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Math Help - Problem

  1. #1
    Junior Member rednest's Avatar
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    Problem

    Determine for which values of x the following series converge and diverge.

    \Sigma n!x^n and \Sigma \frac{(2x)^n}{n}

    Any suggestions will be appreciated
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  2. #2
    Junior Member
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    Quote Originally Posted by rednest View Post
    Determine for which values of x the following series converge and diverge.

    \Sigma n!x^n and \Sigma \frac{(2x)^n}{n}

    Any suggestions will be appreciated
    \Sigma n!x^n is biger
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by rednest View Post
    Determine for which values of x the following series converge and diverge.

    \Sigma n!x^n and \Sigma \frac{(2x)^n}{n}

    Any suggestions will be appreciated
    \sum_{n=0}^{\infty}n!x^n

    You can either apply the Root or Ratio test to find that the interval of covnergence is \{0\}, or you can use the fact that \forall{x}\in\mathbb{R}-\{0\}~\lim_{n\to\infty}n!x^n=\infty


    \sum_{n=1}^{\infty}\frac{2^nx^n}{n}

    Once again the Root, Ratio, or integral tests would work. Or you could note that \sum\frac{2^n}{n} is divergent and the only thing that will make it converge in this case is a "dominant" geometric series provided by x^n, so \lim_{n\to\infty}2^nx^n=0\implies{x\in\left(\frac{-1}{2},\frac{1}{2}\right)}
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