Skip to main content

Code Scholars Tutoring · AP Calculus BC

Alternating-series error bounds and approximation intervals

Unit 10 · Topics 10.10 · Detailed notes, a worked example, and a self-check with reasoning.

Understand the idea

For an alternating series whose positive magnitudes decrease to zero, the error after a partial sum is no larger than the magnitude of the first omitted term. The sign of that omitted term also indicates which side of the true sum the partial sum lies on.

Identify the last included index and the first omitted index explicitly. The error bound uses the omitted term, not the last term that was added. To meet a tolerance, solve the corresponding inequality for a whole-number index and respect whether the tolerance is strict.

The error bound is a guarantee, not the exact error. Consecutive partial sums bracket the limit under the alternating-series conditions. This gives an interval containing the sum without requiring a decimal approximation of the exact value.

∣S−SN∣≤bN+1when bn>0 decreases to 0|S-S_N|\le b_{N+1}\quad\text{when }b_n>0\text{ decreases to }0

A method you can apply

  1. Verify alternating signs and decreasing magnitudes tending to zero.
  2. Calculate the first omitted magnitude after the requested partial sum.
  3. Use the omitted sign to locate the sum between consecutive partial sums.

Worked example

Use four terms to approximate ln⁡2=∑n=1∞(−1)n−1/n\displaystyle \ln2=\sum_{n=1}^{\infty}(-1)^{n-1}/n, and give a guaranteed interval.

  1. The fourth partial sum is S4=1−1/2+1/3−1/4=7/12\displaystyle S_4=1-1/2+1/3-1/4=7/12.
  2. The next term is positive and has magnitude 1/5\displaystyle 1/5.
  3. Therefore the true sum lies above S4\displaystyle S_4 and below S5=7/12+1/5=47/60\displaystyle S_5=7/12+1/5=47/60.
The approximation is 7/12\displaystyle 7/12, the absolute error is at most 1/5\displaystyle 1/5, and 7/12≤ln⁡2≤47/60\displaystyle 7/12\le\ln2\le47/60. This bound is valid but relatively coarse.

Try it yourself

After ten terms of ∑n=1∞(−1)n−1/n2\displaystyle \sum_{n=1}^{\infty}(-1)^{n-1}/n^2, give an error bound and an interval in terms of S10\displaystyle S_{10}.

Show solution and reasoning

The first omitted magnitude is 1/112=1/121\displaystyle 1/11^2=1/121, and its sign is positive. Hence ∣S−S10∣≤1/121\displaystyle |S-S_{10}|\le1/121 and S10≤S≤S10+1/121\displaystyle S_{10}\le S\le S_{10}+1/121.

AP® is a trademark registered by the College Board, which is not affiliated with, and does not endorse, this material.