# Comparison tests for convergence

> AP Calculus BC · Unit 10: Infinite Sequences and Series
> Source: https://www.owlsprep.com/study/ap-calculus-bc-u10-comparison-tests-for-convergence/

This module covers the Direct Comparison Test and Limit Comparison Test for infinite series convergence, including extensions to series with negative terms via absolute convergence, with worked examples and exam tips for AP Calculus BC.

**Prerequisites:** Convergence and divergence definitions for infinite series; Convergence rules for p-series and geometric series; Basic limit laws for sequences

## Learning objectives

- State the conditions required for the Direct and Limit Comparison Tests
- Apply comparison tests to determine convergence of series with non-negative terms
- Extend comparison tests to test absolute convergence of series with mixed terms
- Identify and avoid common mistakes when applying comparison tests

## Overview of Comparison Tests

Comparison tests for convergence are a pair of related methods for determining whether an infinite series converges or diverges, by comparing an unknown series to a second series with known convergence behavior. This topic accounts for 17-18% of the total AP Calculus BC exam score, appearing regularly in both multiple-choice and free-response sections.

Unlike the Integral Test, which requires integrating 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 are initially designed for series with non-negative terms, with extensions to other series via absolute convergence.

## The Direct Comparison Test

**Direct Comparison Test** — For two non-negative series, if $\sum b_n$ converges, then $\sum a_n$ also converges. If $\sum a_n$ diverges, then $\sum b_n$ also diverges. The inequality only needs to hold for all sufficiently large $n$, as early terms do not change convergence behavior.

*Notation:* $0 \leq a_n \leq b_n$ for all $n \geq N$, for some finite $N$

*Example:* Best used when you can easily bound the unknown series terms above by a convergent series or below by a divergent series.

The intuition for this rule is straightforward: if a larger upper bound series does not diverge, then a smaller series bounded above by it cannot diverge either. Conversely, if the smaller lower bound series already diverges, any larger series must also diverge.

> **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.

**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 $n \geq 1$, which satisfies the test's core requirement.
2. For all $n \geq 1$, $n^2 + 5 > n^2$, so taking reciprocals preserves the inequality:
3. $$0 < \frac{1}{n^2 + 5} < \frac{1}{n^2}$$
4. The comparison series $\sum_{n=1}^{\infty} \frac{1}{n^2}$ is a p-series with $p=2 > 1$, which is known to converge.
5. 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 check inequality direction before drawing a conclusion

## The Limit Comparison Test

**Limit Comparison Test** — If $0 < L < \infty$, $\sum a_n$ and $\sum b_n$ have the same convergence behavior. For edge cases: if $L=0$ and $\sum b_n$ converges, $\sum a_n$ converges; if $L = \infty$ and $\sum b_n$ diverges, $\sum a_n$ diverges. All other edge cases are inconclusive.

*Notation:* $\lim_{n \to \infty} \frac{a_n}{b_n} = L$, for non-negative $a_n, b_n$

*Example:* Ideal for rational functions where bounding for Direct Comparison is algebraically messy.

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 perfect for rational functions of $n$, where the leading terms of the numerator and denominator dominate the behavior for large $n$.

> **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.

**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 $n \geq 3$ (after the first two terms, which do not affect convergence), so the test applies.
3. Compute the limit of the ratio:
4. $$\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{(2n + 1)/(n^3 - 4n + 2)}{1/n^2} = \lim_{n \to \infty} \frac{2n^3 + n^2}{n^3 - 4n + 2} = 2$$
5. 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.

## 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.

> **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.

**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 $n \geq 1$, $2^n + 1 > 2^n$, so $\frac{n}{2^n + 1} < \frac{n}{2^n}$. We use the Direct Comparison Test with $\sum \frac{n}{2^n}$, a known convergent series.
3. By Direct Comparison, $\sum |a_n|$ converges, so the original series converges (absolutely). The comparison approach confirms convergence.

## AP-Style Concept Check

**Check your understanding**

Test your understanding of comparison test rules with this multiple-choice question:

1. 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

   *Answer:* A. The series converges by the Direct Comparison Test with the convergent geometric series $\sum \left(\frac{4}{5}\right)^n$

   *Why:* Correct: $ \frac{4^n}{3^n + 5^n} < \left(\frac{4}{5}\right)^n $, which is a convergent geometric series. The other options are incorrect: $\sum (4/3)^n$ diverges, $\sum (4/5)^n$ converges, and the limit of the general term is 0.

## Common pitfalls

- **Wrong:** 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 it fails: Students memorize that comparison to p-series works, but forget the inequality direction is wrong for their desired conclusion.
  - Correct: 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:** Applying the Limit Comparison Test when $L=0$ and the comparison series diverges, then concluding the original series diverges.
  - Why it fails: Students generalize the $0 < L < \infty$ rule incorrectly to edge cases.
  - Correct: 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:** Applying comparison tests directly to an alternating series without taking absolute values first.
  - Why it fails: Students forget the non-negative term requirement for all comparison methods.
  - Correct: Any series with mixed positive/negative terms must be tested by applying comparison to the series of absolute values of its terms first.
- **Wrong:** Claiming divergence because your original series is larger than a convergent comparison series.
  - Why it fails: Students mix up what information each comparison gives.
  - Correct: 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:** Comparing $\sum \frac{1}{\sqrt{n}(n + 3)}$ to $\sum \frac{1}{n}$ instead of $\sum \frac{1}{n^{3/2}}$.
  - Why it fails: Students forget to add the exponents of $n$ in the denominator to find the correct p-series.
  - Correct: 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.

## Cheatsheet

| Category | Rule / Details | Notes |
| --- | --- | --- |
| Direct Comparison Test (non-negative terms) | $0 \leq a_n \leq b_n$ for all $n \geq N$ | If $\sum b_n$ converges, $\sum a_n$ converges; if $\sum a_n$ diverges, $\sum b_n$ diverges. Only needs to hold for large $n$. |
| Limit Comparison Test (non-negative terms) | $\lim_{n \to \infty} \frac{a_n}{b_n} = L$, $0 < L < \infty$ | Both series have the same convergence behavior. |
| Limit Comparison Edge 1: $L=0$ | Only conclude convergence if $\sum b_n$ converges | No conclusion if $\sum b_n$ diverges. |
| Limit Comparison Edge 2: $L=\infty$ | Only conclude divergence if $\sum b_n$ diverges | No conclusion if $\sum b_n$ converges. |
| All-negative terms | Apply tests to $\sum \|a_n\|$ | Convergence is identical to $\sum \|a_n\|$. |
| Mixed positive/negative terms | Test convergence of $\sum \|a_n\|$ | If $\sum \|a_n\|$ converges via comparison, original series converges. If $\sum \|a_n\|$ diverges, test is inconclusive. |
| Common convergent comparison series | p-series: $p>1$; geometric: $\|r\|<1$ | Go-to options for most AP problems. |
| Common divergent comparison series | p-series: $p \leq 1$; geometric: $\|r\| \geq 1$ | Use these to prove divergence. |

## What's next

Comparison tests are the foundational tool for analyzing the behavior of series with 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. 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 test for convergence](https://www.owlsprep.com/study/ap-calculus-bc-u10-alternating-series-test-for-convergence/)
- [Ratio Test for Convergence](https://www.owlsprep.com/study/ap-calculus-bc-u10-ratio-test-for-convergence/)
- [Determining absolute or conditional convergence](https://www.owlsprep.com/study/ap-calculus-bc-u10-determining-absolute-or-conditional-convergence/)

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