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Math Help - convergence problems

  1. #1
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    convergence problems

    Hey all I got two convergence problems I am stuck on

    1) Determine whether 1/(x((lnx)^2)) from 2 to infinity converges or diverges. I've thought about the integral test and the comparison test but I'm still stuck.

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    2) Determine whether (-1)^(n-1)*(n/((n^2)+1)) from 1 to infinity converges or diverges and if it converges what's its limit. I tried direct comparison but that broke down at n=4 so I'm not sure what to do know.

    Thanks for any help
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  2. #2
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    The integral test should work for the first...

    Note that \displaystyle \frac{1}{n\ln^2{n}} is a decreasing function for \displaystyle n \geq 2, so

    \displaystyle \sum_{n = 2}^{\infty}{\frac{1}{n\ln^2{n}}} \leq \int_2^{\infty}{\frac{1}{x\ln^2{x}}\,dx}.

    Let \displaystyle u = \ln{x} so that \displaystyle \frac{du}{dx} = \frac{1}{x}.

    When \displaystyle x = 2, u = \ln{2} and when \displaystyle x \to \infty, u \to \infty, so


    \displaystyle \int_2^{\infty}{\frac{1}{x\ln^2{x}}\,dx} = \int_{\ln{2}}^{\infty}{u^{-2}\,du}

    \displaystyle = \lim_{\epsilon \to \infty}\left[-u^{-1}\right]_{\ln{2}}^{\epsilon}

    \displaystyle = \lim_{\epsilon \to \infty}\left(\frac{1}{\ln{2}} - \frac{1}{\epsilon}\right)

    \displaystyle = \frac{1}{\ln{2}} - 0

    \displaystyle = \frac{1}{\ln{2}}.


    Therefore \displaystyle \sum_{n = 2}^{\infty}{\frac{1}{n\ln^2{n}}} \leq \frac{1}{\ln{2}} and thus the series is convergent.
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