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Math Help - Test for convergence

  1. #1
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    Test for convergence

    (1) \Sigma(1 to infinity) (-1)^{n+1} \displaystyle \frac{1-cos(1/n)}{sin(1/n)}



    (2) ) \Sigma(2 to infinity) (-1)^n8^{-n/2}(\displaystyle \frac{n+1}{n})^{n^2}

    (3) \Sigma(2 to infinity) \displaystyle \frac{(-1)^n}{nlnn(lnlnn)^2}

    knowing 3 shud use the cauchy condensation test, how do i apply it, and what test shud i use for the other two?

    need to specify if it's absolutely convergent
    Last edited by wopashui; February 26th 2011 at 01:31 PM.
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  2. #2
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by wopashui View Post
    (1) \Sigma(1 to infinity) (-1)^{n+1} \displaystyle \frac{1-cos(1/n)}{sin(1/n)}
    Remembering the identity \displaystyle \tan \frac{\theta}{2} = \frac{1-\cos \theta}{\sin \theta} we 'discover' that the general term is \displaystyle (-1)^{n+1}\ \tan \frac{1}{2n} so that the series converges for the Leibnitz's rule...

    Kind regards

    \chi \sigma
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