Here is the definition from the textbook I am using.

Let $\{a_n\}$ be a sequence of positive terms. Suppose that $a_n=f(n)$, where $f$ is a continuous, positive, decreasing function of $x$ for all $x\ge N$ ($N$ a positive integer). Then the series $\sum_{n=N}^{\infty}a_n$ and the integral $\int_{N}^{\infty}f(x)dx$ both converge or both diverge.

I understand that in order to apply the integral test to a series that the series must satisfy all three conditions (continuous, positive, and decreasing). I know how to check if those conditions are satisfied. The part of the test I do not understand is how to determine if the series either converges or diverges. I assume you represent the series as an improper integral, and evaluate it? If the limit of the improper integral is $\infty$, it diverges, else it converges? Is this a correct assumption or is there more?


1 Answer 1


You don't represent the series with an improper integral; you just compute an improper integral that is related to the series. The convergence status of that improper integral will be the same as the convergence status of the series. Both will converge, or both will diverge (to infinity, since the terms are positive).

As an aside, $f$ needs to be a continuous function so that we can even talk about integrals at all. And $f$ needs to be decreasing or because there is a step in the proof that requires that in its logic.

  • $\begingroup$ Would I test for convergence of the improper integral using either the Direct Comparison Test or the Limit Comparison Test or is there a simpler method? There is a section on Comparison tests a few sections ahead of the one I am currently on, would that be related in any way? $\endgroup$
    – Kot
    Commented Dec 2, 2013 at 23:17
  • $\begingroup$ Usually you compute something like $\int_1^\infty f(x)\,dx$ as $\lim_{t\to\infty}\int_1^tf(x)\,dx$. First you have a definite integral to compute in terms of $t$. Then you have a limit to take with respect to $t$. There is no special test involved, just direct computation. $\endgroup$
    – 2'5 9'2
    Commented Dec 2, 2013 at 23:24
  • $\begingroup$ So after you compute the limit of the improper integral, how would you conclude whether it converges or diverges? $\endgroup$
    – Kot
    Commented Dec 2, 2013 at 23:26
  • $\begingroup$ If the limit exists and is a real number, that is the definition of "it converges". The definition of "it diverges" is simply that it does not converge. $\endgroup$
    – 2'5 9'2
    Commented Dec 2, 2013 at 23:28
  • $\begingroup$ I don't remember reading that in my book but I just found the exact same thing online. Thanks! $\endgroup$
    – Kot
    Commented Dec 2, 2013 at 23:31

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