1. Limited time only! Sign up for a free 30min personal 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!

Series - Integral Test

  1. Apr 4, 2009 #1
    1. The problem statement, all variables and given/known data
    Determine whether this series converges or diverges :
    http://mathbin.net/equations/7548_0.png [Broken]

    2. Relevant equations
    Integral Test

    3. The attempt at a solution
    http://mathbin.net/equations/7548_1.png [Broken] from that it shows to diverge but book says it converges. What did I do wrong?
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Apr 4, 2009 #2


    User Avatar
    Science Advisor
    Homework Helper

    Where did you get the ln| ... | from?
    Try [itex]1/e^{n^2}[/itex] as a primitive.
  4. Apr 4, 2009 #3
    I don't understand what you mean.
  5. Apr 4, 2009 #4
    Just use a substitution for n2:

    Let u = n2 and du/2n = dn. The n's will cancel leaving you with an easy integral to evaluate. Then use the fundamental theorem to get a (convergent) answer. There shouldn't be any Ln's in your answer!
  6. Apr 4, 2009 #5
    Oh, so the integral is [tex]\frac{-1e^{-n^2}}{2}[/tex]. Okay, now it converges. Thanks!
  7. Apr 5, 2009 #6


    User Avatar
    Science Advisor
    Homework Helper

    In general there are some rules which can help you determine whether your answer is correct. For example:
    • in polynomials, the highest power wins. For example, x/x^2 converges, [itex]6 x^3 \sqrt{x} / (12 x^4)^{5/6}[/itex] diverges because it goes like [itex]x^{7/2}/x^{10/3} = x^{1/6}[/itex] which has positive power.
    • logarithms converge slower than polynomials, e.g. x/log(x) will diverge.
    • exponentials go faster than any polynomial, for example [itex]x^{999999}/e^x[/itex] converges while [itex]x^{-9999999} e^x [/itex] diverges.

    You will find more of these as you go, they can often help you make an estimate.
    In this case, the terms in the sum converge to 0 much faster than 1/n, so in general you expect a convergent sum.
  8. Apr 5, 2009 #7
    My text book left those out. They probably assumed the student would figure them out themselves. Thank you for pointing them out!
  9. Apr 5, 2009 #8


    User Avatar
    Science Advisor
    Homework Helper

    Just to stress my point, they will not always help you.
    First of all, you need to take into account that the sum will get an extra n (for example, if you can bound all terms in the sum from above by M, your sum will be like M n). For example: even though 1/n converges to zero, the sum over n = 1 to infinity of 1/n will go like n / n = 1 and will diverge.
    Even then it is not guaranteed, for example, it is not clear if the sum over 1/(n log n) converges (for n > 1 for example; even if it does, it does so very slowly) although n/(n log n) goes to zero.
    So you shouldn't use my general rules as more than rules of thumb, and always use a rigorous method to prove convergence.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Series - Integral Test
  1. Series Test (Replies: 1)

  2. Test series (Replies: 49)

  3. Series test ? (Replies: 5)