convergence tests | Teaching Calculus

In last Friday’s post I really didn’t answer this question. Rather, I tried to show that there is not only one convergence test that must be used on a given series. Nevertheless, the form of a series suggests a test that is likely to work. In this post, I’ll try to give some suggestions as to what test to try first based on the form of the series.

For reference, click here for a table summarizing the common convergence tests.

The goal is for students to be able to decide which test to start with at a glance.

Start with the n^th-term test for divergence. If the limit of the general term as n goes to infinity is not zero, the sequence will diverge. The $\underset{{n\to \infty }}{\mathop{{\lim }}}\,{{a}_{n}}=0$ is a necessary condition for convergence. It is not sufficient; if the limit is zero then the series may converge. Look for a convergence test.

If the series alternates plus and minus signs, it is an alternating series and if it satisfies the other hypotheses use the Alternating Series Test. If the series contains positive and negative signs that do not alternate, or one of the other hypotheses is not met, then a different test must be used.

If the series is geometric then the Geometric Series Test may be used. If the common ratio (the number multiplied by each term to get the next term) is between –1 and 1 the series converges. If the common ratio is greater than or equal to 1, or less than or equal to –1, the series diverges.

The remaining tests are for series with all positive terms. They are tests for absolute convergence. If you series has negative terms then you may ignore the signs and try one of the following tests. If your series is absolutely convergent, then it is convergent. (If not, it may still be convergent.)

If the general term (written with x’s) looks like something that you can integrate, use the Integral Test.

The Direct Comparison Test and the Limit Comparison Test are used if you can find a test to compare them with.

A p-series, $\sum\limits_{{n=1}}^{\infty }{{\frac{1}{{{{n}^{p}}}}}}$ converges if $p>1$ and diverges if $p\le 1$ . A p-series is often a good test to use for comparison in the next two tests. However, any series whose convergence you are sure of may be used.

The Direct Comparison Test is used with fraction expressions. “Extra” factors in the denominator can often be ignored. Some examples

$\displaystyle \sum\limits_{{n=1}}^{\infty }{{\frac{1}{{{{5}^{n}}\sqrt{n}}}}}$ would be a geometric series except for the radical. Compare it with the geometric series $\displaystyle \sum\limits_{{n=1}}^{\infty }{{\frac{1}{{{{5}^{n}}}}}}$
$\displaystyle \sum\limits_{{n=1}}^{\infty }{{\frac{{{{n}^{2}}}}{{{{n}^{4}}+2n+1}}}}$ can be compared with the p-series $\displaystyle \sum\limits_{{n=1}}^{\infty }{{\frac{1}{{{{n}^{2}}}}}}$ . The hint here is that ignoring the lower power terms in the denominator and reducing we see that the original series looks like $\displaystyle \sum\limits_{{n=1}}^{\infty }{{\frac{1}{{{{n}^{2}}}}}}$ . Both series converge. But be careful $\displaystyle \sum\limits_{{n=1}}^{\infty }{{\frac{{{{n}^{2}}}}{{{{n}^{4}}-2n-1}}}}$ while similar, has terms greater than the terms of $\displaystyle \sum\limits_{{n=1}}^{\infty }{{\frac{1}{{{{n}^{2}}}}}}$ .)
The terms of the series $\displaystyle \sum\limits_{{n=1}}^{\infty }{{\frac{1}{{{{{\left( {{{n}^{2}}+2} \right)}}^{{1/3}}}}}}}$ are larger than the harmonic series $\displaystyle \sum\limits_{{n=1}}^{\infty }{{\frac{1}{n}}}$ a divergent p-series, so this series diverges.

The Limit Comparison Test may be used with the same kinds of series that are messy to use with direct comparison.

Returning to $\displaystyle \sum\limits_{{n=1}}^{\infty }{{\frac{{{{n}^{2}}}}{{{{n}^{4}}+2n+1}}}}$ , try the limit comparison test with $\displaystyle \sum\limits_{{n=1}}^{\infty }{{\frac{1}{{{{n}^{2}}}}}}$ . The limit is 1, so both series converge.
$\displaystyle \sum\limits_{{n=1}}^{\infty }{{\frac{1}{{\sqrt{{{{n}^{2}}+3}}}}}}$ Series with radicals also are candidates for the limit comparison test. Since the general terms is approximately $\displaystyle {\frac{1}{n}}$ Compare this with $\displaystyle \sum\limits_{{n=1}}^{\infty }{{\frac{1}{n}}}$ . Both series diverge.

More complicated series, perhaps with exponential factors and/or factorials can be examined with the Ratio Test.

$\displaystyle \sum\limits_{{n=1}}^{\infty }{{\frac{{{{3}^{n}}}}{{n!}}}}$ or $\displaystyle \sum\limits_{{n=1}}^{\infty }{{\frac{{{{n}^{3}}}}{{{{5}^{n}}}}}}$ are candidates for the Ratio Test. Both Converge.
$\displaystyle \sum\limits_{{n=1}}^{\infty }{{{{{\left( {-1} \right)}}^{n}}\frac{{n!}}{{{{{500}}^{n}}}}}}$ appears to be a candidate for the alternating series test. However, for large values of n > 530 the terms increase in absolute vale, so the alternating series test cannot be applied. The ratio test works here, but since the terms do not approach 0 as n increases, the n^th-term test for divergence also works. This series diverges.

Practice, Practice, Practice

The AP Calculus BC exams rarely, if ever, specify which test to use. Often these are multiple-choice questions. If students can see whether the series converges or diverges, that is enough. But here again the key is practice, practice, practice.

As you teach the various tests, pause to look at the form of the series in the exercises for each test that your book provides. Most books also have mixed sets of exercises where tests other than the one in that section are needed. One of the things you can do is assign these entire sets with the directions that students should determine what test they would try, and, for their comparison tests, to which series they would compare it. Discuss their opinions especially if there is more than one suggested or suggest others. Work only those those students are confused about or those for which they have divergent opinions; try to converge on a good test for each.

Revised July 18, 2021, January 29, 2023

One common question from students first learning about series is how to know which convergence test to use with a given series. The first answer is: practice, practice, practice. The second answer is that there is often more than one convergence test that can be used with a given series.

I will illustrate this point with a look at one series and the several tests that may be used to show it converges. This will serve as a review of some of the tests and how to use them. For a list of convergence tests that are required for the AP Calculus BC exam click here.

To be able to use these tests the students must know the hypotheses of each test and check that they are met for the series in question. On multiple-choice questions students do not need to how their work, but on free-response questions (such as checking the endpoints of the interval of convergence of a Taylor series) they should state them and say that the series meets them.

For our example we will look at the series $\displaystyle 1-\frac{1}{3}+\frac{1}{9}-\frac{1}{{27}}+\frac{1}{{81}}-+\ldots =\sum\limits_{{n=1}}^{\infty }{{{{{\left( {-\frac{1}{3}} \right)}}^{{n-1}}}}}$

Spoiler: Except for the first two tests to be considered, the other tests are far more work than is necessary for this series. The point is to show that several tests may be used for a given series, and to practice the other tests.

The Geometric Series Test is the obvious test to use here, since this is a geometric series. The common ratio is (–1/3) and since this is between –1 and 1 the series will converge.

The Alternating Series Test (the Leibniz Test) may be used as well. The series alternates signs, is decreasing in absolute value, and the limit of the n^th term as n approaches infinity is 0, therefore the series converges.

The Ratio Test is used extensively with power series to find the radius of convergence, but it may be used to determine convergence as well. To use the test, we find

$\displaystyle \underset{{n\to \infty }}{\mathop{{\lim }}}\,\frac{{\left| {{{{\left( {-\frac{1}{3}} \right)}}^{{n+1}}}} \right|}}{{\left| {{{{\left( {-\frac{1}{3}} \right)}}^{n}}} \right|}}=\frac{1}{3}$ Since the limit is less than 1, we conclude the series converges.

Absolute Convergence

A series, $\sum\limits_{{n=1}}^{\infty }{{{{a}_{n}}}}$ , is absolutely convergent if, and only if, the series $\sum\limits_{{n=1}}^{\infty }{{\left| {{{a}_{n}}} \right|}}$ converges. In other words, if you make all the terms positive, and that series converges, then the original series also converges. If a series is absolutely convergent, then it is convergent. (A series that converges but is not absolutely convergent is said to be conditionally convergent.)

The advantage of going for absolute convergence is that we do not have to deal with the negative terms; this allows us to use other tests.

Applied to our example, if the series $\sum\limits_{{n=1}}^{\infty }{{{{{\left( {\frac{1}{3}} \right)}}^{{n-1}}}}}$ converges, then our series $\sum\limits_{{n=1}}^{\infty }{{{{{\left( {-\frac{1}{3}} \right)}}^{{n-1}}}}}$ will converge absolutely and converge.

The Geometric Series Test can be used again as above.

The Integral Test says if the improper integral $\displaystyle {{\int_{1}^{\infty }{{\left( {\frac{1}{3}} \right)}}}^{x}}dx$ converges, then our original series will converge absolutely.

$\displaystyle \int\limits_{1}^{\infty }{{{{{\left( {\frac{1}{3}} \right)}}^{x}}}}dx=\underset{{n\to \infty }}{\mathop{{\lim }}}\,\int\limits_{1}^{n}{{{{{\left( {\frac{1}{3}} \right)}}^{x}}}}dx=\underset{{n\to \infty }}{\mathop{{\lim }}}\,\left( {\frac{{{{{\left( {\frac{1}{3}} \right)}}^{n}}}}{{\ln \left( {1/3} \right)}}-\frac{{{{{\left( {\frac{1}{3}} \right)}}^{1}}}}{{\ln \left( {1/3} \right)}}} \right)=0-\frac{{1/3}}{{\ln \left( {1/3} \right)}}$

$\displaystyle =-\frac{{1/3}}{{\ln \left( {1/3} \right)}}>0$ since ln(1/3) < 0.

The limit is finite, so our series converges absolutely, and therefore converges.

The Direct Comparison Test may also be used. We need to find a positive convergent series whose terms are term-by-term greater than the terms of our series. The geometric series $\sum\limits_{{n=1}}^{\infty }{{{{{\left( {\frac{1}{2}} \right)}}^{{n-1}}}}}$ meets these two requirements. Therefore, the original series converges absolutely and converges.

The Limit Comparison Test is another possibility. Here we need a positive series that converges; we can use $\sum\limits_{{n=1}}^{\infty }{{{{{\left( {\frac{1}{2}} \right)}}^{{n-1}}}}}$ again. We look at

$\displaystyle \underset{{n\to \infty }}{\mathop{{\lim }}}\,\frac{{{{{\left( {1/3} \right)}}^{n}}}}{{{{{\left( {1/2} \right)}}^{n}}}}=\underset{{n\to \infty }}{\mathop{{\lim }}}\,{{\left( {\frac{2}{3}} \right)}^{n}}=0$ and since the series in the denominator converges, our series converges absolutely.

So, for this example all the convergences that may be tested on the AP Calculus BC exam may be used with the single exception of the p-series Test which cannot be used with this series.

Teaching suggestions

While the convergence of the series used here can be done all these ways, other series lend themselves to only one. Stress the form of the series that works with each test. For example, the Limit Comparison Test is most often used for rational expressions with the numerator of lower degree than the denominator and for expressions involving radicals of polynomials. The comparison is made with a p-series of whatever degree will make the numerator and denominator the same degree allowing the limit to be found.
Most textbooks, after explaining each test and giving exercises on them, include a series of mixed exercises that require all the test covered up to that point. A good way to use this set is to assign students to state which test they would try first on each series. Discuss the opinions of the class and work any questions that students are unsure of or on which several ways are suggested.
Give your students the series above, or a similar one, and have them prove its convergence using each of the convergence tests as was done above.
Divide your class into groups and assign each group the series and one of the convergence tests. Ask them to use the test to prove convergence and then discuss the results as a group.

Of course, I didn’t really answer the question, did I? Check What Convergence Test Should I use Part 2

Updated February 23, 2013

Teaching Calculus

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Tag Archives: convergence tests

What Convergence Test Should I Use? Part 2

Which Convergence Test Should I Use? Part 1

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