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Math Help - Radius of convergence

  1. #1
    CowPizza
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    Radius of convergence

    Hey, I'm supposed to find out the interval of convergence for this power series:

    \sum_{n = 0}^\infty\frac{(-1)^n \ast (x-5)^n}{n5^n}

    Any thoughts?
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  2. #2
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    Quote Originally Posted by CowPizza
    Hey, I'm supposed to find out the interval of convergence for this power series:

    \sum_{n = 0}^\infty\frac{(-1)^n \ast (x-5)^n}{n5^n}

    Any thoughts?
    First of all n\not =0 thus maybe you mean \sum_{n = 1}^\infty\frac{(-1)^n \ast (x-5)^n}{n5^n}
    Well, I think it would be easier to consider,
    \sum^{\infty}_{k=1}\frac{(1-\frac{x}{5})^k}{k}, which is equivalent to the one you gave.
    Then let u=1-\frac{x}{5}, thus,
    \sum^{\infty}_{k=1}\frac{u^k}{k}.
    Notice that this is a power series, thus use the generalized ratio test, where a_k=\frac{1}{k}.
    Thus,
    \lim_{k\to\infty}\left|\frac{a_{k+1}}{a_k}\right|=  \left|\frac{k}{k+1}<br />
\right|=1, thus the radius of convergence is the reciprocal of this which is also 1.

    Let us now find the interval of convergence, since -1<u<1 the series is absolutely convergent we have that -1<1-\frac{x}{5}<1, thus,
    0<x<10.
    Notice when x=0 the series diverges because it forms a harmonic series and when x=10 the series converges conditionally because it form and alternating series. Thus,
    0<x\leq 10 is the interval of convergence.
    Q.E.D.
    Last edited by ThePerfectHacker; February 25th 2006 at 04:41 PM.
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