Alternating series test for convergence — AP Calculus BC Study Guide

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: Identification of alternating series, statement and application of the Alternating Series Test (Leibniz Test), conditions for convergence, alternating series remainder estimation, and classifying alternating series convergence as absolute or conditional.

You should already know: Limits of infinite sequences. nth term test for divergence. Convergence tests for positive-term series (p-test, integral test).

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 Alternating series test for convergence?

An alternating series is any infinite series where consecutive terms alternate in sign, most commonly written as ${\sum }_{n=1}^{\infty }(-1{)}^n{a}_n$ or ${\sum }_{n=1}^{\infty }(-1{)}^{n+1}{a}_n$ where $a_n>0$ for all $n$. The Alternating Series Test (often called Leibniz’s Test for Alternating Series) is the primary convergence test for this class of series, and it is a required topic in the AP Calculus BC Course and Exam Description (CED). It accounts for approximately 1-2% of the total AP exam score, which translates to 1-2 multiple-choice questions or one weighted part of a free-response question. It appears on both MCQ and FRQ sections of the exam, most frequently paired with concepts like error estimation, convergence classification, or endpoint testing for power series intervals of convergence. Unlike tests for positive-term series, the Alternating Series Test only applies to sign-alternating series, and it can only confirm convergence (not divergence, except indirectly via the nth term test). It relies on the intuitive idea that alternating positive and negative terms cancel each other, leading to shrinking partial sum oscillations if the magnitude of terms approaches zero.

2. Statement and Conditions of the Alternating Series Test

To apply the Alternating Series Test (AST), you must first rewrite any alternating series in the standard form ${\sum }_{n=k}^{\infty }(-1{)}^n{a}_n$ or ${\sum }_{n=k}^{\infty }(-1{)}^{n+1}{a}_n$, where $a_n>0$ for all $n\ge k$. This means $a_n$ is always the positive magnitude of the nth term of the series, not the full signed term. The test states that the alternating series converges if and only if both of the following conditions are satisfied for all sufficiently large $n$ (i.e., after some finite starting index $N$):

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

The intuition behind these conditions is straightforward: partial sums of an alternating series alternate above and below the true total sum of the series. If the terms are strictly decreasing in magnitude, each oscillation is smaller than the last, so the distance between consecutive partial sums shrinks. If the terms also approach zero, the oscillations shrink to zero, forcing the partial sums to converge to a finite limit. If the second condition fails (i.e., ${\lim }_{n\to \infty }{a}_n\ne 0$), the nth term test for divergence applies, and the series definitely diverges. If the first condition fails but the second holds, the test is inconclusive, and you must use another test to determine convergence.

Worked Example

Problem: 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 $a_n$: The series is ${\sum }_{n=1}^{\infty }(-1{)}^{n+1}{a}_n$ where $a_n=\frac{1}{3n+2}>0$ for all $n\ge 1$, so it meets the definition of an alternating series.
2. Check the decreasing condition: Compare $a_{n+1}$ to $a_n$: $a_{n+1}=\frac{1}{3(n+1)+2}=\frac{1}{3n+5}<\frac{1}{3n+2}=a_n$ for all $n$, so the sequence is strictly decreasing for all terms.
3. Check the limit condition: Evaluate ${\lim }_{n\to \infty }{a}_n={\lim }_{n\to \infty }\frac{1}{3n+2}=0$, so the second condition is satisfied.
4. Conclusion: Both conditions of the Alternating Series Test are met, so the series converges.

Exam 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.

3. Alternating Series Remainder Estimation

Once you confirm an alternating series converges via the Alternating Series Test, the AP exam frequently asks you to bound the error (called the remainder) when approximating the total sum $S$ by a finite partial sum $S_n$. For any convergent alternating series that satisfies the two AST conditions, the Alternating Series Remainder Theorem gives a simple, exam-friendly bound on this error: $$|R_n|=|S-S_n|\le a_{n+1}$$ where $a_{n+1}$ is the first term omitted from the partial sum $S_n$. Additionally, the sign of the error matches the sign of the first omitted term, which tells you if your partial sum is an overestimate or underestimate: if the first omitted term is positive, $S_n<S$ (underestimate), and if it is negative, $S_n>S$ (overestimate). This theorem is heavily tested because it assesses both understanding of convergence and practical estimation skills, which is a key CED learning objective. It is also much simpler to apply than integral remainder bounds for positive-term series.

Worked Example

Problem: 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 $n\ge 1$, and ${\lim }_{n\to \infty }{a}_n=0$, so the remainder theorem applies.
2. Set up the error inequality: We need $|R_n|\le a_{n+1}<0.002=\frac{1}{500}$. Substitute $a_{n+1}=\frac{1}{(n+1{)}^4}$ to get $\frac{1}{(n+1{)}^4}<\frac{1}{500}$.
3. Solve for $n$: Reverse the inequality (since all terms are positive): $(n+1{)}^4>500$. Calculate powers: $4^4=256<500$, $5^4=625>500$, so $n+1=5$ which means $n=4$.
4. Conclusion: 4 terms are required to get an error less than $0.002$.

Exam 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.

4. Classifying Convergence: Absolute vs Conditional

After confirming an alternating series converges, the next step (frequently tested on the AP exam) is to classify its convergence as either absolute or conditional. A series ${\sum }_{n=1}^{\infty }{b}_n$ is called 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 called conditionally convergent. The Alternating Series Test is the primary tool for establishing conditional convergence, because many common alternating series (like the alternating harmonic series) fall into this category. For example, the alternating harmonic series ${\sum }_{n=1}^{\infty }\frac{(-1{)}^{n+1}}{n}$ converges by AST, but the harmonic series ${\sum }_{n=1}^{\infty }\frac{1}{n}$ diverges, so it is conditionally convergent. This distinction is important for power series, where the interval of convergence often includes conditionally convergent endpoints.

Worked Example

Problem: 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 $n\ge 1$. $a_{n+1}=\frac{1}{(n+1{)}^{1/3}}<\frac{1}{n^{1/3}}=a_n$, and ${\lim }_{n\to \infty }\frac{1}{n^{1/3}}=0$, so by AST the original series converges.
2. Check convergence of the series of absolute values: ${\sum }_{n=1}^{\infty }\left|\frac{(-1{)}^{n+1}}{n^{1/3}}\right|={\sum }_{n=1}^{\infty }\frac{1}{n^{1/3}}$, which is a p-series with $p=\frac{1}{3}<1$. All p-series with $p\le 1$ diverge, so the series of absolute values diverges.
3. Conclusion: The original series converges, but the series of absolute values diverges, so the series is conditionally convergent.

Exam 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.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Applying the Alternating Series Test 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 $\sum \frac{(-1{)}^n}{n-0.2n^2}$ diverges because $a_1>a_2$ but $a_2<a_3<...$ for all $n\ge 2$. 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 ${\lim }_{n\to \infty }{a}_n\ne 0$, 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. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

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 ${\lim }_{n\to \infty }\frac{n}{e^{n^2}}\ne 0$

Worked Solution: First, rewrite the series in standard AST form with $a_n=\frac{n}{e^{n^2}}>0$ for all $n\ge 1$. We calculate ${\lim }_{n\to \infty }\frac{n}{e^{n^2}}=0$, since exponential growth dominates polynomial growth, so option D is eliminated. Next, to check if $a_n$ is decreasing, we take the derivative of $f(x)=\frac{x}{e^{x^2}}$, which gives $f^{\prime }(x)=\frac{1-2x^2}{e^{x^2}}<0$ for all $x\ge 1$, so $a_n$ is decreasing, and AST confirms the series converges, eliminating option A. Now check the series of absolute values ${\sum }_{n=1}^{\infty }\frac{n}{e^{n^2}}$, which converges by the integral test (the integral from 1 to infinity of $\frac{x}{e^{x^2}}dx$ evaluates to a finite value). Since the series of absolute values converges, the original series converges absolutely. The correct answer is B.

Question 2 (Free Response)

Consider the series ${\sum }_{n=2}^{\infty }\frac{(-1{)}^n}{n\ln n}$. (a) Show that the series converges by the Alternating Series Test. (b) Classify the convergence of the series as absolute or conditional. Justify your answer. (c) Estimate the sum of the series with an error less than $0.1$. How many terms are required? State the error bound.

Worked Solution: (a) First, write $a_n=\frac{1}{n\ln n}>0$ for all $n\ge 2$, so the series is a valid alternating series. For all $n\ge 2$, $(n+1)\ln (n+1)>n\ln n$, so $a_{n+1}=\frac{1}{(n+1)\ln (n+1)}<\frac{1}{n\ln n}=a_n$, so the sequence is strictly decreasing. The limit ${\lim }_{n\to \infty }\frac{1}{n\ln n}=0$, so both conditions of AST are satisfied, and the series converges.

(b) The series of absolute values is ${\sum }_{n=2}^{\infty }\frac{1}{n\ln n}$. By the integral test, ${\int }_2^{\infty }\frac{1}{x\ln x}dx={\lim }_{b\to \infty }\ln (\ln x){|}_2^b=\infty$, so the series of absolute values diverges. Therefore, the original series is conditionally convergent.

(c) We need $|R_n|\le a_{n+1}<0.1$, so $\frac{1}{(n+1)\ln (n+1)}<0.1$, which rearranges to $(n+1)\ln (n+1)>10$. Testing values: $5\ln 5\approx 8.05<10$, $6\ln 6\approx 10.75>10$, so $n+1=6$ and $n=5$. We need 5 terms total (from $n=2$ to $n=6$), and the maximum error is less than $0.1$.

Question 3 (Application / Real-World Style)

In signal processing, the approximation of a periodic alternating square wave with peak amplitude 8 volts is given by the infinite series $V(0)=\frac{32}{\pi }{\sum }_{k=0}^{\infty }\frac{(-1{)}^k}{2k+1}$ at time $t=0$. Use the Alternating Series Remainder Theorem to find the minimum number of terms needed to approximate $V(0)$ to within $0.05$ volts of the true value.

Worked Solution: Confirm the series satisfies AST conditions: $a_k=\frac{1}{2k+1}>0$, $a_{k+1}<a_k$, and ${\lim }_{k\to \infty }{a}_k=0$, so the remainder bound applies. We require the total error to be less than $0.05$ volts, so $\frac{32}{\pi }|R_n|\le \frac{32}{\pi }{a}_n<0.05$. Substitute $a_n=\frac{1}{2n+1}$ to get $\frac{32}{\pi (2n+1)}<0.05$. Rearranging gives $2n+1>\frac{32}{0.05\pi }\approx 203.7$, so $2n>202.7$, so $n>101.35$. Therefore, we need 102 terms to meet the error requirement. In context, this means approximating the voltage of the square wave at $t=0$ with 102 terms gives a value within 0.05 volts of the true voltage.

7. Quick Reference Cheatsheet

8. What's Next

Mastering the Alternating Series Test is a critical prerequisite for the remaining topics in Unit 10 (Infinite Sequences and Series). You will next apply AST to classify conditional convergence when using the 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 FRQ 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 will cost you significant points on the exam.

Conditional and absolute convergence Ratio test for convergence Power series interval of convergence Integral test for convergence