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Math Help - Another radius of convergence problem

  1. #1
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    Another radius of convergence problem

    Again, I know I have to use the ratio test, but this just complicates things even further, it seems like I'm not left with anything to cancel out or anything or whatnot unless I'm just going about it wrong, could someone help me out?
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  2. #2
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    Are you familiar with the ratio test?. That is what we usually use to test convergences.

    Anyway:

    \sum_{n=1}^{\infty}\frac{(-1)^{n}\cdot x^{n+8}}{n+4}

    When we apply the ratio test, we can note that

    |(-1)^{n}|=|(-1)^{n+1}|=1

    \lim_{n\to \infty}\left|\frac{x^{n+9}}{n+5}\cdot\frac{n+4}{x^  {n+8}}\right|

    =|x|

    The ratio test implies the series converges if |x|<1 and diverges if

    |x|>1.

    The ratio test does not provide info for |x|=1 or |x|=-1, so we can test

    them separately by subbing into the original series.

    x=1:

    \sum_{n=1}^{\infty}\frac{(-1)^{n}}{n+4}=\frac{-1}{5}+\frac{1}{6}-\frac{1}{7}+...........

    This is a conditionally convergent harmonic series.

    For x=-1, we get:

    \sum_{n=1}^{\infty}\frac{(-1)^{2n+8}}{n+4}=\frac{1}{5}+\frac{1}{6}+\frac{1}{7  }+.......................

    This is a divergent harmonic series.

    Therefore, the interval of convergence for the series is (-1,1] and the radius of convergence is R=1.
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