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Thread: Series, is this reasonable?

  1. #1
    Senior Member Twig's Avatar
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    Series, is this reasonable?

    Hi!

    $\displaystyle \sum_{k=3}^{\infty} ln(cos(\frac{\pi}{k})) $ , convergent or divergent?

    I tried:

    $\displaystyle a_{k}=ln(1+(cos(\frac{\pi}{k}) -1 ) $

    Now using the Taylor expansion of $\displaystyle ln(1+x) = x-\frac{x^2}{2}+...$ with $\displaystyle x=(cos(\frac{\pi}{k})-1) $ AND using the taylor expansion of $\displaystyle cos(x)=1-\frac{x^2}{2}+... $ I got

    $\displaystyle a_{k}=\frac{-\pi^{2}}{2k^{2}}+O(\frac{1}{k^{4}}) $

    To be able to use the comparison critera, $\displaystyle \lim_{k\to \infty} \frac{a_{k}}{b_{k}} = A > 0 $ , I applied it with $\displaystyle b_{k}=\frac{1}{k^{2}} $ on $\displaystyle - \sum_{k=3}^{\infty} ln(cos(\frac{\pi}{k})) $

    And hence $\displaystyle a_{k}=\frac{\pi^{2}}{2k^{2}}-O(\frac{1}{k^{4}}) $

    And $\displaystyle \lim_{k\to \infty} \frac{a_{k}}{b_{k}} = \frac{\pi^2}{2}>0 $

    And $\displaystyle \sum_{k=1}^{\infty} \frac{1}{k^2} $ converges.

    So the original series converges.

    I assume if $\displaystyle \sum_{k=3}^{\infty} ln(cos(\frac{\pi}{k})) =s $ then $\displaystyle - \sum_{k=3}^{\infty} ln(cos(\frac{\pi}{k})) =-s $ ?

    Basically what I want, is someone to check the work.

    Thx!
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  2. #2
    MHF Contributor matheagle's Avatar
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    I ignored your Taylor work, but I tested your $\displaystyle 1/n^2$ idea

    I just tried the limit comp test with $\displaystyle \sum{\pi^2\over n^2}<\infty$

    $\displaystyle {\ln\cos (\pi/n)\over{\pi^2\over n^2}}\to -.5$ by letting $\displaystyle w=\pi/n\to 0^+$

    By that, I mean $\displaystyle \lim_{w\to 0^+} {\ln\cos w\over w^2} = \lim_{w\to 0^+} {-\tan w\over 2w} =-.5$

    So, $\displaystyle -\sum \ln\cos (\pi/n)$ is a positive series and it can be compared to the p=2 series.
    Last edited by matheagle; Aug 11th 2009 at 11:18 PM.
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