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Math Help - Radius and interval of convergence

  1. #1
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    Radius and interval of convergence

    May someone check my work?

    \sum_{n}^{\infty}(n3^{-n})(x-2)^n

    To find R
    \lim_{n \rightarrow \infty} \left| \frac{n3^{-n}}{(n+1)3^{1-n}} \right| = \frac{1}{3}\lim_{n \rightarrow \infty} \left| \frac{n}{n+1} \right|=\frac{1}{3}


    To find the interval of convergence, C
    \left| x-2 \right|<\frac{1}{3} \Rightarrow \frac{5}{3}<x<\frac{7}{3}

    For x=\frac{5}{3}:
    \sum_{n}^{\infty}(n3^{-n})\left\(\frac{-1}{3}\right\)^n=\sum_{n}^{\infty}n(-1)^n\left\(\frac{1}{3}\right\)^{2n} which converges by alternating series test.

    For x=\frac{7}{3}:
    \sum_{n}^{\infty}(n3^{-n})\left\(\frac{1}{3}\right\)^n=\sum_{n}^{\infty} \frac{n}{9^n} which converges by the root test.

    So C= \frac{5}{3}\leq x \leq \frac{7}{3}\right]
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  2. #2
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    Re: Radius and interval of convergence

    Hi,
    No, you made an algebraic error. Given $a_n=n3^{-n}$, then $a_{n+1}=(n+1)3^{-n-1}$ and so the quotient
    $${a_n\over a_{n+1}}={n3^{-n}\over (n+1)3^{-n-1}}=3\,{n\over n+1}$$
    So your interval of convergence is $|x-2|<3$ or $-1<x<5$.

    Aside. When you find the interval of convergence of a power series, the ratio test or root test will always fail at the endpoints; the interval of convergence is found by one of these tests.
    Thanks from MadSoulz
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  3. #3
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    Re: Radius and interval of convergence

    Quote Originally Posted by johng View Post
    When you find the interval of convergence of a power series, the ratio test or root test will always fail at the endpoints; the interval of convergence is found by one of these tests.
    I'm not sure I understand what you're saying. If they always fail at the endpoints then why should we use them?
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  4. #4
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    Re: Radius and interval of convergence

    Hi again,
    Back to basics:

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