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  1. #1
    Junior Member rednest's Avatar
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    Problem

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

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

    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 $\displaystyle x$ the following series converge and diverge.

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

    Any suggestions will be appreciated
    $\displaystyle \Sigma n!x^n$ 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 $\displaystyle x$ the following series converge and diverge.

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

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

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


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

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