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Math Help - [SOLVED] Possibly simple series question

  1. #1
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    [SOLVED] Possibly simple series question




    code: \sum_{n=1}^{\infty}\frac{(-1)^{n}}{\log (e^{n}+e^{-n})}

    (Evaluate)
    Last edited by CaptainBlack; January 15th 2010 at 02:00 AM.
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  2. #2
    Senior Member Shanks's Avatar
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    It is a alternating series and \frac{1}{log(e^n+e^{-n})}\to 0\text{ as }n\to \infty, thus converges.
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    Thanks, good point. What about absolutely?
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  4. #4
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    Quote Originally Posted by student1 View Post



    code: \sum_{n=1}^{\infty}\frac{(-1)^{n}}{\log (e^{n}+e^{-n})}

    (Evaluate)
    It would appear that the stuff in red is NOT what you wanted to do.

    It would be a big help if you actually said clearly and correctly what you did want to do with this series. eg. Find whether it converges or diverges. etc. We are not mind readers.
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    Quote Originally Posted by student1 View Post
    Thanks, good point. What about absolutely?

    \frac{1}{\ln(e^n+e^{-n})}\geq \frac{1}{\ln(2e^n)}=\frac{1}{n+\ln 2} and thus the series doesn't converge absolutely.

    Tonio
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