1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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]


    [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


    User Avatar
    Science Advisor
    Gold Member

    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


    User Avatar
    Science Advisor
    Homework Helper

    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
    is divergent.
  6. Mar 11, 2005 #5


    User Avatar
    Science Advisor
    Homework Helper

    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]
  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


    User Avatar
    Science Advisor
    Homework Helper

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: The integral test
  1. Test for integrability (Replies: 5)

  2. Integral test HELP! (Replies: 4)