Good Question 14

Good Question 14 – The Integral Test

I have no criteria for what constitutes a “Good Question” for this series of occasional posts. They are just questions that I found interesting, or that seem more than usually instructive, or that I learn something from. I cannot quote this question (2016 BC 92) since it is on a secure exam. What made it interesting is that to answer it students pretty much needed to know the proof of the Integral Test and the figures that go with it.

I recall only one AP question from many years ago that asked students to “prove” something – usually students are asked to show that a result was true by citing the theorem that applied and showing the hypotheses were met. The directions are often “justify your answer.”

Doing an original proof is not, in my opinion, a fair question and proving some known theorem is just a matter of memorization. For these reasons, students are not asked to prove things on the exams. So, should you prove things in class? Probably, yes.

Here is the usual proof of the integral test. Afterwards I’ll discuss the question from the exam.

The Integral Test

Hypotheses: Let f\left( x \right) be a function that is positive, decreasing, and continuous for x\ge 1 ; and let {{a}_{n}}=f\left( n \right) for x\ge 1

In the first drawing the rectangles have a height of an and a width of 1. The area is of each is an, and the sum of their areas the series is \sum\limits_{{n=1}}^{\infty }{{{{a}_{n}}}}.

Part 1: Notice that \sum\limits_{{n=1}}^{\infty }{{{{a}_{n}}}}>\int_{1}^{\infty }{{f\left( x \right)dx}} Assume that the improper integral  \int_{1}^{\infty }{{f\left( x \right)dx}} diverges.

  • Conclusion 1: If the improper integral \int_{1}^{\infty }{{f\left( x \right)dx}}  diverges, then the series \sum\limits_{{n=1}}^{\infty }{{{{a}_{n}}}} diverges.
  • Conclusion 2: (The contrapositive of conclusion 1) If the series \sum\limits_{{n=1}}^{\infty }{{{{a}_{n}}}} converges, then the improper integral \int_{1}^{\infty }{{f\left( x \right)dx}} converges.

 

Part 2: In the second drawing below, Assume that the improper integral \int_{1}^{\infty }{{f\left( x \right)dx}} converges. The sum of the areas of the rectangles is \sum\limits_{{n=2}}^{\infty }{{{{a}_{n}}}}. (NB: this series starts at n = 2.) Since \sum\limits_{{n=2}}^{\infty }{{{{a}_{n}}}} is less than the convergent improper integral it will also converge. Adding {{a}_{1}} to this gives the original series, \sum\limits_{{n=1}}^{\infty }{{{{a}_{n}}}}; this series also converges.

  • Conclusion 3: If the improper integral \int_{1}^{\infty }{{f\left( x \right)dx}} converges, then the series \sum\limits_{{n=1}}^{\infty }{{{{a}_{n}}}} converges.
  • Conclusion 4: (The contrapositive of conclusion 3) If the series   \sum\limits_{{n=1}}^{\infty }{{{{a}_{n}}}} diverges, then the improper integral \int_{1}^{\infty }{{f\left( x \right)dx}} diverges.

Putting the four conclusions together is the Integral Test: If the hypotheses above are met, then the series and the improper integral will both converge, or both diverge.


To answer the multiple-choice question on the exams students were told that improper integral, and therefore the associated series, converged. They had to determine whether the series or the improper integral has the greater value. Stop here and see if you can figure that out.


Return to the first figure above only this time assume that the improper integral and the series converge. It is pretty obvious that \sum\limits_{{n=1}}^{\infty }{{{{a}_{n}}}}>\int_{1}^{\infty }{{f\left( x \right)dx}}.

So, even though students were not asked to prove anything, a familiarity with the proof and its figures is necessary to answer the question. That’s why I liked it,

On the other hand, it is kind of an obscure point and I’m not sure it has any practical value.


 

 

 

 

 

 

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2 thoughts on “Good Question 14

  1. Why couldn’t you refer to your second figure, saying that because the improper integral converged, then the series converged, and therefore the integral > series? It sounds like the question was worded this way. Is it because the first term is omitted from the series in your second figure?

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    • Yes, it is because in the second figure the first rectangle on the left is \displaystyle {{a}_{2}}, so the entire series is not shown. The rectangle for \displaystyle {{a}_{1}} extends to the left between x = 1 and the y-axis. You cannot tell (for sure) from this that the series is greater than the integral. From the first figure (with the new assumption that the series and integral both converge) you can see that the area representing the series is greater than the area represented by the integral.

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