Alternating series test for convergence

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

1. The Alternating Series Test: Definition and Conditions★★☆☆☆⏱ 4 min

An alternating series is any infinite series where consecutive terms alternate in sign. The Alternating Series Test (AST, also called Leibniz's Test) is the standard convergence test for this series type, and it is frequently tested on AP Calculus BC, appearing in both multiple-choice and free-response sections.

📘 Definition

Standard Form of an Alternating Series

An alternating series is written as $\sum_{n = k}^{\infty} (- 1)^{n} a_{n}$ or $\sum_{n = k}^{\infty} (- 1)^{n + 1} a_{n}$ , where $a_{n} > 0$ for all $n \geq k$ . $a_{n}$ is always the positive magnitude of the nth term, not the full signed term.

The Alternating Series Test states that an alternating series converges if both of the following conditions hold for all sufficiently large $n$ (after some finite starting index $N$ ):

The sequence of positive terms ${a_{n}}$ is eventually strictly decreasing: $a_{n + 1} < a_{n}$ for all $n > N$
The limit of the positive terms approaches zero: $lim_{n \to \infty} a_{n} = 0$

📐 Worked Example

Does the series $\sum_{n = 1}^{\infty} \frac{( - 1 ) ^{n + 1}}{3 n + 2}$ converge or diverge? Justify your answer using the Alternating Series Test.

1
Rewrite the series in standard form to isolate $a_{n}$ :
2
$a_n = \frac{1}{3n + 2} > 0 \text{ for all } n \geq 1$
3
This confirms we have a valid alternating series eligible for AST.
4
Check the decreasing condition by comparing $a_{n + 1}$ to $a_{n}$ :
5
$a_{n+1} = \frac{1}{3(n+1) + 2} = \frac{1}{3n + 5} < \frac{1}{3n + 2} = a_n$
6
The sequence is strictly decreasing for all terms, so the first condition is satisfied.
7
Check the limit condition:
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$\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{1}{3n + 2} = 0$
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The second condition is satisfied. Conclusion: both conditions of AST are met, so the series converges.

2. Alternating Series Remainder Estimation★★★☆☆⏱ 4 min

After confirming an alternating series converges via AST, the AP exam often asks to bound the error (called the remainder) when approximating the total sum $S$ with a finite partial sum $S_{n}$ . The Alternating Series Remainder Theorem gives a simple, exam-friendly bound for this error.

📘 Definition

Alternating Series Remainder Theorem

|R_n| = |S - S_n|

For any convergent alternating series that meets both AST conditions, the absolute error of approximation with $n$ terms is bounded by the magnitude of the first omitted term: $∣ R_{n} ∣ \leq a_{n + 1}$ . The sign of the error matches the sign of the first omitted term, so you can determine if $S_{n}$ is an overestimate or underestimate.

📐 Worked Example

How many terms of the convergent series $\sum_{n = 1}^{\infty} \frac{( - 1 ) ^{n + 1}}{n ^{4}}$ must be used to estimate the sum with an error less than $0.002$ ?

1
Confirm the series meets AST conditions: $a_{n} = \frac{1}{n ^{4}} > 0$ , $a_{n + 1} < a_{n}$ for all $n \geq 1$ , and $lim_{n \to \infty} a_{n} = 0$ , so the remainder theorem applies.
2
Set up the error inequality: we require $∣ R_{n} ∣ \leq a_{n + 1} < 0.002 = \frac{1}{500}$ :
3
$\frac{1}{(n+1)^4} < \frac{1}{500}$
4
Solve for $n$ : reverse the inequality (all terms are positive) to get $(n + 1)^{4} > 500$ . We calculate $4^{4} = 256 < 500$ and $5^{4} = 625 > 500$ , so $n + 1 = 5$ , which means $n = 4$ .
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Conclusion: 4 terms are required to get an error less than $0.002$ .

3. Classifying Convergence: Absolute vs Conditional★★★☆☆⏱ 4 min

After confirming an alternating series converges, the AP exam almost always asks to classify its convergence as either absolute or conditional. This distinction is especially important for finding the interval of convergence of power series, where endpoints are often conditionally convergent.

📘 Definition

Absolute vs Conditional Convergence

A series $\sum_{n = 1}^{\infty} b_{n}$ is absolutely convergent if the series of absolute values $\sum_{n = 1}^{\infty} ∣ b_{n} ∣$ converges. If the original alternating series converges, but the series of absolute values diverges, the series is conditionally convergent.

📐 Worked Example

Classify the convergence of the series $\sum_{n = 1}^{\infty} \frac{( - 1 ) ^{n + 1}}{3 n}$ . Justify your answer.

1
Check convergence of the original alternating series: write $a_{n} = \frac{1}{n ^{1/3}} > 0$ for all $n \geq 1$ . We confirm $a_{n + 1} = \frac{1}{( n + 1 ) ^{1/3}} < \frac{1}{n ^{1/3}} = a_{n}$ , and $lim_{n \to \infty} \frac{1}{n ^{1/3}} = 0$ . By AST, the original series converges.
2
Check convergence of the series of absolute values:
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$\sum_{n=1}^\infty \left| \frac{(-1)^{n+1}}{n^{1/3}} \right| = \sum_{n=1}^\infty \frac{1}{n^{1/3}}$
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This is a p-series with $p = \frac{1}{3} < 1$ . All p-series with $p \leq 1$ diverge, so the series of absolute values diverges.
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Conclusion: The original series converges, but the series of absolute values diverges, so the series is conditionally convergent.

4. Exam-Style Concept Check★★★★☆⏱ 2 min

✓ Quick check

Test your understanding of AST with this AP-style multiple choice question:

Which of the following statements about the series $\sum_{n = 1}^{\infty} \frac{( - 1 ) ^{n} n}{e ^{n^{2}}}$ is true?
- A) The series diverges by the Alternating Series Test
- B) The series converges absolutely
- C) The series converges conditionally but not absolutely
- D) The series diverges because $lim_{n \to \infty} \frac{n}{e ^{n^{2}}} \neq = 0$
Reveal answer
B —
First, confirm $a_{n} = \frac{n}{e ^{n^{2}}} > 0$ , $lim_{n \to \infty} a_{n} = 0$ (eliminating D), and $a_{n}$ is decreasing for all $n \geq 1$ , so AST confirms convergence (eliminating A). The series of absolute values converges by the integral test, so the original series converges absolutely.

5. Common Pitfalls

Wrong move:

Applying AST to a series with non-positive $a_{n}$ , e.g. testing $\sum \frac{( - 1 ) ^{n} n}{n + 1}$ and claiming $a_{n} = \frac{( - 1 ) ^{n} n}{n + 1}$ then checking $lim a_{n} = 0$ .

Why:

Students confuse the signed overall terms with the $a_{n}$ defined in AST, which must be the positive magnitude of each term.

Correct move:

Always factor out $(- 1)^{n}$ or $(- 1)^{n + 1}$ first to isolate $a_{n}$ , and confirm $a_{n} > 0$ for all sufficiently large $n$ before applying AST.

Wrong move:

Concluding divergence because $a_{n}$ is not decreasing for the first 2-3 terms, e.g. claiming $\sum \frac{( - 1 ) ^{n}}{n - 0.2 n ^{2}}$ diverges because $a_{1} > a_{2}$ but $a_{2} < a_{3} < ...$ for all $n \geq 2$ .

Why:

Students forget convergence only depends on the behavior of terms for large $n$ , and early terms do not affect convergence.

Correct move:

If $a_{n}$ is not decreasing at $n = 1$ , check if it becomes decreasing after some finite $N$ —if it does, the first condition is still satisfied.

Wrong move:

Concluding convergence by AST when $lim_{n \to \infty} a_{n} \neq = 0$ , even if $a_{n}$ is decreasing.

Why:

Students memorize the decreasing condition but forget the limit condition is required for convergence.

Correct move:

Always check $lim_{n \to \infty} a_{n}$ first—if the limit is non-zero, stop and conclude divergence via the nth term test.

Wrong move:

Using the alternating series remainder bound when the series does not satisfy both AST conditions.

Why:

The bound only applies to convergent alternating series, not divergent alternating series or positive-term series.

Correct move:

Only apply the alternating series error bound after you confirm both AST conditions are satisfied and the series converges.

Wrong move:

Classifying an alternating series as absolutely convergent just because it converges by AST.

Why:

Students confuse convergence of the alternating series with convergence of the series of absolute values, which is required for absolute convergence.

Correct move:

After confirming the alternating series converges, always test the series of absolute values with a positive-term test (p-test, integral test, ratio test) to classify convergence.

6. Quick Reference Cheatsheet

Category	Key Rule	Notes
Standard Form	$\sum (- 1)^{n} a_{n}$ / $\sum (- 1)^{n + 1} a_{n}$	$a_{n} > 0$ for all $n \geq k$ by definition
AST Conditions	$a_{n}$ eventually strictly decreasing 2. $lim_{n \to \infty} a_{n} = 0$	Both conditions required; non-zero limit implies divergence
Remainder Bound	$∣ S - S_{n} ∣ = ∣ R_{n} ∣ \leq a_{n + 1}$	Only applies to convergent alternating series meeting AST conditions
Over/Underestimate	Error matches sign of first omitted term	Positive omitted term = $S_{n}$ is underestimate; negative = overestimate
Absolute Convergence	Original converges if $\sum ∣ a_{n} ∣$ converges	Absolute convergence always implies original series converges
Conditional Convergence	Original converges, $\sum ∣ a_{n} ∣$ diverges	Most conditionally convergent series are proven via AST
Alternating p-series	$\sum_{n = 1}^{\infty} \frac{( - 1 ) ^{n + 1}}{n ^{p}}$	Converges for $p > 0$ ; absolutely convergent if $p > 1$ , conditional if $0 < p \leq 1$

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.

2022 · BC
Endpoint convergence classification
2023 · BC FRQ
Error bound calculation
2021 · BC MCQ
Convergence type identification

What's Next

Mastering the Alternating Series Test is a critical prerequisite for all remaining topics in Unit 10 of AP Calculus BC. You will next use AST to confirm conditional convergence when working with ratio and root tests, which are used to test absolute convergence for general series. Most importantly, AST is required to test convergence of endpoints of the interval of convergence for power series, which is a common full free-response question on the AP exam. Without mastering AST and its remainder bound, you will not be able to correctly identify the full interval of convergence or justify endpoint convergence, which can cost you significant points on the exam. Building a solid understanding of AST now will make more advanced series topics much easier to master.