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How do you bound the error when approximating an alternating series by a partial sum?

Topic 10.10 Alternating Series Error Bound: bound the error of a partial-sum approximation of an alternating series by the magnitude of the first omitted term (BC).

A focused answer to AP Calculus BC Topic 10.10, bounding the error of approximating a convergent alternating series by a partial sum using the magnitude of the first omitted term, with worked examples.

Generated by Claude Opus 4.88 min answer

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  1. What this topic is asking
  2. The bound
  3. Why the next term bounds the error
  4. A worked error estimate
  5. Choosing how many terms for a given accuracy
  6. Limits of the bound

What this topic is asking

The College Board (Topic 10.10, BC only) gives the alternating series error bound, a remarkably simple way to control the error when you approximate the sum of a convergent alternating series by stopping after finitely many terms. The error is no larger than the first term you left out.

The bound

Why the next term bounds the error

The bound comes straight from the oscillation that makes alternating series converge (Topic 10.7). Because the partial sums overshoot and undershoot the true sum SS in ever-smaller steps, the true sum is always trapped between two consecutive partial sums SnS_n and Sn+1S_{n+1}. The gap between those partial sums is exactly the next term bn+1b_{n+1}, so SS sits within bn+1b_{n+1} of SnS_n. That is the whole proof in one sentence: the limit lies between consecutive partial sums, which differ by bn+1b_{n+1}. This geometric trapping is why no derivative bounds or factorials are needed, unlike the general Taylor error bound.

A worked error estimate

Choosing how many terms for a given accuracy

A common reverse question asks how many terms you need so the approximation is accurate to within a stated tolerance ε\varepsilon. Since the error after nn terms is at most bn+1b_{n+1}, you require bn+1εb_{n+1} \le \varepsilon and solve for the smallest such nn. For (1)n+1n\sum\frac{(-1)^{n+1}}{n} to be accurate within 0.010.01, you need 1n+10.01\frac{1}{n+1} \le 0.01, so n+1100n + 1 \ge 100, meaning n99n \ge 99 terms. This direction is just the bound rearranged, and it is a standard free-response setup. The work expected is the inequality bn+1εb_{n+1} \le \varepsilon and the resulting value of nn.

Limits of the bound

The alternating series error bound is only valid when the series genuinely alternates and satisfies the test (decreasing magnitudes tending to zero). For a non-alternating series, or for the error of a general Taylor polynomial whose series is not alternating at the point in question, you must use the Lagrange error bound (Topic 10.12) instead. Many Taylor approximations do happen to be alternating (for example sinx\sin x, cosx\cos x, and ln(1+x)\ln(1+x) at suitable points), and there the simpler alternating bound applies, which is why these often appear together. Recognizing whether the series you are truncating is alternating decides which error tool to reach for.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2022 (BC, style)1 marksSection I (multiple choice, no calculator). For n=1(1)n+1n\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}, the error in approximating the sum by the first 4 terms is at most (A) 15\frac{1}{5} (B) 14\frac{1}{4} (C) 120\frac{1}{20} (D) 12\frac{1}{2}
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The correct answer is (A), 15\frac{1}{5}.

For a convergent alternating series, the error after nn terms is at most the magnitude of the first omitted term. The 5th term has magnitude 15\frac{1}{5}, so the error is at most 15\frac{1}{5}.

AP 2024 (BC, style)4 marksSection II (free response, no calculator). Consider n=0(1)n(2n+1)!\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} (which equals sin1\sin 1). (a) Approximate the sum using the first two terms. (b) Bound the error of this approximation.
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A 4-point error-bound problem.

(a) (2 points) First two terms (n=0,1n = 0, 1): 11!13!=116=56\frac{1}{1!} - \frac{1}{3!} = 1 - \frac{1}{6} = \frac{5}{6}.
(b) (2 points) The first omitted term (n=2n = 2) has magnitude 15!=1120\frac{1}{5!} = \frac{1}{120}, so the error is at most 1120\frac{1}{120}.

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