Comparison tests for convergence

AP Calculus BC· AP Calculus BC CED — Infinite Sequences and Series· 14 min read

1. Overview of Comparison Tests★★☆☆☆⏱ 2 min

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.

2. The Direct Comparison Test★★☆☆☆⏱ 4 min

📘 Definition

Direct Comparison Test

$0 \leq a_{n} \leq b_{n}$ for all $n \geq N$ , for some finite $N$

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.

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.

📐 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

3. The Limit Comparison Test★★★☆☆⏱ 5 min

📘 Definition

Limit Comparison Test

$lim_{n \to \infty} \frac{a _{n}}{b _{n}} = L$ , for non-negative $a_{n}, b_{n}$

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.

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 $n$ .

📐 Worked Example

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

1
For large $n$ , the leading term of the numerator is $2 n$ , and the leading term of the denominator is $n^{3}$ , so $a_{n} \approx \frac{2 n}{n ^{3}} = \frac{2}{n ^{2}}$ . 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.

4. Extending Comparison Tests to Series with Negative Terms★★★☆☆⏱ 3 min

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 $n \geq 1$ , $2^{n} + 1 > 2^{n}$ , 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.

5. AP-Style Concept Check★★☆☆☆⏱ 2 min

✓ Quick check

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

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 $\sum (\frac{4}{5})^{n}$
- B. The series converges by the Direct Comparison Test with the convergent geometric series $\sum (\frac{4}{3})^{n}$
- C. The series diverges by the Direct Comparison Test with the divergent geometric series $\sum (\frac{4}{5})^{n}$
- D. The series diverges by the nth Term Test, because $lim_{n \to \infty} \frac{4 ^{n}}{3 ^{n} + 5 ^{n}}$ does not exist

6. Common Pitfalls

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 $\sum \frac{1}{n ( n + 3 )}$ 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.

7. Quick Reference 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.

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · BC MCQ
Convergence conclusion via comparison
2021 · BC FRQ
Justify convergence with comparison

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.