What Convergence Test Should I Use? Part 2

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

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What Convergence Test Should I Use? Part 2

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