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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

    CompuChip

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    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

    CompuChip

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    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

    CompuChip

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    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.
     
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