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Math Help - Radius and Interval of Convergence

  1. #1
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    Radius and Interval of Convergence

    Determine the radius and interval of convergence of the series ∑(n=0, ∞) [n(x+2)^n]/[3^(n+1)]
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  2. #2
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    Hello, schnek!

    Determine the radius and interval of convergence of the series"

    . . . \sum^{\infty}_{n=0} \frac{n(x+2)^n}{3^{n+1}}
    Use the Ratio Test . . .

    R \;=\;\frac{a_{n+1}}{a_n} \;=\; \frac{(n+1)(x+2)^{n+1}}{3^{n+2}}\cdot \frac{3^{n+1}}{n(x+2)^n} \;=\;\frac{n+1}{n}\cdot\frac{(x+2)^{n+1}}{(x+2)^n}  \cdot\frac{3^{n+1}}{3^{n+2}}

    . . . =\frac{n+1}{n}\cdot(x+2)\cdot\frac{1}{3} \;=\;\frac{x+2}{3}\cdot\frac{n+1}{n}\;=\;\frac{x+2  }{3}\left(1 + \frac{1}{n}\right)

    Then: . \lim_{n\to\infty} |R| \;=\;\lim_{n\to\infty} \left|\frac{x+2}{3}\cdot\left(1 + \frac{1}{n}\right)\right| \;=\;\left|\frac{x+2}{3}\right|

    So we have: . \left|\frac{x+2}{3}\right| \:<\:1 \quad\Rightarrow\quad |x + 2| \:<\:3 \quad\hdots\quad \boxed{\text{ radius of convergence: 3 }}

    . . -3 \:<\:x + 2 \:<\:3 \quad\Rightarrow\quad -5 \:<\:x \:<\:1


    At x = \text{-}5, we have: . \sum^{\infty}_{n=0}\frac{n(\text{-}3)^n}{3^{n+1}} \;=\;\sum^{\infty}_{n=0}\frac{(\text{-}1)^n n}{3}\;=\;\tfrac{1}{3}\sum^{\infty}_{n=0}(\text{-}1)^n n . . . diverge

    At x = 1, we have: . \sum^{\infty}_{n=0}\frac{n\!\cdot\!3^n}{3^{n+1}} \;=\; \sum^{\infty}_{n=0}\frac{n}{3} \;=\;\tfrac{1}{3}\sum^{\infty}_{n=0}n . . . diverge


    . . \boxed{\text{ Interval of convergence: }\;(-5,\:1)\;}

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