Integral Test for Convergence (College Board AP® Calculus BC): Revision Note

Roger B

Written by: Roger B

Reviewed by: Mark Curtis

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Integral test

What is the integral test?

  • The integral test is a method of determining whether an infinite series converges or diverges

  • The integral test states that:

    • Given that an=f(n), where f is a continuous, positive, decreasing function on [c, )

    • If the improper integral cf(x) dx exists

    • Then the series n=can converges

      • If the improper integral doesn't exist then the series diverges

  • For example, you can use the function  f(x)=1x3 to test whether n=11n3 converges

    •  f is continuous, positive, and decreasing for x1

Examiner Tips and Tricks

For an FRQ, be sure to state that your function is continuous, positive, and decreasing.

How does the integral test work?

  • Each term in the infinite series a1+a2+a3+... can be represented by the area of a rectangle of width 1 and height an

  • The integral 1f(x) dx=A represents the area Aunder the curve f(x) from x=1 to

  • In the first image, A is an underestimate of the rectangles a1+a2+a3+...

    • so A<n=1an

Graph showing a curve y=f(x) with marked points at intervals. Grey rectangles overestimate the area under the graph and the areas a1 to a5 represent the infinite series.
  • In the next image, A is an overestimate of the rectangles a2+a3+...

    • Adding a1 to both sides gives a1+A>n=1an

Graph showing a curve y=f(x) with decreasing step-like shaded rectangular areas under it representing the infinite series. Points (1,a1) to (5,a5) are marked. Axes labelled x and y.
  • So A<n=1an<A+a1 which means

    • if A is finite, then n=1an ⁣must have a finite value (the series converges)

    • if A is infinite, then n=1an must also be infinite (the series diverges)

Worked Example

Use the integral test to determine whether each of the following series converges or diverges.

(a) n=11n

(b) n=11n2

Answer:

(a)

Note that this series is the harmonic series 1+12+13+14+...

 f(x)=1x is continuous, positive and decreasing for x1

Evaluate the improper integral 11x dx

11x dx=limp1p1x dx=limp[ln|x|]1p=limp(lnpln1)=limplnp=

The improper integral diverges to infinity (as logarithmic growth tends to infinity, ln x as x), so the series diverges

The integral 11x dx diverges to infinity, so by the integral test the series is divergent

(b)

Note that this is a p-series with p=2

 f(x)=1x2 is continuous, positive and decreasing for x1

Evaluate the improper integral 11x2 dx

11x2 dx=limp1px2 dx=limp[x1]1p=limp[1x]1p=limp(1p(11))=limp(11p)=10=1

The improper integral converges to a finite value, so the series converges

The integral 11x2 dx exists with a finite value, so by the integral test the series is convergent

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Roger B

Author: Roger B

Expertise: Development Editor

Roger's teaching experience stretches all the way back to 1992, and in that time he has taught students at all levels between Year 7 and university undergraduate. Having conducted and published postgraduate research into the mathematical theory behind quantum computing, he is more than confident in dealing with mathematics at any level the exam boards might throw at you.

Mark Curtis

Reviewer: Mark Curtis

Expertise: Maths Content Creator

Mark graduated twice from the University of Oxford: once in 2009 with a First in Mathematics, then again in 2013 with a PhD (DPhil) in Mathematics. He has had nine successful years as a secondary school teacher, specialising in A-Level Further Maths and running extension classes for Oxbridge Maths applicants. Alongside his teaching, he has written five internal textbooks, introduced new spiralling school curriculums and trained other Maths teachers through outreach programmes.