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The integral test

  1. Mar 11, 2005 #1
    How many terms of the series [tex]\sum^{\infty}_{n=2} \frac{1}{n(ln\;n)^{2}} [/tex] would you need to add to find its sum to within 0.01?

    Here's what i got:

    let [tex] f(n) = \frac{1}{n(ln\;n)^{2}} [/tex]. Since [tex] f(n) [/tex] is continuous, positive and decreasing for all n over the interval [tex] [2,\infty] [/tex], we can use the integral test to evaluate the series.

    [tex]\sum^{\infty}_{n=2} \frac{1}{n(ln\;n)^{2}} = \lim_{b \to \infty} \int_{2}^{b} \frac{1}{n(ln\;n)^{2}} = \frac{1}{ln\;n} [/tex]

    thus,

    [tex] R_{n} \leq \frac{1}{ln\;n} [/tex]

    since we want [tex] R_{n} \leq 0.01 [/tex],

    [tex] \frac{1}{ln\;n} \leq 0.01 [/tex]

    implying [tex] n = e^{100}[/tex].


    but that can't be right...e^100 is way too big, isn't it? thanks in advance.
     
  2. jcsd
  3. Mar 11, 2005 #2

    mathman

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    It may not be. The series you have is just barely convergent. Without the (ln n)2 term, it would diverge.
     
  4. Mar 11, 2005 #3
    So it's right?
     
  5. Mar 11, 2005 #4

    NateTG

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    I haven't followed the math, but I don't see why it should be. You can have sequences that converge *very* slowly like, apparently, the one you cite, and the series definitely looks like a slowly converging one. Consider the sequence:
    [tex]\sum_{n=2}^{\infty}\frac{1}{n \ln n}[/tex]
    which converges significantly more slowly. And that makes sense considering that [itex]\ln x[/itex] grows very slowly, and
    [tex]\sum_{n=2}^{\infty}\frac{1}{n}[/tex]
    is divergent.
     
  6. Mar 11, 2005 #5

    shmoe

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    This actually diverges by the integral test, [tex]\int_{2}^{b}\frac{1}{t \ln t}dt=\ln\ln b-\ln\ln 2[/tex], which goes to infinity as b does (though obviously very slowly).


    For the OP,the result is fine, but the process seems garbled:

    [tex]\sum^{\infty}_{n=2} \frac{1}{n(ln\;n)^{2}} = \lim_{b \to \infty} \int_{2}^{b} \frac{1}{n(ln\;n)^{2}} = \frac{1}{ln\;n} [/tex]

    This isn't correct, the sum and the integral are not equal, and the evaluation of the limit and/or integral is off. This may be a translation error from paper to screen though?
     
  7. Mar 11, 2005 #6
    it should be [tex]\sum^{\infty}_{n=2} \frac{1}{n(ln\;n)^{2}} = \lim_{b \to \infty} \int_{n}^{b} \frac{1}{n(ln\;n)^{2}} = \frac{1}{ln\;n} [/tex]
    right?
     
  8. Mar 11, 2005 #7
    (noting that the lower limit is n now, instead of 2)
     
  9. Mar 11, 2005 #8
    what you have above is wrong.



    [tex]\sum^{\infty}_{n=2} \frac{1}{n(ln\;n)^{2}} \neq \lim_{b \to \infty} \int_{2}^{b} \frac{1}{n(ln\;n)^{2}} dn= \frac{1}{ln\;2} [/tex]

    if you want it to be equal to [tex]\frac{1}{ln\;n} [/tex]
    then write


    [tex]\lim_{b \to \infty} \int_{n}^{b} \frac{1}{t(ln\;t)^{2}} dt= \frac{1}{ln\;n} [/tex]
     
    Last edited: Mar 11, 2005
  10. Mar 11, 2005 #9

    shmoe

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    3.14..., do you know how to use Zone Ranger's correction to get an estimate for [tex]R_n[/tex]? Remember that [tex]R_n[/tex] is the 'tail' of your series.

    edit-What I'm hoping is that you can justify why [tex]R_n[/tex] can be bounded using that integral.
     
    Last edited: Mar 11, 2005
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