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Thread: Partial Sums

  1. #1
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    Partial Sums

    http://math.rutgers.edu/~greenfie/we...fstuff/w4H.pdf

    I know for a), I can show it converges with integral test, but how can I find a specific partial sum?

    Do I just take the integral from 2 to n of 1/(x*(ln x)^2) and set it <10^-5 so that I get 5 digit accuracy, and then solve for N?

    For b), I know how to show divergence, but how do I find the specific partial sum? Same idea?
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    I don't know exactly what resources you can use but for example if an alternating series $\displaystyle u_1-u_2+u_3-\ldots$ satisfies the hypothesis of the Leibnitz's rule test then, $\displaystyle |S_n-S|\leq u_{n+1}$ .
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  3. #3
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by twittytwitter View Post
    http://math.rutgers.edu/~greenfie/we...fstuff/w4H.pdf

    I know for a), I can show it converges with integral test, but how can I find a specific partial sum?

    Do I just take the integral from 2 to n of 1/(x*(ln x)^2) and set it <10^-5 so that I get 5 digit accuracy, and then solve for N?

    For b), I know how to show divergence, but how do I find the specific partial sum? Same idea?
    a) the series...

    $\displaystyle \displaystyle \sum_{n=2}^{\infty} \frac{(-1)^{n}}{n\ \ln^{2} n}$ (1)

    ... is 'alternate sign' and the absolute value of the genral term decreases monotonically and tends to 0 with n, so that the series converges. For this type of series the truncation error is inferior, in absolute value, to the first 'discharged term' and that happens for...

    $\displaystyle \displaystyle n\ \ln^{2} n \sim 10^{5} \implies n \sim 1800$ (2)

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  4. #4
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by twittytwitter View Post
    http://math.rutgers.edu/~greenfie/we...fstuff/w4H.pdf

    I know for a), I can show it converges with integral test, but how can I find a specific partial sum?

    Do I just take the integral from 2 to n of 1/(x*(ln x)^2) and set it <10^-5 so that I get 5 digit accuracy, and then solve for N?

    For b), I know how to show divergence, but how do I find the specific partial sum? Same idea?
    b) the series...

    $\displaystyle \displaystyle \sum_{n=2}^{\infty} \frac{1}{n\ \sqrt{\ln n}}$ (2)

    ... diverges because is...

    $\displaystyle \displaystyle I(x)= \int_{2}^{x} \frac{dt}{t\ \sqrt{\ln t}} = 2\ \sqrt {\ln x} - 2\ \sqrt{\ln 2}$ (2)

    ... and I(x) tends to infinity with x. The condition I(x)= 100 is verified for $\displaystyle x \sim 5.5\ 10^{1085}$ ...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  5. #5
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    Ok thanks, but that part for b) isn't what I had trouble with. Obviously if you take something crazy like 5.5*10^1085, it will work, but how can I find an n for which it will work and the partial sum will exceed 100 using paper and pencil (i.e. no computer programs).
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