Unit 10.9 – Determining Absolute or Conditional Convergence BC ONLY

AP® Calculus BC | Understanding the Strength of Convergence

Why This Matters: Not all convergent series are created equal! Absolute convergence is the strongest form of convergence—series that converge absolutely behave nicely and can be rearranged freely. Conditional convergence is weaker—series converge but only because positive and negative terms cancel carefully. Understanding this distinction is crucial for Unit 10!

🎯 Absolute vs. Conditional Convergence

Fundamental Definitions

ABSOLUTE CONVERGENCE

A series \(\sum a_n\) converges absolutely if:

\[ \sum_{n=1}^{\infty} |a_n| \text{ converges} \]

The series of absolute values converges

CONDITIONAL CONVERGENCE

A series \(\sum a_n\) converges conditionally if:

\(\sum a_n\) converges, BUT
\(\sum |a_n|\) diverges

Converges, but NOT absolutely

⭐ The Fundamental Theorem

Absolute Convergence Implies Convergence

THE KEY THEOREM

\[ \text{If } \sum |a_n| \text{ converges} \Rightarrow \sum a_n \text{ converges} \]

Absolute convergence is STRONGER than regular convergence!

⚠️ IMPORTANT: The converse is NOT true! \(\sum a_n\) can converge without \(\sum |a_n|\) converging (that's conditional convergence).

📝 Logic Flow:

\[ \text{Absolute Convergence} \Rightarrow \text{Convergence} \]

\[ \text{Convergence} \not\Rightarrow \text{Absolute Convergence} \]

📊 Convergence Hierarchy

Strength of Convergence

From Strongest to Weakest:

1. Absolute Convergence (STRONGEST)

\(\sum |a_n|\) converges
Can rearrange terms freely
Most robust form of convergence

2. Conditional Convergence

\(\sum a_n\) converges but \(\sum |a_n|\) diverges
CANNOT rearrange terms!
Convergence depends on cancellation

3. Divergence (WEAKEST)

\(\sum a_n\) diverges
No convergence at all

🔍 How to Determine Type of Convergence

Step-by-Step Procedure:

Test \(\sum |a_n|\) first (absolute values)
- Use ratio test, comparison test, p-series, etc.
If \(\sum |a_n|\) converges:
- Series converges ABSOLUTELY
- Done! No need to test original series
If \(\sum |a_n|\) diverges:
- Now test original series \(\sum a_n\)
If original series converges:
- Series converges CONDITIONALLY
If original series also diverges:
- Series DIVERGES

💡 Pro Tip: Always test absolute convergence FIRST. If it converges absolutely, you're done—that implies regular convergence too!

📖 Comprehensive Worked Examples

Example 1: Absolute Convergence

Problem: Determine if \(\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}\) converges absolutely, conditionally, or diverges.

Solution:

Step 1: Test absolute convergence

\[ \sum_{n=1}^{\infty} \left|\frac{(-1)^n}{n^2}\right| = \sum_{n=1}^{\infty} \frac{1}{n^2} \]

This is a p-series with \(p = 2 > 1\)

Therefore \(\sum \frac{1}{n^2}\) CONVERGES

Since \(\sum |a_n|\) converges, the series converges ABSOLUTELY

(This also means the original series converges)

Example 2: Conditional Convergence (Classic!)

Problem: Determine if \(\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\) converges absolutely, conditionally, or diverges.

Step 1: Test absolute convergence

\[ \sum_{n=1}^{\infty} \left|\frac{(-1)^{n+1}}{n}\right| = \sum_{n=1}^{\infty} \frac{1}{n} \]

This is the harmonic series, which DIVERGES

So the series does NOT converge absolutely

Step 2: Test original series

\(\sum \frac{(-1)^{n+1}}{n} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots\)

This is alternating harmonic series

Check AST: \(b_n = \frac{1}{n}\) is positive, decreasing, and \(\lim b_n = 0\) ✓

By Alternating Series Test, it CONVERGES

Series converges but NOT absolutely → CONDITIONALLY CONVERGENT

(This is the classic example of conditional convergence!)

Example 3: Divergent Series

Problem: Classify \(\sum_{n=1}^{\infty} \frac{(-1)^n n}{n+1}\).

Check nth-term test:

\[ \lim_{n \to \infty} \frac{(-1)^n n}{n+1} \text{ DNE (oscillates)} \]

Since limit doesn't go to zero, series DIVERGES by nth-term test

Series DIVERGES

Example 4: Using Ratio Test

Problem: Classify \(\sum_{n=1}^{\infty} \frac{(-1)^n 3^n}{n!}\).

Test absolute convergence using Ratio Test:

\[ a_n = \left|\frac{(-1)^n 3^n}{n!}\right| = \frac{3^n}{n!} \]

\[ \frac{a_{n+1}}{a_n} = \frac{3^{n+1}/(n+1)!}{3^n/n!} = \frac{3}{n+1} \]

\[ L = \lim_{n \to \infty} \frac{3}{n+1} = 0 < 1 \]

By Ratio Test, \(\sum |a_n|\) converges

Series converges ABSOLUTELY by Ratio Test

Example 5: Alternating p-Series

Problem: For what values of \(p\) does \(\sum_{n=1}^{\infty} \frac{(-1)^n}{n^p}\) converge:
(a) Absolutely?
(b) Conditionally?

Part (a): Absolute convergence

\[ \sum \left|\frac{(-1)^n}{n^p}\right| = \sum \frac{1}{n^p} \]

This is p-series: converges when \(p > 1\)

Answer (a): Converges absolutely for \(p > 1\)

Part (b): Conditional convergence

Need: original converges BUT absolute doesn't

Original \(\sum \frac{(-1)^n}{n^p}\) converges by AST when \(p > 0\)

Absolute \(\sum \frac{1}{n^p}\) diverges when \(p \leq 1\)

Answer (b): Converges conditionally for \(0 < p \leq 1\)

📊 Complete Classification Guide

ptionion>Series Classification

Condition	Classification	Can Rearrange?
\(\sum \|a_n\|\) converges	Absolutely Convergent	YES ✓
\(\sum a_n\) conv., \(\sum \|a_n\|\) div.	Conditionally Convergent	NO ✗
\(\sum a_n\) diverges	Divergent	N/A

⭐ Famous Series Classifications

Classic Series Examples
Series	Classification	Reason
\(\sum \frac{(-1)^{n+1}}{n}\)	Conditional	AST works, harmonic diverges
\(\sum \frac{(-1)^n}{n^2}\)	Absolute	p-series p=2 converges
\(\sum \frac{(-1)^n}{\sqrt{n}}\)	Conditional	AST works, p=1/2 diverges
\(\sum \frac{(-1)^n}{n^{3/2}}\)	Absolute	p-series p=3/2 converges
\(\sum \frac{(-1)^n \cdot n!}{2^n}\)	Diverges	Ratio test L > 1

💡 Essential Tips & Strategies

✅ Success Strategies:

Test absolute FIRST: If converges, you're done!
Remove absolute value signs: When testing \(\sum |a_n|\)
Alternating series often conditional: Check both tests
Ratio test gives absolute: If it shows convergence
State type clearly: "Absolutely" or "conditionally"
Remember hierarchy: Absolute → Conditional → Divergent
Can't rearrange conditional: Important property!
p-series boundary: p > 1 absolute, 0 < p ≤ 1 conditional

🔥 Quick Decision Tree:

Does \(\sum |a_n|\) converge? → YES = Absolute, done!
Does \(\sum |a_n|\) converge? → NO = not absolute
Does \(\sum a_n\) converge? → YES = Conditional
Does \(\sum a_n\) converge? → NO = Divergent

❌ Common Mistakes to Avoid

Mistake 1: Thinking convergence implies absolute convergence (NO!)
Mistake 2: Not testing \(\sum |a_n|\) for absolute convergence
Mistake 3: Saying "conditionally divergent" (no such thing!)
Mistake 4: Forgetting to remove absolute value when testing \(\sum |a_n|\)
Mistake 5: Testing original series when absolute convergence found
Mistake 6: Not stating which type of convergence on exams
Mistake 7: Thinking conditional convergence is weaker than divergence
Mistake 8: Rearranging conditionally convergent series
Mistake 9: Not recognizing alternating harmonic as conditional
Mistake 10: Confusing which implication works (absolute → convergence only)

📝 Practice Problems

Classify each series as absolutely convergent, conditionally convergent, or divergent:

\(\sum_{n=1}^{\infty} \frac{(-1)^n}{n^3}\)
\(\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{\sqrt{n}}\)
\(\sum_{n=1}^{\infty} \frac{(-1)^n n}{n^2+1}\)
\(\sum_{n=1}^{\infty} \frac{(-1)^n 2^n}{n!}\)
\(\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n \ln n}\) (n ≥ 2)

Answers:

Absolutely convergent (p-series p=3 converges)
Conditionally convergent (AST works, p=1/2 diverges)
Conditionally convergent (AST works, absolute diverges by LCT with 1/n)
Absolutely convergent (Ratio Test: L=0)
Conditionally convergent (AST works, integral test shows absolute diverges)

✏️ AP® Exam Success Tips

What AP® Graders Look For:

Test absolute convergence first: Show \(\sum |a_n|\)
State test used: "By p-series test..." or "By Ratio Test..."
Show work for absolute test: Don't skip steps
If absolute diverges: Then test original series
State conclusion clearly: "Absolutely convergent" or "Conditionally convergent"
Justify both parts: Why absolute diverges AND why original converges
Use proper notation: \(\sum |a_n|\) with absolute value bars
Don't say "conditionally divergent": Not a thing!

💯 Exam Strategy:

Always test absolute convergence FIRST
Write \(\sum |a_n|\) explicitly
Apply appropriate convergence test to absolute series
If converges: state "absolutely convergent" and STOP
If diverges: now test original series \(\sum a_n\)
If original converges: "conditionally convergent"
If original diverges: "divergent"
State which tests you used for each step

⚡ Quick Reference Guide

ABSOLUTE VS CONDITIONAL CONVERGENCE

Absolute Convergence:

\(\sum |a_n|\) converges
Implies \(\sum a_n\) converges
Strongest form
Can rearrange terms

Conditional Convergence:

\(\sum a_n\) converges
\(\sum |a_n|\) diverges
Weaker form
CANNOT rearrange!

Testing Procedure:

Test \(\sum |a_n|\) first
If converges → absolutely convergent
If diverges → test \(\sum a_n\)
If converges → conditionally convergent
If diverges → divergent

Remember:

Always test absolute FIRST!
Absolute ⇒ Convergent (not reverse!)
Alternating harmonic = classic conditional!

Master Absolute vs Conditional Convergence! Absolute convergence: \(\sum |a_n|\) converges (strongest form)—implies regular convergence and allows term rearrangement. Conditional convergence: \(\sum a_n\) converges but \(\sum |a_n|\) diverges (weaker)—convergence depends on cancellation, cannot rearrange terms. Key theorem: absolute convergence ⇒ convergence (but NOT reverse!). Testing procedure: (1) test \(\sum |a_n|\) first; if converges → absolutely convergent (done!); if diverges → (2) test \(\sum a_n\); if converges → conditionally convergent; if diverges → divergent. Classic example: alternating harmonic \(\sum \frac{(-1)^{n+1}}{n}\) converges conditionally (to ln 2) because AST works but \(\sum \frac{1}{n}\) diverges. Alternating p-series: absolutely convergent for p > 1, conditionally for 0 < p ≤ 1, divergent for p ≤ 0. This distinction is crucial—appears on every BC exam! Master the testing hierarchy! 🎯✨