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Math Help - Limit Comparison test

  1. #1
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    Limit Comparison test

    This was on my test today and when we were done me and another guy had a discussion about it. We both did it differently so I was to see who did it correctly.

    use limit comparison to find whether the series converges or diverges

     \sum_{k=1}^{\infty}\frac{4}{7+4nln^2n}
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Limit Comparison test

    Quote Originally Posted by delgeezee View Post
     \sum_{k=1}^{\infty}\frac{4}{7+4nln^2n}
    I suppose you mean  \sum_{n=2}^{\infty}\frac{4}{7+4n\ln^2n} . The series  \sum_{n=2}^{\infty}\frac{1}{n\ln^2n} is convergent (use the integral test) and  \lim_{n\to \infty}\left(\frac{4}{7+4n\ln^2n}:\frac{1}{n\ln^2n  }\right)=4\neq 0, then the given series is convergent.
    Last edited by FernandoRevilla; March 13th 2012 at 09:21 PM.
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