Power Series

    How would you find the series radius and interval of convergence? and for what value of x does the series converge absolutely, conditionally?

    series n=1 to infinity ((-1)^n(x-2)^n)/n
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  2. #2
    MHF Contributor chisigma's Avatar
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    The series...

    \sum_{n=1}^{\infty}(-1)^{n}\cdot \frac{(x-2)^{n}}{n}

    ... converges absolutely if |x-2|<1, so that the circle of convergence has center in x=2 and radius r=1...

    Kind regards

    \chi \sigma
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  3. #3
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    Use the ratio test: A series, \sum a_n, converges absolutely if the limit of the ratio [tex]|\frac{a_{n+1}}{a_n}|[/itex] is less than 1. For a power series, \sum b_n(x-a)^n that ratio becomes |\frac{b_{n+1}(x-a)^{n+1}}{b_n(x-a)^n}|= |\frac{b_{n+1}}{b_n}||x- a|. The "interval of convergence" is the interval in which x makes that less than 1.

    In your example, b_n(x-a)^n= \frac{(-1)^n}{n}(x-2)^n and b_{n+1}(x-a)^{n+1}= \frac{(-1)^{n+1}}{n+1}(x-2)^{n+1} so that ratio is |\frac{(-1)^{n+1}(x-2)^{n+1}}{n+1}\frac{n}{(-1)^n(x-2)^n|= \frac{n+1}{n}|x-2|.

    As n goes to infinity, [tex]\frac{n+1}{n}[tex] goes to 1 so that ratio goes to |x-2|. The series will converge if |x-2|< 1 or if -1< x- 2< 1 so that 1< x< 3. The radius of convergence is 1 and the interval of convergence is from 2-1= 1 to 2+1= 3.

    That is on the real line. If x is allowed to be a complex number, then the "interval of convergence" is a disk in the complex plane with center 2 and radius 1. That is what Chisigma was referring to.
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  4. #4
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    Don't you have to test the end points for convergence too though since the interval of absolute convergence is finite? The end points for this problem are geometric right?
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  5. #5
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    provided that 1 and 3 are correct (I didn't check any work, just looked at the post above), then yeah, you need to check the endpoints, so..


    ((-1)^n(x-2)^n)/n

    plugging in 1..

    [(-1)^n][(-1)^n]/n

    or,

    1/n
    which diverges by the p-series test

    and plugging in 3..

    [(-1)^n][(1)^n]/n

    or

    [(-1)^n]/n

    which converges by the alternating series test
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