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Homework Help: Integral convergence and limit comparison test

  1. Aug 13, 2005 #1
    Hi, I've been thinking about the comparison test for integrals. Usually when I have an integrand where the denominator is f(x) + something positive I can usually find a suitable bound without much trouble. However when the denomoninator of the integrand is f(x) - something positive finding a bound is generally more difficult. This is because I would generally have no idea as to whether the integral converges or not.

    So one day I was thinking about the limit comparison test which comes in handy when I have series' where the denominator is something like 2^n - 1. Long story short I used the limit comparison test with integrals by considering the integral as a series(in the mathematically incorrect way) and for a few examples it seemed to work. For example if I have a integral where the integrand is of the form 1/(f(x) - c), c > 0 I just wrote the series(I've omitted the sigma etc) 1/(f(n) - c) and used the limit comparison test for series to determine whether or not the series 1/(f(n)-1) converges. In the few examples I tried, if the corresponding series converges I've also found that the integral converges.

    Can the limit comparison test for series be used as a test for integrals in this way? I know that there is an integral test for series but I'm not sure if there is anything the other way around.
  2. jcsd
  3. Aug 14, 2005 #2
    I think you're a bit confused here...

    The integral test for a series determines if a series converges/diverges. The limit comparison test does the same thing to a series. At times, it is preferred to use one over the other if the series is complicated or simple.

    So if the question you're asking is whether they can "replace" one another, then I guess the answer is yes.
  4. Aug 14, 2005 #3


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    I think what is meant is does the integral test for series work the other way. The answer is yes, the theorem can be formulated as the integral and the series converge or diverge together. Then oncce the question of does this integral coverge is exchanged for does this sum converge any method can be used, including comparison.
  5. Aug 14, 2005 #4
    Thanks for clearing that up.
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