Alternating series error bound — AP Calculus BC Study Guide

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: Alternating Series Remainder Theorem (Leibniz error bound), pre-requisite conditions from the Alternating Series Test, maximum error calculation, bounding the exact sum, and finding the number of terms needed for a given error tolerance.

You should already know: Alternating series definition and convergence via the Alternating Series Test, partial sums of infinite series, and basic summation notation.

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 error bound?

Alternating series error bound (also called the Alternating Series Remainder Theorem or Leibniz error bound) is a simple, powerful theorem that estimates the maximum possible error when you approximate the sum of a convergent alternating series with a finite partial sum. This topic is explicitly required by the AP Calculus BC Course and Exam Description (CED), and counts toward the 17-18% exam weight for Unit 10 (Infinite Sequences and Series). It appears regularly in both multiple-choice (MCQ) and free-response (FRQ) sections, and is often paired with Taylor polynomial approximation on FRQs. When an alternating series converges to an exact sum (S), the nth partial sum (S_n) (the sum of the first (n) terms) is an approximation of (S), but it will never be exactly equal for non-trivial series. Unlike other error bounds (such as the integral test remainder), the alternating series error bound requires no integration or complex calculation, relying only on the magnitude of the next unused term in the series. It is a high-yield concept that is straightforward to master with practice.

2. The Alternating Series Remainder Theorem: Statement and Conditions

To apply the alternating series error bound, we first write any alternating series in the standard form: $$\sum _{k=1}^{\infty }(-1{)}^{k+1}{a}_k\text{where }{a}_k>0\text{ for all }k$$ For the error bound to be valid, the series must satisfy two conditions (the same conditions required for convergence via the Alternating Series Test):

1. (\lim_{k \to \infty} a_k = 0) (the nth term goes to zero)
2. The sequence ({a_k}) is strictly decreasing for all (k \geq n) (each term is smaller than the previous one, out to infinity)

If both conditions are met, the series converges to an exact sum (S), with nth partial sum (S_n = \sum_{k=1}^n (-1)^{k+1}a_k). The error (remainder) (R_n = S - S_n) satisfies the alternating series error bound: $$|R_n|=|S-S_n|\le a_{n+1}$$ Intuition: Partial sums of a convergent alternating series oscillate around the exact sum (S), getting closer with each new term. After stopping at (S_n), the next term moves you closer to (S), so the error can never be larger than the size of that next (first neglected) term.

Worked Example

Problem: Consider the alternating series (\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^2}). Does the alternating series error bound apply to an approximation using the first 4 terms? If yes, what is the maximum possible error?

Solution:

1. Rewrite the series in standard form: (a_k = \frac{1}{k^2}), which is positive for all (k \geq 1).
2. Check the first condition: (\lim_{k \to \infty} a_k = \lim_{k \to \infty} \frac{1}{k^2} = 0), so the first condition is satisfied.
3. Check the second condition: (a_{k+1} = \frac{1}{(k+1)^2} < \frac{1}{k^2} = a_k) for all (k \geq 1), so ({a_k}) is strictly decreasing, and the second condition is satisfied.
4. For (n=4) terms, the first neglected term is (a_5 = \frac{1}{5^2} = 0.04). By the error bound, (|R_4| \leq 0.04).

Conclusion: The error bound applies, and the maximum error is 0.04.

Exam tip: Always explicitly confirm the two conditions (decreasing sequence, limit zero) before applying the error bound on FRQs—AP exam readers require this justification to award full credit.

3. Constructing an Interval for the Exact Sum

Once you have the partial sum (S_n) and the error bound, you can construct an interval that is guaranteed to contain the exact sum (S), which is a common AP exam question. In addition to the magnitude of the error, we know one extra useful fact: the error (R_n) has the same sign as the first neglected term. This lets us construct a tighter interval than the symmetric interval (S_n - a_{n+1} \leq S \leq S_n + a_{n+1}), which is what exam questions almost always expect.

For example: if the first neglected term is positive, then (R_n > 0), so (S = S_n + R_n > S_n), and since (|R_n| \leq a_{n+1}), we get (S_n < S \leq S_n + a_{n+1}). If the first neglected term is negative, we get (S_n - a_{n+1} \leq S < S_n). This interval is half the width of the symmetric interval, and it is always correct when the error bound conditions are met.

Worked Example

Problem: For the series (\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{2^k k}), find the tightest possible interval that contains the exact sum (S) using the first 3 terms.

Solution:

1. Calculate the 3rd partial sum: $$S_3=\frac{1}{2^{1}1}-\frac{1}{2^{2}2}+\frac{1}{2^{3}3}=\frac{1}{2}-\frac{1}{8}+\frac{1}{24}=\frac{10}{24}\approx 0.4167$$
2. Confirm error bound conditions: (a_k = \frac{1}{2^k k}) is positive, strictly decreasing, and (\lim_{k\to\infty}a_k=0), so the bound applies. The first neglected term is (a_4 = \frac{1}{2^4 4} = 0.015625).
3. Find the sign of the first neglected term: The 4th term of the series is ((-1)^{4+1}a_4 = -a_4), so it is negative, meaning (R_3 = S - S_3) is also negative, so (S < S_3).
4. Combine the magnitude and sign: (-a_4 \leq R_3 < 0), so adding (S_3) to all parts gives (S_3 - a_4 < S < S_3). Substituting values: $$0.4167-0.015625<S<0.4167⟹0.401<S<0.417$$

Conclusion: The tightest interval containing (S) is approximately ((0.401, 0.417)).

Exam tip: If an AP question asks for an interval containing the exact sum, always use the sign of the first neglected term to get a tight interval—this matches the expected answer 9 times out of 10, and avoids unnecessarily wide intervals that will be marked incorrect.

4. Finding the Number of Terms for a Given Error Tolerance

A common AP exam problem asks for the minimum number of terms needed to approximate the sum of an alternating series to within a given maximum error (E). The alternating series error bound makes this problem straightforward: we need the error (|R_n|) to be less than (E), and since (|R_n| \leq a_{n+1}), we just need to find the smallest integer (n) such that (a_{n+1} < E). Remember that (n) is the number of terms, and (n+1) is the index of the first neglected term—this is the source of the most common mistake on this type of problem.

Worked Example

Problem: How many terms of the alternating series (\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k!}) are needed to approximate the sum to within 0.001?

Solution:

1. Confirm conditions: (a_k = \frac{1}{k!}) is positive, strictly decreasing for all (k \geq 1), and (\lim_{k\to\infty}a_k=0), so the error bound applies. We need (|R_n| < 0.001), which requires (a_{n+1} < 0.001).
2. Set up the inequality: $$\frac{1}{(n+1)!}<0.001⟹(n+1)!>1000$$
3. Calculate factorials to find the smallest (n+1): (6! = 720 < 1000) and (7! = 5040 > 1000), so (n+1 = 7).
4. Solve for (n): (n = 7 - 1 = 6). Verify: (a_7 = \frac{1}{5040} \approx 0.0002 < 0.001), which meets the requirement, while (n=5) gives (a_6 = \frac{1}{720} \approx 0.0014 > 0.001), which does not.

Conclusion: 6 terms are needed to meet the error tolerance.

Exam tip: Always double-check what the question is asking for: if it asks for the number of terms, the answer is (n), not (n+1) (the index of the first neglected term)—this is the most common careless error on this problem type.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Applying the alternating series error bound to an alternating series where ({a_k}) is not decreasing for all (k \geq n). For example, applying it to (n=2) for the series (\sum_{k=1}^{\infty} (-1)^{k+1} \frac{k}{10^k}), where ({a_k}) only becomes decreasing after (k=3). Why: Students forget the decreasing condition applies to all terms after (n), not just the series as a whole. Correct move: Always confirm (a_{k+1} < a_k) for all (k \geq n) before applying the bound, and only use the bound for (n \geq N), where (N) is the first index where the sequence becomes decreasing.
• Wrong move: For a minimum terms problem, solving (a_n < E) instead of (a_{n+1} < E), leading to an answer that is one term too few. Why: Students confuse the index of the last term used with the index of the first neglected term. Correct move: Explicitly label your terms: if the last term you use is term (n), the first neglected term is term (n+1), so the inequality is always (a_{n+1} < E).
• Wrong move: Constructing a symmetric interval (S_n - a_{n+1} < S < S_n + a_{n+1}) when a question asks for the tightest possible interval. Why: Students memorize the magnitude bound but forget the sign rule that gives a tighter interval. Correct move: Always check the sign of the first neglected term and adjust the interval to be one-sided, as shown in Section 3.
• Wrong move: Applying the alternating series error bound to a divergent alternating series. For example, applying the bound to (\sum_{k=1}^{\infty} (-1)^{k+1} \frac{k}{k+1}), which diverges by the nth term test. Why: Students assume all alternating series are convergent. Correct move: Always confirm (\lim_{k\to\infty}a_k=0) (the first AST condition) before applying the error bound, since the theorem only applies to convergent series.
• Wrong move: Using the alternating series error bound for the remainder of a positive-term Taylor series. Why: Students confuse alternating series error bound with Lagrange error bound. Correct move: Only use alternating series error bound for convergent alternating series meeting the AST conditions; use Lagrange error bound for general Taylor series with non-alternating signs.

6. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

How many terms of the convergent alternating series (\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^3}) are needed to approximate the sum to within 0.01 of the exact value? A) 3 B) 4 C) 5 D) 10

Worked Solution: First, confirm (a_k = 1/k^3) meets the error bound conditions: it is positive, strictly decreasing, and has limit 0. We need the maximum error (|R_n| < 0.01), which by the error bound requires (a_{n+1} < 0.01). Set up the inequality: (\frac{1}{(n+1)^3} < 0.01 \implies (n+1)^3 > 100). Testing integer values: (4^3 = 64 < 100) and (5^3 = 125 > 100), so (n+1 = 5), which means (n = 4). The correct answer is B.

Question 2 (Free Response)

Consider the series (S = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{3^k k}). (a) Show that the series converges, and confirm that the alternating series error bound applies for any (n \geq 1). (b) Approximate (S) using the first 2 terms, and find the maximum error of this approximation. (c) Find the tightest possible interval that contains the exact value of (S) using your work from part (b).

Worked Solution: (a) Rewrite the series as (\sum_{k=1}^{\infty} (-1)^{k+1} a_k) where (a_k = \frac{1}{3^k k} > 0) for all (k \geq 1). Check AST conditions: 1. (\lim_{k \to \infty} \frac{1}{3^k k} = 0), since exponential growth dominates the linear term. 2. (a_{k+1} = \frac{1}{3^{k+1}(k+1)} < \frac{1}{3^k k} = a_k) for all (k \geq 1), so ({a_k}) is strictly decreasing. Both conditions are satisfied, so the series converges, and the error bound applies.

(b) For (n=2) terms, the partial sum is: $$S_2=\frac{1}{3^{1}1}-\frac{1}{3^{2}2}=\frac{1}{3}-\frac{1}{18}=\frac{5}{18}\approx 0.2778$$ The first neglected term is (a_3 = \frac{1}{3^3 3} = \frac{1}{81} \approx 0.0123). By the error bound, the maximum error is (|R_2| \leq 0.0123).

(c) The first neglected term (the 3rd term of the series) is ((-1)^{3+1}a_3 = +a_3), so it is positive, meaning (R_2 = S - S_2 > 0). We know (0 < R_2 \leq 0.0123), so adding (S_2) to all parts gives: $$\frac{5}{18}<S\le \frac{5}{18}+\frac{1}{81}=\frac{47}{162}\approx 0.2901$$ The tightest interval is approximately ((0.278, 0.290)).

Question 3 (Application / Real-World Style)

The voltage across a discharging capacitor in an RC circuit can be approximated by the alternating series: $$V(t)=V_0\sum _{k=1}^{\infty }\frac{(-1{)}^{k+1}(t/RC{)}^k}{k!}$$ for (t < RC), where (V_0 = 10) V, (RC = 1) s, and (t = 0.5) s. Electrical engineers need to approximate (V(t)) to within 0.01 V of the exact value. What is the minimum number of terms of the series needed, and what is the resulting approximation to the nearest hundredth of a volt? Interpret your result.

Worked Solution: Substitute the given values to get (V = \sum_{k=1}^{\infty} (-1)^{k+1} a_k), where (a_k = 10 \cdot \frac{(0.5)^k}{k!}). We need (|R_n| < 0.01), so (a_{n+1} < 0.01), which simplifies to (\frac{(0.5)^{n+1}}{(n+1)!} < 0.001). Testing values: for (n+1=4), (\frac{(0.5)^4}{4!} \approx 0.0026 > 0.001), and for (n+1=5), (\frac{(0.5)^5}{5!} \approx 0.00026 < 0.001), so (n+1=5) meaning (n=4) terms are needed. Calculate the 4th partial sum: $$S_4=10\left(\frac{0.5}{1!}-\frac{0.5^2}{2!}+\frac{0.5^3}{3!}-\frac{0.5^4}{4!}\right)\approx 3.93\text{ V}$$ Interpretation: With 4 terms, we get an approximate voltage of 3.93 V after 0.5 seconds, and we can guarantee the true voltage is within 0.01 V of this value, which meets the engineer's accuracy requirement.

7. Quick Reference Cheatsheet

8. What's Next

Alternating series error bound is a critical prerequisite for approximating sums of convergent series, and specifically for error analysis of alternating Taylor series, a commonly tested topic on the AP Calculus BC exam. After mastering this topic, you will next apply alternating series error bound to find the error in alternating Taylor polynomial approximations; for these series, it is much simpler to apply than the Lagrange error bound, and it is a frequent source of points on both MCQ and FRQ sections. Without a solid understanding of this chapter, you will struggle to justify error bounds for Taylor approximations, a high-weight FRQ skill. This topic also feeds into the broader study of series convergence and approximation, which makes up ~17% of the AP Calculus BC exam.

• Alternating series convergence test
• Lagrange error bound for Taylor polynomials
• Taylor polynomials and approximations
• Ratio test for absolute convergence