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AP Calculus BC · Unit 10: Infinite Sequences & Series

10.10 Alternating Series Error Bound

What the Error Bound Says

For a converging alternating series satisfying the Alternating Series Test conditions, if you approximate the sum using the partial sum S_n, the truncation error is bounded by the absolute value of the first omitted term: |S − S_n| ≤ |a_(n+1)|. This gives a quick way to measure accuracy without computing the exact infinite sum.

Requirements to Apply It

The bound only applies when the series is alternating, terms decrease in absolute value, and terms approach zero — the same conditions as the Alternating Series Test. If these hold, the partial sum's error is smaller than the magnitude of the next term you would add, and the true sum lies between consecutive partial sums.

Using It to Guarantee Accuracy

To approximate a sum within a given tolerance (like 0.001), find the smallest n such that |a_(n+1)| is less than that tolerance. This tells you how many terms are needed. It's commonly used with series like the alternating harmonic series or Taylor series remainders for alternating expansions (e.g., sin x, cos x, ln(1+x)).

Key terms

Alternating Series Error Bound
|S − S_n| ≤ |a_(n+1)|; the error is at most the absolute value of the first omitted term.
Conditions to use the error bound
Series must be alternating, terms must decrease in absolute value, and terms must approach zero (AST conditions).
Why does the bound work?
Because consecutive partial sums bracket the true sum for a valid alternating series, the true value is trapped between S_n and S_(n+1).
Finding n for a desired accuracy
Find smallest n such that |a_(n+1)| is less than the required error tolerance.
Common application
Used to bound the remainder/error in Taylor series approximations when the series is alternating, e.g., for sin x, cos x, or ln(1+x).

Try an AP-style question

Using the first 4 terms of Σ (−1)ⁿ⁺¹/n, what is the error bound?

  1. 1/5
  2. 1/16
  3. 1/25
  4. 1/4
Show the answer

A. |error| ≤ the first omitted term, b₅ = 1/5.

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