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Math Help - Series, is this reasonable?

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

    Hi!

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

    I tried:

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

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

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

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

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

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

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

    So the original series converges.

    I assume if \sum_{k=3}^{\infty} ln(cos(\frac{\pi}{k})) =s then - \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 1/n^2 idea

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

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

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

    So, -\sum \ln\cos (\pi/n) is a positive series and it can be compared to the p=2 series.
    Last edited by matheagle; August 12th 2009 at 12:18 AM.
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