Comparison tests for convergence — AP Calculus BC Study Guide

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: Direct Comparison Test, Limit Comparison Test, convergence/divergence criteria, edge cases for limit results, extension to series with negative terms, and testing for absolute convergence with comparison methods.

You should already know: Convergence and divergence definitions for infinite series, convergence rules for p-series and geometric series, basic limit laws for sequences.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus BC style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Comparison tests for convergence?

Comparison tests for convergence are a pair of related methods for determining whether an infinite series converges or diverges, by comparing the unknown series to a second series with known convergence behavior. This topic is a core component of Unit 10: Infinite Sequences and Series, which accounts for 17-18% of the total AP Calculus BC exam score. Comparison questions appear regularly in both multiple-choice (MCQ) and free-response (FRQ) sections, typically with 1-2 MCQ questions and occasionally as a graded part of a larger FRQ focusing on series convergence. Unlike the Integral Test, which requires you to integrate the general term, comparison tests rely on algebraic simplification and knowledge of standard series (p-series, geometric series), making them faster to apply for most rational, radical, and exponential series. All comparison tests have a core requirement: they are initially designed for series with non-negative terms, with extensions to other series via absolute convergence.

2. The Direct Comparison Test

The Direct Comparison Test is the simpler of the two comparison methods, and applies only to series with non-negative terms. For two infinite series ${\sum }_{n=1}^{\infty }{a}_n$ and ${\sum }_{n=1}^{\infty }{b}_n$, suppose that for all $n$ greater than some finite integer $N$ (meaning after the first few finite terms), $0\le a_n\le b_n$. The test gives two possible conclusions:

1. If $\sum b_n$ converges, then $\sum a_n$ must also converge.
2. If $\sum a_n$ diverges, then $\sum b_n$ must also diverge.

The intuition for this rule is straightforward: if a larger "upper bound" series does not grow without bound (diverge), then a smaller series bounded above by it cannot grow without bound either. Conversely, if the smaller "lower bound" series already diverges, any larger series must also diverge. A key detail that is easy to miss is that the inequality only needs to hold for all sufficiently large $n$, not for the first term or early terms. Finite numbers of terms do not change the convergence behavior of a series, so early violations of the inequality do not invalidate the test.

Worked Example

Determine if ${\sum }_{n=1}^{\infty }\frac{1}{n^2+5}$ converges using the Direct Comparison Test.

1. Confirm all terms are non-negative for $n\ge 1$, which satisfies the test's core requirement.
2. For all $n\ge 1$, $n^2+5>n^2$, so taking reciprocals preserves the inequality: $0<\frac{1}{n^2+5}<\frac{1}{n^2}$.
3. The comparison series ${\sum }_{n=1}^{\infty }\frac{1}{n^2}$ is a p-series with $p=2>1$, which is known to converge.
4. By the Direct Comparison Test, since the original series is smaller than a convergent series with non-negative terms, the original series converges.

Exam tip: Always confirm your inequality direction matches the conclusion you want: to prove convergence, your unknown series must be smaller than a known convergent series; to prove divergence, it must be larger than a known divergent series. If your inequality is the wrong direction, switch to the Limit Comparison Test instead.

3. The Limit Comparison Test

The Limit Comparison Test is a more flexible alternative to the Direct Comparison Test, designed for cases where the inequality does not work out to the direction you need, or where bounding the terms algebraically is messy. Like the Direct Comparison Test, it applies exclusively to series with non-negative terms. For two series $\sum a_n$ and $\sum b_n$ (with $b_n>0$ for all sufficiently large $n$), compute the limit of the ratio of the general terms: $$\underset{n\to \infty }{\lim }\frac{a_n}{b_n}=L$$ If $0<L<\infty$, meaning the limit is a positive finite number, then $\sum a_n$ and $\sum b_n$ have the same convergence behavior: both converge or both diverge. Special cases for edge limits: if $L=0$ and $\sum b_n$ converges, then $\sum a_n$ converges; if $L=\infty$ and $\sum b_n$ diverges, then $\sum a_n$ diverges.

The intuition here is that for very large $n$, $a_n\approx L{b}_n$, so the overall series grow at the same rate. This makes it ideal for rational functions of $n$, where the leading terms of the numerator and denominator dominate the behavior for large $n$.

Worked Example

Determine if ${\sum }_{n=1}^{\infty }\frac{2n+1}{n^3-4n+2}$ converges using the Limit Comparison Test.

1. For large $n$, the leading term of the numerator is $2n$, and the leading term of the denominator is $n^3$, so $a_n\approx \frac{2n}{n^3}=\frac{2}{n^2}$. We choose $b_n=\frac{1}{n^2}$, a known convergent p-series term.
2. Confirm all $a_n,b_n$ are positive for $n\ge 3$ (after the first two terms, which do not affect convergence), so the test applies.
3. Compute the limit: $$\underset{n\to \infty }{\lim }\frac{a_n}{b_n}=\underset{n\to \infty }{\lim }\frac{(2n+1)/(n^3-4n+2)}{1/n^2}=\underset{n\to \infty }{\lim }\frac{2n^3+n^2}{n^3-4n+2}=2$$
4. The limit $L=2$ is positive and finite, so the series have the same behavior. $\sum \frac{1}{n^2}$ converges, so the original series converges.

Exam tip: For rational $a_n$, always drop all lower-degree terms when choosing $b_n$: only the highest power of $n$ in the numerator and denominator matters for the limit as $n\to \infty$, so this shortcut will always give you the correct comparison series.

4. Extending Comparison Tests to Series with Negative Terms

The core comparison tests only work for series with all non-negative (or all non-positive) terms, because they rely on bounding to draw conclusions. To apply comparison methods to series with mixed positive and negative terms, we use the link between absolute convergence and convergence: if the series of absolute values $\sum |a_n|$ converges, then the original series $\sum a_n$ converges absolutely, and therefore converges. Since $\sum |a_n|$ has all non-negative terms, we can apply direct or limit comparison to this transformed series to test for convergence.

The key limitation to remember: if $\sum |a_n|$ diverges, comparison tests cannot tell you anything about the convergence of the original alternating/mixed series. You will need another test (like the Alternating Series Test) to check for conditional convergence in that case.

Worked Example

Does the comparison test approach confirm convergence of ${\sum }_{n=1}^{\infty }\frac{(-1{)}^{n}n}{2^n+1}$?

1. This is a series with mixed terms, so we test absolute convergence by analyzing $\sum |a_n|=\sum \frac{n}{2^n+1}$.
2. For all $n\ge 1$, $2^n+1>2^n$, so $\frac{n}{2^n+1}<\frac{n}{2^n}$. We use the Direct Comparison Test with $\sum \frac{n}{2^n}$.
3. The series $\sum \frac{n}{2^n}$ is a convergent power series (it can be shown to converge to 2 via geometric series differentiation), so it converges.
4. By Direct Comparison, $\sum |a_n|$ converges, so the original series converges (absolutely). The comparison approach confirms convergence.

Exam tip: If an FRQ asks you to "use a comparison test" to determine convergence of an alternating series, they almost always want you to test for absolute convergence via comparison to get full credit.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Using the Direct Comparison Test to claim $\sum \frac{1}{n^2-1}$ converges because $\frac{1}{n^2-1}>\frac{1}{n^2}$ and $\sum \frac{1}{n^2}$ converges. Why: Students memorize that comparison to p-series works, but forget the inequality direction is wrong for their desired conclusion. Correct move: Always confirm: to prove convergence, your unknown series must be smaller than a known convergent; reverse for divergence. Use Limit Comparison instead if your inequality is the wrong direction.
• Wrong move: Applying the Limit Comparison Test when $L=0$ and the comparison series diverges, then concluding the original series diverges. Why: Students generalize the $0<L<\infty$ rule incorrectly to edge cases. Correct move: Only draw conclusions for edge cases: if $L=0$, only conclude convergence if the comparison series converges; if $L=\infty$, only conclude divergence if the comparison series diverges. For all other combinations, the test is inconclusive.
• Wrong move: Applying comparison tests directly to an alternating series without taking absolute values first. Why: Students forget the non-negative term requirement for all comparison methods. Correct move: Any series with mixed positive/negative terms must be tested by applying comparison to the series of absolute values of its terms first.
• Wrong move: Claiming divergence because your original series is larger than a convergent comparison series. Why: Students mix up what information each comparison gives. Correct move: If your original series is larger than a convergent series, you learn nothing: the original series could still converge or diverge, so you need to try a new comparison series or switch tests.
• Wrong move: Comparing $\sum \frac{1}{\sqrt{n}(n+3)}$ to $\sum \frac{1}{n}$ instead of $\sum \frac{1}{n^{3/2}}$. Why: Students forget to add the exponents of $n$ in the denominator to find the correct p-series. Correct move: For any product of powers/radicals of $n$, simplify the total exponent of $n$ in the denominator minus the exponent in the numerator to get $p$ for your comparison p-series.

6. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

Which of the following statements about the series ${\sum }_{n=1}^{\infty }\frac{4^n}{3^n+5^n}$ is true? A. The series converges by the Direct Comparison Test with the convergent geometric series $\sum {\left(\frac{4}{5}\right)}^n$ B. The series converges by the Direct Comparison Test with the convergent geometric series $\sum {\left(\frac{4}{3}\right)}^n$ C. The series diverges by the Direct Comparison Test with the divergent geometric series $\sum {\left(\frac{4}{5}\right)}^n$ D. The series diverges by the nth Term Test, because ${\lim }_{n\to \infty }\frac{4^n}{3^n+5^n}$ does not exist.

Worked Solution: For all $n\ge 1$, $3^n+5^n>5^n$, so $\frac{4^n}{3^n+5^n}<\frac{4^n}{5^n}={\left(\frac{4}{5}\right)}^n$. The series $\sum {\left(\frac{4}{5}\right)}^n$ is a geometric series with common ratio $r=4/5$, so $|r|<1$, meaning it converges. Option B is incorrect because $\sum (4/3{)}^n$ is divergent. Option C is incorrect because $\sum (4/5{)}^n$ converges, not diverges. Option D is incorrect because ${\lim }_{n\to \infty }\frac{4^n}{3^n+5^n}=0$. The correct answer is A.

Question 2 (Free Response)

Consider the infinite series ${\sum }_{n=1}^{\infty }\frac{n^3+n^2-1}{n^5+4n^3-2n}$. (a) Identify an appropriate comparison series to use with the Limit Comparison Test, and explain why you chose it. (b) Use the Limit Comparison Test to determine whether the series converges or diverges. (c) Can the Direct Comparison Test be used to reach the same conclusion? Justify your answer.

Worked Solution: (a) For large $n$, the highest-degree terms dominate the behavior of the general term: numerator leading term is $n^3$, denominator leading term is $n^5$, so $a_n\approx \frac{n^3}{n^5}=\frac{1}{n^2}$. The appropriate comparison series is ${\sum }_{n=1}^{\infty }\frac{1}{n^2}$, which is a known convergent p-series with $p=2>1$. (b) Let $a_n=\frac{n^3+n^2-1}{n^5+4n^3-2n}$ and $b_n=\frac{1}{n^2}$. All terms are positive for $n\ge 1$, so the Limit Comparison Test applies: $$\underset{n\to \infty }{\lim }\frac{a_n}{b_n}=\underset{n\to \infty }{\lim }\frac{(n^3+n^2-1)n^2}{n^5+4n^3-2n}=\underset{n\to \infty }{\lim }\frac{n^5+n^4-n^2}{n^5+4n^3-2n}=1$$ Since $0<1<\infty$, the series share the same convergence behavior. $\sum \frac{1}{n^2}$ converges, so the original series converges. (c) The Direct Comparison Test cannot easily give this conclusion. For all $n\ge 1$, $n^5+4n^3-2n<n^5+4n^3<5n^5$ for $n\ge 1$, so $\frac{n^3+n^2-1}{n^5+4n^3-2n}>\frac{n^3}{5n^5}=\frac{1}{5n^2}$. This means the original series is larger than a convergent series, which gives no information about its own convergence, so Direct Comparison is not useful here.

Question 3 (Application / Real-World Style)

An environmental engineer models the total amount of nutrient runoff from a growing watershed over time as the infinite series $R={\sum }_{n=1}^{\infty }\frac{100\sqrt{n}}{n^2+2}$ units of kg per year. Does the total long-term nutrient runoff converge to a finite value? If it converges, find an upper bound for the total runoff using the Direct Comparison Test.

Worked Solution: For all $n\ge 1$, $\sqrt{n}/(n^2+2)<\sqrt{n}/n^2=1/n^{3/2}$, so: $$R=100\sum _{n=1}^{\infty }\frac{\sqrt{n}}{n^2+2}<100\sum _{n=1}^{\infty }\frac{1}{n^{3/2}}$$ The comparison series $\sum 1/n^{3/2}$ is a convergent p-series with $p=3/2>1$, so by Direct Comparison, the total runoff converges. The sum of the p-series is approximately 2.612, so the upper bound is $100\times 2.612=261.2$ kg. In context: the total long-term nutrient runoff from the watershed approaches a finite value of at most 261.2 kg.

7. Quick Reference Cheatsheet

8. What's Next

Comparison tests are the foundational tool for analyzing the behavior of series with non-elementary general terms that cannot be easily evaluated with the nth Term Test or Integral Test. Mastery of comparison tests is an absolute prerequisite for the next topics in Unit 10, starting with absolute and conditional convergence, followed by the Ratio and Root Tests used to find intervals of convergence for power series. Without understanding how to compare the asymptotic behavior of two series, you will struggle to interpret inconclusive test results and justify convergence conclusions on FRQs. This topic also forms the basis for bounding the error of partial sums of convergent Taylor series, a common FRQ question on the AP exam.

Alternating Series Convergence Ratio and Root Tests for Convergence Absolute and Conditional Convergence Taylor and Maclaurin Series