Unit 10.10 – Alternating Series Error Bound BC ONLY
AP® Calculus BC | Estimating Sums with Guaranteed Accuracy
Why This Matters: The Alternating Series Error Bound is UNIQUE among all convergence tests—it not only tells you a series converges, but gives you an EXACT bound on how accurate your approximation is! This is incredibly powerful and appears on virtually every BC exam. You can guarantee your estimate is within any desired accuracy!
🎯 The Alternating Series Error Bound
The Error Bound Theorem
If an alternating series \(\sum_{n=1}^{\infty} (-1)^{n+1} b_n\) or \(\sum_{n=1}^{\infty} (-1)^n b_n\) satisfies the conditions of the Alternating Series Test:
- \(b_n > 0\)
- \(b_n\) is decreasing
- \(\lim_{n \to \infty} b_n = 0\)
And the series converges to sum \(S\), then:
The error is at most the FIRST OMITTED TERM!
What This Means:
- \(S\) = actual sum of infinite series
- \(S_n\) = partial sum (first n terms)
- \(b_{n+1}\) = first term NOT included in \(S_n\)
- Error = \(|S - S_n|\) ≤ \(b_{n+1}\)
✅ When Does the Error Bound Apply?
Requirements for Error Bound
⚠️ CRITICAL: The error bound ONLY works for alternating series that satisfy ALL three AST conditions!
- Alternating signs: \((-1)^n\) or \((-1)^{n+1}\)
- Positive terms: \(b_n > 0\)
- Decreasing: \(b_{n+1} \leq b_n\)
- Limit to zero: \(\lim b_n = 0\)
📝 Important: This error bound does NOT apply to non-alternating series, even if they converge!
📋 How to Use the Error Bound
Two Main Applications:
Question: How many terms needed for error < E?
Method:
- Set up inequality: \(b_{n+1} < E\)
- Solve for n
- Need n terms in sum
Question: What's the error using n terms?
Method:
- Calculate \(b_{n+1}\)
- Error ≤ \(b_{n+1}\)
👁️ Visual Understanding
WHY THE ERROR BOUND WORKS
The Bouncing Pattern:
Alternating series partial sums "bounce" back and forth, getting closer to S:
- \(S_1\) is on one side of S
- \(S_2\) overshoots to other side
- \(S_3\) comes back, closer to S
- \(S_4\) overshoots again, even closer
- Each "bounce" is smaller than the previous
The next term \(b_{n+1}\) represents the maximum distance \(S_n\) can be from S!
📝 Key Insight: The partial sum \(S_n\) is BETWEEN the true sum S and the next partial sum \(S_{n+1}\), so the error can't exceed \(b_{n+1}\)!
📖 Comprehensive Worked Examples
Example 1: Finding Number of Terms
Problem: How many terms of \(\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}\) are needed to approximate the sum within 0.01?
Solution:
Step 1: Verify AST conditions
\(b_n = \frac{1}{n^2}\): positive ✓, decreasing ✓, limit = 0 ✓
Series converges by AST, so error bound applies
Step 2: Set up error bound inequality
Need: \(|S - S_n| \leq b_{n+1} < 0.01\)
Step 3: Solve for n
ANSWER: Need 10 terms
\(S_{10}\) approximates S within 0.01
Example 2: Estimating the Error
Problem: If you use the first 5 terms of \(\sum_{n=1}^{\infty} \frac{(-1)^n}{n!}\) to estimate the sum, what's the maximum possible error?
Verify AST applies:
\(b_n = \frac{1}{n!}\): positive, decreasing, limit = 0 ✓
Apply error bound:
Using 5 terms means \(S_5\)
Error ≤ \(b_6 = \frac{1}{6!} = \frac{1}{720}\)
Maximum Error: \(\frac{1}{720} \approx 0.00139\)
Example 3: Alternating Harmonic Series
Problem: How many terms of \(\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\) are needed to guarantee the approximation is within 0.001 of the actual sum?
Setup:
\(b_n = \frac{1}{n}\), need \(b_{n+1} < 0.001\)
ANSWER: Need 1000 terms
📝 Note: This shows alternating harmonic series converges VERY slowly!
Example 4: With Exponential
Problem: For \(\sum_{n=0}^{\infty} \frac{(-1)^n}{3^n}\), how many terms guarantee error < 0.0001?
Solve:
\(b_n = \frac{1}{3^n}\), need \(b_{n+1} < 0.0001\)
ANSWER: Need 8 terms
Example 5: Estimating the Sum
Problem: Estimate \(\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^3}\) using 4 terms and find the error bound.
Calculate \(S_4\):
Find error bound:
Estimate: \(S \approx 0.896 \pm 0.008\)
True sum is between 0.888 and 0.904
📊 Quick Reference Examples
| Series | \(b_n\) | For Error < 0.01 |
|---|---|---|
| \(\sum \frac{(-1)^{n+1}}{n}\) | \(\frac{1}{n}\) | n ≥ 100 |
| \(\sum \frac{(-1)^n}{n^2}\) | \(\frac{1}{n^2}\) | n ≥ 10 |
| \(\sum \frac{(-1)^{n+1}}{n^3}\) | \(\frac{1}{n^3}\) | n ≥ 5 |
| \(\sum \frac{(-1)^n}{n!}\) | \(\frac{1}{n!}\) | n ≥ 5 |
💡 Essential Tips & Strategies
✅ Success Strategies:
- Verify AST first: Error bound only applies if AST conditions met
- First omitted term: If using n terms, error ≤ \(b_{n+1}\)
- Solve inequalities carefully: \(b_{n+1} < E\), then solve for n
- Round up: If you get n = 9.5, need n = 10
- Don't confuse: \(b_n\) (last included) vs \(b_{n+1}\) (first omitted)
- State clearly: "By alternating series error bound..."
- Check starting index: n = 0 vs n = 1 affects counting
- Unique property: Only test that gives numerical error bounds!
🔥 Quick Problem-Solving Steps:
- Verify series satisfies AST (alternating, decreasing, limit = 0)
- Identify \(b_n\) (positive terms)
- For "how many terms": solve \(b_{n+1} < E\)
- For "what's the error": calculate \(b_{n+1}\)
- State answer clearly with units/context
❌ Common Mistakes to Avoid
- Mistake 1: Using error bound on non-alternating series
- Mistake 2: Not verifying AST conditions before applying
- Mistake 3: Confusing \(b_n\) (last term in sum) with \(b_{n+1}\) (error bound)
- Mistake 4: Using \(b_n\) instead of \(b_{n+1}\) for error
- Mistake 5: Forgetting to round up when solving for n
- Mistake 6: Not checking if series starts at n=0 or n=1
- Mistake 7: Saying "need n+1 terms" when answer is n terms
- Mistake 8: Applying to conditionally convergent series that don't satisfy AST
- Mistake 9: Inequality direction errors when solving
- Mistake 10: Not stating "by alternating series error bound"
📝 Practice Problems
Solve the following:
- How many terms of \(\sum_{n=1}^{\infty} \frac{(-1)^n}{n^4}\) for error < 0.001?
- What's the error using 8 terms of \(\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{\sqrt{n}}\)?
- How many terms of \(\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}\) for error < 0.00001?
- Estimate \(\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}\) using 5 terms with error bound
- For \(\sum_{n=1}^{\infty} \frac{(-1)^n n}{n^2+1}\), how many terms for error < 0.01?
Answers:
- Need 6 terms (since \(\frac{1}{7^4} < 0.001\))
- Error ≤ \(\frac{1}{\sqrt{9}} = \frac{1}{3} \approx 0.333\)
- Need 4 terms (since \(\frac{1}{8!} < 0.00001\))
- \(S \approx 0.822 \pm 0.028\) (error ≤ \(\frac{1}{36}\))
- Need 10 terms (solve \(\frac{n+1}{(n+1)^2+1} < 0.01\))
✏️ AP® Exam Success Tips
What AP® Graders Look For:
- State the theorem: "By alternating series error bound..."
- Verify AST applies: Show series satisfies conditions
- Identify \(b_n\) clearly: Show the positive terms
- Set up correct inequality: \(b_{n+1} < E\) or \(|S - S_n| \leq b_{n+1}\)
- Show algebraic work: Solving for n step-by-step
- Justify answer: Explain why n is sufficient
- Round appropriately: Can't use fractional terms!
- State conclusion: "Need n terms" or "Error ≤ [value]"
💯 Exam Strategy:
- Read problem carefully (find n? or find error?)
- Verify alternating series with AST conditions
- Identify \(b_n\) (positive terms without \((-1)^n\))
- For finding n: solve \(b_{n+1} < \text{given error}\)
- For finding error: calculate \(b_{n+1}\)
- Show all algebra clearly
- Box or circle final answer
- Include units or context as appropriate
⚡ Quick Reference Guide
ALTERNATING SERIES ERROR BOUND
The Formula:
Error ≤ first omitted term!
When It Applies:
- Alternating series: \((-1)^n\) or \((-1)^{n+1}\)
- Satisfies ALL three AST conditions
- Series converges
Two Uses:
- Find n: Solve \(b_{n+1} < E\)
- Find error: Calculate \(b_{n+1}\)
Remember:
- ONLY for alternating series!
- First OMITTED term = \(b_{n+1}\)!
- UNIQUE: gives numerical bounds!
Master the Alternating Series Error Bound! For alternating series satisfying AST (alternating, positive, decreasing, limit=0), the error bound states: \(|S - S_n| \leq b_{n+1}\), where \(S\) is true sum, \(S_n\) is partial sum of n terms, and \(b_{n+1}\) is first omitted term. This is THE ONLY convergence test that provides numerical error estimates! Two applications: (1) Find number of terms: solve \(b_{n+1} < \text{desired error}\) for n; (2) Find error: calculate \(b_{n+1}\) for given n. Why it works: partial sums oscillate around S, each bounce smaller than previous; next term bounds maximum distance. Classic examples: alternating harmonic converges slowly (1000 terms for 0.001 accuracy); factorial terms converge fast (5 terms for 0.001 accuracy). Must verify ALL AST conditions first. This appears on EVERY BC exam—master finding n for given accuracy! Round up fractional answers. State "by alternating series error bound" explicitly. Uniquely powerful tool! 🎯✨