# Alternating series test for convergence

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

This study guide covers identification of alternating series, the Alternating Series Test (Leibniz Test) conditions and application, alternating series error estimation, and classification of absolute vs conditional convergence for AP Calculus BC.

**Prerequisites:** Limits of infinite sequences; nth term test for divergence; Convergence tests for positive-term series (p-test, integral test)

## Learning objectives

- Identify alternating series and isolate positive $a_n$ terms for testing
- Apply the Alternating Series Test to confirm convergence of alternating series
- Use the Alternating Series Remainder Theorem to bound estimation error
- Classify convergence of alternating series as absolute or conditional

## The Alternating Series Test: Definition and Conditions

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.

**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 $N$):

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

**Worked example:** Does the series $\sum_{n=1}^\infty \frac{(-1)^{n+1}}{3n + 2}$ 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:
8. $$\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{1}{3n + 2} = 0$$
9. The second condition is satisfied. Conclusion: both conditions of AST are met, so the series converges.

> **tip**
>
> On FRQ, you must explicitly state and verify both conditions of AST to earn full credit. Failing to mention one condition will cost you a point, even if your final conclusion is correct.

## Alternating Series Remainder Estimation

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.

**Alternating Series Remainder Theorem** — 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.

*Notation:* |R_n| = |S - S_n|

**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 $n = 4$.
5. Conclusion: 4 terms are required to get an error less than $0.002$.

> **tip**
>
> When asked for the maximum possible error for a given partial sum, always write the exact value of $a_{n+1}$ as your bound. AP graders do not accept inequalities in place of the exact bound when the question asks for the maximum error.

## Classifying Convergence: Absolute vs Conditional

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.

**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 $\sum_{n=1}^\infty \frac{(-1)^{n+1}}{\sqrt[3]{n}}$. 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:
3. $$\sum_{n=1}^\infty \left| \frac{(-1)^{n+1}}{n^{1/3}} \right| = \sum_{n=1}^\infty \frac{1}{n^{1/3}}$$
4. 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.
5. Conclusion: The original series converges, but the series of absolute values diverges, so the series is conditionally convergent.

> **tip**
>
> Never skip checking convergence of the original series before classifying. If the original series diverges, it cannot be classified as either absolutely or conditionally convergent.

## Exam-Style Concept Check

**Check your understanding**

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

1. 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 $\lim_{n \to \infty} \frac{n}{e^{n^2}} \neq 0$

   *Why:* 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.

## Common pitfalls

- **Wrong:** 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 it fails: Students confuse the signed overall terms with the $a_n$ defined in AST, which must be the positive magnitude of each term.
  - Correct: 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:** Concluding divergence because $a_n$ is not decreasing for the first 2-3 terms, e.g. claiming $\sum \frac{(-1)^n}{n - 0.2n^2}$ diverges because $a_1 > a_2$ but $a_2 < a_3 < ...$ for all $n \geq 2$.
  - Why it fails: Students forget convergence only depends on the behavior of terms for large $n$, and early terms do not affect convergence.
  - Correct: 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:** Concluding convergence by AST when $\lim_{n \to \infty} a_n \neq 0$, even if $a_n$ is decreasing.
  - Why it fails: Students memorize the decreasing condition but forget the limit condition is required for convergence.
  - Correct: 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:** Using the alternating series remainder bound when the series does not satisfy both AST conditions.
  - Why it fails: The bound only applies to convergent alternating series, not divergent alternating series or positive-term series.
  - Correct: Only apply the alternating series error bound after you confirm both AST conditions are satisfied and the series converges.
- **Wrong:** Classifying an alternating series as absolutely convergent just because it converges by AST.
  - Why it fails: Students confuse convergence of the alternating series with convergence of the series of absolute values, which is required for absolute convergence.
  - Correct: 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.

## 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 | 1. $a_n$ eventually strictly decreasing<br>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 $0 < p \leq 1$ |

## 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.

- [Ratio Test for Convergence](https://www.owlsprep.com/study/ap-calculus-bc-u10-ratio-test-for-convergence/)
- [Integral Test for Convergence](https://www.owlsprep.com/study/ap-calculus-bc-u10-integral-test-for-convergence/)
- [Unit 10: Infinite Sequences and Series Overview](https://www.owlsprep.com/study/ap-calculus-bc-u10-overview/)

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