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AP®MathsCollege BoardCalculus BCStudy GuidesUnit 10: Infinite Sequences and SeriesTests for Divergence & ConvergenceComparison Tests for Convergence

Comparison Tests for Convergence (College Board AP® Calculus BC)

Study Guide

Roger B

Written by: Roger B

Reviewed by: Mark Curtis

Updated on

Comparison test

  • The comparison test is a method for determining whether an infinite series with non-negative terms converges or diverges

    • It does this by comparing the infinite series to a different series whose convergence or divergence is already known

  • Let sum from n equals 1 to infinity of a subscript n and sum from n equals 1 to infinity of b subscript n be two series with non-negative terms

    • If sum from n equals 1 to infinity of b subscript n converges and if a subscript n less or equal than b subscript n for all n

      • then sum from n equals 1 to infinity of a subscript n converges

    • If sum from n equals 1 to infinity of b subscript n diverges and if a subscript n greater or equal than b subscript n for all n

      • then sum from n equals 1 to infinity of a subscript n diverges

  • This should make intuitive sense

    • If every term in a series is less than or equal to the corresponding terms in a convergent series, then the series converges

    • If every term in a series is greater than or equal to the corresponding terms in a divergent series, then the series diverges

Examiner Tips and Tricks

Useful series to use for comparisons are

  • the geometric series sum from n equals 0 to infinity of a r to the power of n with a greater than 0 comma space r greater than 0, which converges for 0 less than r less than 1 and diverges for r greater or equal than 1

  • the p-series sum from n equals 1 to infinity of 1 over n to the power of p, which converges for p greater than 1 and diverges for p less or equal than 1

Worked Example

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

(a) sum from n equals 1 to infinity of fraction numerator 1 over denominator n cubed plus 4 end fraction equals 1 fifth plus 1 over 12 plus 1 over 31 plus...

The comparison here will be with sum from n equals 1 to infinity of 1 over n cubed, which is a convergent p-series

Note that n cubed plus 4 greater than n cubed for n greater or equal than 1 so fraction numerator 1 over denominator n cubed plus 4 end fraction less than 1 over n cubed

For all n greater or equal than 1, fraction numerator 1 over denominator n cubed plus 4 end fraction less than 1 over n cubed

sum from n equals 1 to infinity of 1 over n cubed is a convergent p-series with p equals 3 greater than 1

By the comparison test, sum from n equals 1 to infinity of fraction numerator 1 over denominator n cubed plus 4 end fraction converges

(b) sum from n equals 1 to infinity of fraction numerator n plus 3 over denominator n squared plus 1 end fraction equals 2 plus 1 plus 3 over 5 plus...

The comparison here is a lot less obvious!

Thinking that fraction numerator n plus 3 over denominator n squared plus 1 end fraction almost equal to n over n squared equals 1 over n suggests using the divergent harmonic series sum from n equals 1 to infinity of 1 over n

Method 1

There's a bit of algebra required to get an inequality in the desired form

For all n greater or equal than 1,

table row cell fraction numerator n plus 3 over denominator n squared plus 1 end fraction end cell equals cell fraction numerator n plus 1 over denominator n squared plus 1 end fraction plus fraction numerator 2 over denominator n squared plus 1 end fraction end cell row blank greater than cell fraction numerator n plus 1 over denominator n squared plus 1 end fraction end cell row blank equals cell fraction numerator n plus 1 over denominator n squared plus 1 end fraction times n over n end cell row blank equals cell fraction numerator n squared plus n over denominator n cubed plus n end fraction end cell row blank greater or equal than cell fraction numerator n squared plus 1 over denominator n cubed plus n end fraction end cell row blank equals cell fraction numerator n squared plus 1 over denominator n open parentheses n squared plus 1 close parentheses end fraction end cell row blank equals cell 1 over n end cell end table

So table row blank blank cell fraction numerator n plus 3 over denominator n squared plus 1 end fraction end cell end table greater than 1 over n

And sum from n equals 1 to infinity of 1 over n is the harmonic series, which diverges

By the comparison test, sum from n equals 1 to infinity of table row blank blank cell fraction numerator n plus 3 over denominator n squared plus 1 end fraction end cell end table diverges

You could also describe sum from n equals 1 to infinity of table row blank blank cell 1 over n end cell end table as a divergent p-series with p equals 1

Method 2

Start with the result you want and ask if it is true, i.e. is 1 over n less than fraction numerator n plus 3 over denominator n squared plus 1 end fraction true, where n greater or equal than 1?

Assume it is true and rearrange it to get something that is actually true (this should be done in rough)

Be careful not to multiply both sides of an inequality by something negative (remember that n greater or equal than 1 which is positive)

table row cell 1 over n end cell less than cell fraction numerator n plus 3 over denominator n squared plus 1 end fraction end cell row cell open parentheses n squared plus 1 close parentheses end cell less than cell n open parentheses n plus 3 close parentheses end cell row cell n squared plus 1 end cell less than cell n squared plus 3 n end cell row 1 less than cell 3 n end cell row cell 1 third end cell less than n end table

The last line is actually true, since n greater or equal than 1

Reverse the steps (starting with the line that is actually true) to get the algebraic proof you require

table row cell 1 third end cell less than n row 1 less than cell 3 n end cell row cell n squared plus 1 end cell less than cell n squared plus 3 n end cell row cell open parentheses n squared plus 1 close parentheses end cell less than cell n open parentheses n plus 1 close parentheses end cell row cell 1 over n end cell less than cell fraction numerator n plus 1 over denominator n squared plus 1 end fraction end cell end table

And sum from n equals 1 to infinity of 1 over n is the harmonic series, which diverges

By the comparison test, sum from n equals 1 to infinity of table row blank blank cell fraction numerator n plus 3 over denominator n squared plus 1 end fraction end cell end table diverges

Limit comparison test

What is the limit comparison test?

  • The limit comparison test is a method for determining whether an infinite series with non-negative term converges or diverges

    • It also uses comparison with a series whose convergence or divergence is already known

    • but the limit of a quotient is considered instead of an inequality of terms

  • Let sum from n equals 1 to infinity of a subscript n and sum from n equals 1 to infinity of b subscript n be two series with non-negative terms

    • If limit as n rightwards arrow infinity of a subscript n over b subscript n equals L, where 0 less than L less than infinity

      • then either both series converge

      • or both series diverge

  • I.e. if the limit of that quotient exists and is positive and finite,

    • then if one sequence converges, so does the other one

    • or if one sequence diverges, so does the other one

  • If the terms of a series are expressed as a rational function (i.e., a fraction with polynomials in the numerator and denominator)

    • then considering only the highest powers of the variable in the numerator and denominator (coefficients not needed) can help suggest the comparison series to use for the limit comparison test

    • For example sum from n equals 1 to infinity of fraction numerator 3 n squared plus 2 n plus 7 over denominator 5 n to the power of 4 plus n cubed plus 14 end fraction

      • Assume that fraction numerator 3 n squared plus 2 n plus 7 over denominator 5 n to the power of 4 plus n cubed plus 14 end fraction is going to act approximately like n squared over n to the power of 4 equals 1 over n squared

      • This suggests using the convergent p-series sum from n equals 1 to infinity of 1 over n squared as the comparison series in a limit comparison test

Examiner Tips and Tricks

For series expressed by more complicated rational functions, the limit comparison test can be a lot quicker and simpler to use than the comparison test.

Worked Example

Use the limit comparison test to determine whether the series sum from n equals 1 to infinity of fraction numerator n plus 3 over denominator n squared plus 1 end fraction equals 2 plus 1 plus 3 over 5 plus... converges or diverges.

Considering only the highest powers in the numerator and denominator of fraction numerator n plus 3 over denominator n squared plus 1 end fraction, you get n over n squared equals 1 over n

This suggests using the divergent harmonic series sum from n equals 1 to infinity of 1 over n as the comparison series in the limit comparison test

First set up and rewrite the quotient (it doesn't matter which series term goes on the top, and which goes on the bottom)

table row cell fraction numerator open parentheses fraction numerator n plus 3 over denominator n squared plus 1 end fraction close parentheses over denominator open parentheses 1 over n close parentheses end fraction end cell equals cell fraction numerator n plus 3 over denominator n squared plus 1 end fraction times n over 1 end cell row blank equals cell fraction numerator n squared plus 3 n over denominator n squared plus 1 end fraction end cell end table

Then prepare the resulting expression for taking limits (by dividing the terms on top and bottom by the highest power of n)

table row blank equals cell fraction numerator n squared plus 3 n over denominator n squared plus 1 end fraction times fraction numerator bevelled 1 over n squared over denominator bevelled 1 over n squared end fraction end cell row blank equals cell fraction numerator 1 plus 3 over n over denominator 1 plus 1 over n squared end fraction end cell end table

Now take the limit

table row cell limit as n rightwards arrow infinity of fraction numerator open parentheses fraction numerator n plus 3 over denominator n squared plus 1 end fraction close parentheses over denominator open parentheses 1 over n close parentheses end fraction end cell equals cell limit as n rightwards arrow infinity of fraction numerator 1 plus 3 over n over denominator 1 plus 1 over n squared end fraction equals fraction numerator 1 plus 0 over denominator 1 plus 0 end fraction equals 1 end cell end table

That limit is positive and finite, so either both series converge or both diverge

sum from n equals 1 to infinity of 1 over n is the harmonic series, which diverges

By the limit comparison test, sum from n equals 1 to infinity of table row blank blank cell fraction numerator n plus 3 over denominator n squared plus 1 end fraction end cell end table diverges

You could also describe sum from n equals 1 to infinity of table row blank blank cell 1 over n end cell end table as a divergent p-series with p equals 1

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    Author
    Reviewer
    Roger B

    Author: Roger B

    Expertise: Maths

    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

    Author: Mark Curtis

    Expertise: Maths

    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.