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Math Help - Alternating Series

  1. #1
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    Alternating Series

    Determine the convergence or divergence of the series.
    (E = sigma)

    infinity
    E (-1)^n+1 (2) / e^n - e^-n
    n=1
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  2. #2
    MHF Contributor kalagota's Avatar
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    i assume it is \frac{2}{e^n - e^{-n}}..

    notice that it's a sequence of hyperbolic cosecant.. \frac{2}{e^n - e^{-n}} = \frac{1}{sinh \,\, n} = csch \,\, n

    and from n=1 to infinity, that is monotone and bounded..
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by element View Post
    Determine the convergence or divergence of the series.
    (E = sigma)

    infinity
    E (-1)^n+1 (2) / e^n - e^-n
    n=1

    Lets put this in math notation:

     <br />
\sum_{n=1}^{\infty} (-1)^{n+1} \frac{2}{e^n-e^{-n}}<br />

    Well for large n, \frac{2}{e^n-e^{-n}} is strictly decreasing and goes to zero, and so the alternating series test applies and your series is convergent.

    RonL
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  4. #4
    MHF Contributor kalagota's Avatar
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    yeah, i forgot to mention that \lim_{n\rightarrow \infty} csch \,\, n = 0
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