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Math Help - convergence interval

  1. #1
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    convergence interval

    Find a convergence interval for the following power series
    ((-1)^n(x-2)^n)/3^2nwith the sigma symbol with infinity above and n=0 below.
    (I appologize about the notation)
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  2. #2
    MHF Contributor Calculus26's Avatar
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    In General

    For (-1)^n cn (x-a)^n

    The ratio test yields |x-a|lim{c(n+1)/cn] < 1 where (n+1) is a subscript

    |x-a| < lm{cn/c(n+1}

    lm{cn/c(n+1} = R is called the radius of convergence

    This guarantees convergence on the interval

    |x-a| < R 0r (a-R < x < a+R)

    Then check x =a-R and a +R in the original series to finish the interval of convergence

    In your case cn = 1/3^(2n) ????? or 1/9n ?

    If 1/3^(2n) Then R = 9 you have convergence on (-7,11) check the endpoints
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  3. #3
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    Hello, mathatlast!

    Find the convergence interval for: . \sum^{\infty}_{n=0}\,(\text{-}1)^n\frac{(x-2)^n}{3^{2n}}

    R \;=\;\left|\frac{a_{n+1}}{a_n}\right| \;=\;\left|\frac{(x-2)^{n+1}}{3^{2n+2}} \cdot\frac{3^{2n}}{(x-2)^n}\right| \;=\;\left|\frac{3^{2n}}{3^{2n+2}}\cdot \frac{(x-2)^{n+1}}{(x-2)^n}\right| \;=\;\left|\frac{1}{9}\,(x-2)\right|

    Then: . \left|\frac{x-2}{9}\right| \:< \:1 \quad\Rightarrow\quad -1 \:<\:\frac{x-2}{9} \:<\:1 \quad\Rightarrow\quad -9 \:<\: x-2 \:<\:9


    . . Therefore: . -7 \:<\:x \:<\:11


    Um ... too slow again.
    .
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  4. #4
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    Thanks- I get it all, with the exception of where the 1/9 came about from 3^2n/3^2n+1....at least that is where I think the 1/9 came from. Can you explain it a little? I am sorry if it is obvious, but it has been a long day into night and my brain is melting!! Thanks
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  5. #5
    MHF Contributor Calculus26's Avatar
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    cn = 1/3^(2n)

    if you replace n with n+1 1/3^(2(n+1) = 1/3^(2n+2) = 1/3^(2n)3^2
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