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Math Help - Finding radius and interval of convergence

  1. #1
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    Finding radius and interval of convergence

    The problem: Find the radius and interval of convergence.
    \[\sum_{1}^{\infty}\frac{2^n(x-2^n)}{(n+2)!}\]

    The attempt:
    I basically did the ratio test and it equaled 0. So does R = \infty and I = (-\infty,\infty)?
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: Finding radius and interval of convergence

    Quote Originally Posted by Intrusion View Post
    The problem: Find the radius and interval of convergence.
    \[\sum_{1}^{\infty}\frac{2^n(x-2^n)}{(n+2)!}\]

    The attempt:
    I basically did the ratio test and it equaled 0. So does R = \infty and I = (-\infty,\infty)?
    If there are no errors, the expression can be written as...

    f(x)= x\ \sum_{n=1}^{\infty} \frac{2^{n}}{(n+2)!} - \sum_{n=1}^{\infty} \frac{4^{n}}{(n+2)!}= a x + b (1)

    ... so that the x doesn't play any role in convergence...

    Kind regards

    \chi \sigma
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  3. #3
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    Re: Finding radius and interval of convergence

    chisigma's point is that, the way you wrote it here, that is NOT a power series. I expect that you meant
    \sum_{n=0}^\infty\frac{2^n(x- 2)^n}{(n+2)!}

    Yes, the limit of the ratios is 0 which is less than 1 for all x. The "radius of convergence" is infinite.
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