Short on time? This page is the recap. For the zigzag picture, the proof and worked examples, use the full study guide. BC-only material. Prerequisites (Topics 10.1, 10.7 and 10.9) are on the Calculus BC roadmap.
Recap
- Set-up: Σ (−1)ⁿ⁺¹ aₙ with aₙ > 0, sum S, partial sum Sₙ.
- Conditions (the alternating series test): signs alternate, aₙ decreasing, aₙ → 0.
- Bound: |S − Sₙ| ≤ aₙ₊₁. The error is at most the first term you left out.
- S lies between Sₙ and Sₙ₊₁, because the partial sums zigzag in on S.
- Direction: the error has the sign of the first omitted term.
Key relationships
| Task | Method | Example |
|---|---|---|
| Bound the error of Sₙ | Compute aₙ₊₁ | For S₄ of Σ (−1)ⁿ⁺¹/(n · 2ⁿ): bound a₅ = 1/160 |
| Show error < tolerance | Compute aₙ₊₁, then say it is less than the tolerance | 1/160 = 0.00625 < 0.01 |
| How many terms? | Solve aₙ₊₁ < tolerance for the smallest n | Σ (−1)ⁿ⁺¹/(n³ + 1) within 0.001: n + 1 = 10, so 9 terms |
| Over or under? | Sign of first omitted term: + means Sₙ is too small, − means too big | S₄ above: 5th term is +, so S₄ underestimates |
| Interval for S | Between Sₙ and Sₙ ± aₙ₊₁ (in the direction of the next term) | S₄ ≤ S ≤ S₄ + a₅ |
Assumptions to state
- The series is alternating.
- The term sizes decrease (at least from the point where you stop onwards).
- The term sizes approach 0.
If any of these fails, this bound is not justified. For non-alternating Taylor remainders, Topic 10.12 (Lagrange) is the tool.
Mistakes to avoid
- Using aₙ instead of aₙ₊₁.
- Calling the bound the error. Say “at most”.
- No comparison with the tolerance. Finish the sentence: “… which is less than 0.01”.
- Off-by-one in counting terms, especially when the series starts at n = 0.
- Wrong direction. Look at the sign of the next term, not the last one you added.
- Early rounding of partial sums.
Quick self-check
- For Σ (−1)ⁿ⁺¹/n⁵, give a bound for |S − S₂|. (a₃ = 1/243 ≈ 0.00412)
- For Σ (−1)ⁿ⁺¹/(4n), how many terms guarantee an error below 0.01? (Need 1/(4(n + 1)) < 0.01, so n + 1 > 25: n + 1 = 26, so 25 terms.)
- A convergent alternating series meeting the conditions has S₇ = 2.316, and its 8th term is −0.005. Give an interval for S. (2.311 ≤ S ≤ 2.316: S₇ is an overestimate.)
Next: practice questions.