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Math Help - Absolutely Converges, but why?

  1. #1
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    Absolutely Converges, but why?

    ((-1)ne1/n)/n3 converges absolutely apparently, but I don't see how. When I applied the ratio test, it came out to be one, but obviously that is incorrect. Here is what I did: | [((-1)n+1 e1/n+1)/(n+1)3] * [n3/(-1)n e1/n] | = | [n3/(n+1)3] * [e1/n+1/e1/n] | = 1, as n -> infinity
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  2. #2
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    Re: Absolutely Converges, but why?

    Its easier to apply a limit comparison first.

    \frac{e^{\frac{1}{n}}}{n^3} < \frac{3}{n^3}

    Do you see why the latter converges absolutely? If so, do you understand why the former has to as well?

    The ratio IS indeed 1. But this just means the test is inconclusive. The ratio test for the simplified series on the right side of my inequality would also give 1. They both converge absolutely.
    Last edited by SworD; April 2nd 2013 at 05:52 PM.
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  3. #3
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    Re: Absolutely Converges, but why?

    We can see that \displaystyle e^{\frac{1}{n}} \leq e for all \displaystyle n \geq 1 , so that means

    \displaystyle \begin{align*} \sum_{n = 1}^{\infty} \left| \frac{ (-1)^n \, e^{ \frac{1}{n} } }{n^3} \right| &= \sum_{n = 1}^{\infty} \frac{ e^{ \frac{1}{n} } }{ n^3 } \\ &\leq \sum_{n = 1}^{\infty} \frac{e}{n^3} \\ &= e \sum_{n = 1}^{\infty} \frac{1}{n^3} \end{align*}

    which is a convergent p-series. Since your absolute value series is less than a convergent series, your series is absolutely convergent.
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