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Thread: Some help/input on Limit Comparison Series Ques.

  1. #1
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    Some help/input on Limit Comparison Series Ques.

    $\displaystyle \sum_{n=3}^{\infty} \frac{2n+8}{[n ln(n)]^2 +4}$

    Can anyone give me some input as to what I could compare this to. At first glance, maybe I could do a limit comparison test with
    bn= $\displaystyle \frac{2n}{[n ln(n)]^2}$ , which would simplify to $\displaystyle \frac{2}{n [ln(n)]^2}$

    Since n is the dominant term, would I compare it with the harmonic series of $\displaystyle \frac{1}{n}$ ??

    Any feedback appreciated. Thanks.
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  2. #2
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    Re: Some help/input on Limit Comparison Series Ques.

    Anyone? I have no idea how to break down [n ln(n)] to find a suitable comparison.
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  3. #3
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    Re: Some help/input on Limit Comparison Series Ques.

    ln(n) is larger than 1 for all n> 1 so that "$\displaystyle [n ln(n)]^2+ 4> n^2+ 4$".
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    Re: Some help/input on Limit Comparison Series Ques.

    Quote Originally Posted by HallsofIvy View Post
    ln(n) is larger than 1 for all n> 1 so that "$\displaystyle [n ln(n)]^2+ 4> n^2+ 4$".
    If you replace ln(n) by 1, then the series diverges. On the other hand, $\displaystyle \sum\frac{1}{n(\ln(n))^2}$ converges by the integral test.
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    Re: Some help/input on Limit Comparison Series Ques.

    Thanks for the responses, the problem make much more sense now.
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  6. #6
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    Re: Some help/input on Limit Comparison Series Ques.

    Quote Originally Posted by Beevo View Post
    $\displaystyle \sum_{n=3}^{\infty} \frac{2n+8}{[n ln(n)]^2 +4}$
    Well I will the you that we do know that $\displaystyle \sum {\frac{1}{{n{{\left( {\ln (n)} \right)}^2}}}} $ does converge.

    But I am unsure of the comparison limit.
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