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Part 1: Core Concepts
Alternating Series — The Alternating Series Test
Part 1 of 7 — Foundations
What Is an Alternating Series?
A series whose terms alternate in sign:
∑n=1∞(−1)n+1bn=b1−b2+b3−b4+⋯
or equivalently ∑(−1)nbn where bn.
The Alternating Series Test (AST)
If bn>
The Three Hypotheses
| # | Condition | Why It's Needed |
|---|
| 1 | bn>0 | Terms truly alternate |
| 2 | bn eventually decreasing |
AP Tip: On the AP exam, you must explicitly verify ALL three conditions. Simply stating "by AST" is not sufficient for full credit.
Classic Examples
Alternating Harmonic Series:
∑n=1∞n
Practice: Applying the AST
Summary
- Alternating Series Test: three conditions (bn>0, decreasing, →0)
- Must verify ALL three explicitly on the AP exam
- "Eventually decreasing" is sufficient
- The AST tells you a series converges but does NOT give the sum
Next: Part 2 — Alternating Series Error Bound (Remainder Estimation).
Part 2: Worked Examples
Alternating Series — Error Bound
Part 2 of 7 — The Alternating Series Remainder
The Error Bound Theorem
If ∑(−1)n+1bn satisfies the AST conditions and S is the exact sum, then the error after terms satisfies:
Part 3: Problem-Solving Patterns
Alternating Series — Absolute vs. Conditional Convergence
Part 3 of 7 — Classification of Alternating Series
Review of Definitions
| Classification | Meaning |
|---|
| Absolutely convergent | $\sum |
| Conditionally convergent | ∑an converges but $\sum |
Classification Procedure for Alternating Series
Step 1: Compute (remove the factor).
Part 4: Graphs and Interpretation
Alternating Series — Connections to Taylor Series
Part 4 of 7 — Alternating Series in Taylor/Maclaurin Context
Key Maclaurin Series That Alternate
| Function | Series | Notes |
|---|
| e−x | |
Part 5: Applications
Alternating Series — AP Exam Strategies
Part 5 of 7 — FRQ & MC Techniques
Common AP Question Types
| Type | What They Ask | Key Steps |
|---|
| AST verification | "Show the series converges" | State and verify all 3 conditions |
| Error bound | "Approximate with error < ε" | Find N where bN+1 |
Part 6: Exam Strategy
Alternating Series — Problem-Solving Workshop
Part 6 of 7 — Mixed Practice
Work through these problems systematically. For each, identify whether to apply AST, error bound, or classification.
Warm-Up Review
| Concept | Formula/Rule |
|---|
| AST conditions | bn>0, decreasing, →0 |
| Error bound |
Part 7: Mixed Review
Alternating Series — Comprehensive Review
Part 7 of 7 — Full Topic Review
Complete Reference
| Concept | Key Formula/Rule |
|---|
| AST | bn>0, b, → converges |
>
0
0
,
bn+1
≤
bn
(decreasing), and
n→∞lim
bn
=
0
,
then
∑
(
−
1
)n+1
bn
converges.
| Partial sums "squeeze" toward limit |
| 3 | limbn=0 | Without this, Divergence Test kicks in |
(−1)n+1
=
1−
21+
31−
41+
⋯=
ln2
Check: bn=1/n>0 ✓, 1/(n+1)<1/n ✓, 1/n→0 ✓ → Converges by AST.
∑n=1∞n+1(−1)n+1n
bn=n/(n+1)→1=0. The third condition FAILS. This series diverges by the Divergence Test.
Why "Eventually Decreasing" Suffices
bn only needs to be decreasing for n≥N (some fixed N). A finite number of "bad" terms don't affect convergence.
To show decreasing: verify bn+1<bn, or equivalently bn+1/bn<1, or show f′(x)<0 for the continuous version.
∣S−SN∣≤bN+1
The error is bounded by the absolute value of the first omitted term.
Why This Works
The partial sums of an alternating series "bracket" the true sum:
S1>S>S2,S3>S>S4,…
Each new term overshoots and then corrects, so the error is at most the magnitude of the next term.
Example
∑n=1∞n3(−1)n+1. Approximate S using 4 terms.
S4=1−81+271−641=17281728−216+64−27≈0.8958
Error ≤b5=1/125=0.008. So S≈0.896±0.008.
AP Tip: This error bound appears almost every year on the BC exam, often in FRQ. Know it cold.
Finding N for a Given Accuracy
Problem: How many terms of ∑(−1)n+1/n2 ensure error <0.01?
Solution: Need bN+1<0.01, i.e., (N+1.
(N+1)2>100⟹N+1>10
So 10 terms give accuracy within 0.01.
Important Distinctions
| Feature | Alternating Series Error | Lagrange Error (Taylor) |
|---|
| Formula | $ | R_N |
| Applies to | Any alternating series meeting AST | Taylor polynomial remainders |
| Easy to use? | Very easy | Requires finding M |
| On AP exam | Very common | Also very common |
Summary
- Error ≤∣bN+1∣ = first omitted term
- To find N for accuracy ϵ: solve bN+1<ϵ
- Partial sums alternately overestimate and underestimate
- This is one of the MOST tested concepts on the BC exam
Next: Part 3 — Absolute vs. Conditional Convergence in Alternating Series.
∑∣an∣=∑bn
Step 2: If ∑bn converges → absolutely convergent (done).
Step 3: If ∑bn diverges → check AST conditions on original series.
- If AST conditions met → conditionally convergent
- If AST fails → divergent
The Four Essential Examples
| Series | ∑∣an∣ | ∑an | Classification |
|--------|-------------|-----------|---------------|
| ∑(−1)n/n2 | ∑1/n2 conv. (p=2) | Converges | Absolute |
| ∑(−1)n/n | ∑1/n div. | AST works → conv. | Conditional |
| ∑(−1)n/n | ∑1/n div. (p=1/2) | AST works → conv. | Conditional |
| ∑(−1)n⋅n/(n+1) | Diverges | an→0 → div. | Divergent |
Connection to Interval of Convergence
This classification matters most at endpoints of intervals of convergence for power series.
Example: ∑n=1∞nxn
Ratio test: ∣x∣<1 → converges. At the endpoints:
- x=1: ∑1/n diverges
- x=−1: ∑ converges (conditionally)
So the interval of convergence is [−1,1).
At endpoints, alternating series often converge conditionally while the positive version diverges.
AP Tip: When finding intervals of convergence, ALWAYS test endpoints separately. Expect at least one endpoint to involve an alternating series.
Summary
- Test ∑∣an∣ first; if it converges, you have absolute convergence
- If ∑∣an∣ diverges but ∑an converges (via AST), it's conditional
- Endpoint testing for power series frequently involves this classification
- ∑(−1)n/np: absolute if p>1, conditional if
Next: Part 4 — Alternating Series and Taylor Polynomials.
∑n=0∞n!(−x)n=∑n!(−1)nxn
| Alternates for x>0 |
| sinx | ∑n=0∞(2n+1)!(−1)nx2n+1 | Alternates for all x>0 |
| cosx | ∑n=0∞(2n)!(−1)nx2n | Alternates for all x>0 |
| ln(1+x) | ∑n=1∞n(−1)n+1xn | Alternates for 0<x≤1 |
| arctanx | ∑n=0∞2n+1(−1)nx2n+1 | Alternates for 0<x≤1 |
When a Taylor series alternates, you can use the alternating series error bound for the remainder.
AP Tip: The alternating series error bound is often EASIER to apply than the Lagrange error bound. Use it whenever the series alternates.
Example: Estimating cos(0.5)
cosx=1−2!x2+4!x4−6!x6+⋯
Using 3 terms (n=0,1,2):
P4(0.5)=1−2
Error ≤∣first omitted term∣=720(0.5)
Compare: cos(0.5)=0.877583…, so actual error ≈0.000021 ✓
Alternating vs. Lagrange Error Bound
For alternating Taylor series, both bounds work:
- Alternating: ∣R∣≤∣next term∣ — easy!
- Lagrange: ∣Rn — need to find
The alternating bound is usually tighter and easier. Use it when available!
Summary
- Many important Taylor series alternate for certain x values
- When alternating, use AST error bound: easier than Lagrange
- Key functions: sinx, cosx, e−x, ln(1+x), arctanx
- On the AP exam, choose the simpler error bound when both apply
Next: Part 5 — AP Exam Strategies for Alternating Series.
<
ϵ
| Classification | "Absolutely, conditionally, or diverges?" | Test $\sum |
| Endpoint analysis | "Find interval of convergence" | Test each endpoint separately |
| Taylor connection | "Use AST error bound for Pn(x)" | Identify alternating structure |
FRQ Template: AST Verification
When the AP exam says "show the series converges using the Alternating Series Test":
- Identify: "This is an alternating series with bn=…"
- Positive: "bn>0 for all n≥1" ✓
- Decreasing: "bn+1≤bn because f" ✓
- Limit: "limn→∞bn=…=0" ✓
- Conclude: "Therefore, ∑(−1)n+1bn converges by the AST." ✓
AP Tip: Omitting any of the three verifications costs points. Even if one seems "obvious," state it explicitly.
Common AP Mistakes to Avoid
Mistake 1: Forgetting to check bn→0
∑(−1)nn+1n: Students assume AST applies because it alternates. But bn=n/(n+1)→1=0. Diverges!
Mistake 2: Using an instead of bn
bn is the POSITIVE part. Don't check if (−1)nbn→ — check if .
Mistake 3: Not showing "decreasing"
Must show bn+1<bn or use f. Don't just assert it.
Mistake 4: Confusing AST error bound with Lagrange
| Alternating Error | Lagrange Error |
|---|
| $ | R |
| No M needed | Must bound f(n+1) |
| Only for alternating series | For any Taylor remainder |
Exam Strategy Summary
- Always verify ALL three AST conditions explicitly
- Error bound: ∣S−SN∣≤bN+1
- Overestimate vs. underestimate: depends on parity of N and sign of first term
- Odd N + positive first term → overestimate
- Even N + positive first term → underestimate
Next: Part 6 — Problem-Solving Workshop.
| Conditional conv. | ∑an conv., $\sum |
Workshop Takeaways
- Check AST conditions systematically
- Error bound problems: solve bN+1<ϵ
- Classification: test ∑∣an∣ first
- Factorial denominators converge fast, harmonic-type converge slowly
Next: Part 7 — Comprehensive Review.
n+1
≤
bn
| Over/Under | Odd N + positive first → over; Even N → under |
| Conditional | ∑an conv. but $\sum |
| Rearrangement | Conditionally conv. → rearrange to any sum |
Alternating series: check conditions, bound the error, classify the convergence.
Final Classification Drill
Final Error Bound Problem
Alternating Series — Complete Summary
You've mastered:
- Alternating Series Test — the three conditions and verification
- Error Bound — first omitted term bounds the error
- Over/Underestimate — parity of partial sum count
- Absolute vs. Conditional — classification procedure
- Taylor Series Connection — AST error bound as an alternative to Lagrange
- AP Exam Strategies — full justification requirements
Key Fact: Alternating series and error bounds appear on virtually every BC exam. This is one of the highest-yield topics for your score.
Up Next: Power Series — representation, convergence, and manipulation.
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