Unit 10.1 – Defining Convergent and Divergent Infinite Series BC ONLY

AP® Calculus BC | Introduction to Series

Why This Matters: Infinite series are one of the most powerful concepts in mathematics! They allow us to represent functions as infinite sums, calculate π to billions of digits, and solve differential equations. Understanding convergence is THE foundation for all series work. This begins a major BC-only unit that appears on EVERY exam!

🔢 Sequences vs. Series

THE FUNDAMENTAL DISTINCTION

Sequence:

An ordered list of numbers: \(a_1, a_2, a_3, \ldots, a_n, \ldots\)

\[ \{a_n\} = \{a_1, a_2, a_3, \ldots\} \]

Example: \(\{1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots\}\)

Series:

The SUM of the terms of a sequence:

\[ \sum_{n=1}^{\infty} a_n = a_1 + a_2 + a_3 + \cdots \]

Example: \(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots\)

📝 Key Point: A sequence is a LIST, a series is a SUM!

📊 Partial Sums

Partial Sum Definition

THE N-TH PARTIAL SUM:

\[ S_n = \sum_{k=1}^{n} a_k = a_1 + a_2 + a_3 + \cdots + a_n \]

The sum of the first \(n\) terms

Examples:

\(S_1 = a_1\)
\(S_2 = a_1 + a_2\)
\(S_3 = a_1 + a_2 + a_3\)
\(S_n = a_1 + a_2 + \cdots + a_n\)

✅ Convergence and Divergence

Convergence Definition

A SERIES CONVERGES IF:

\[ \lim_{n \to \infty} S_n = L \]

where \(L\) is a finite number (the sum of the series)

We write: \(\sum_{n=1}^{\infty} a_n = L\)

A SERIES DIVERGES IF:

The limit of partial sums does not exist, OR
The limit is \(\pm\infty\), OR
The partial sums oscillate without approaching a limit

📝 Important: Convergence means the infinite sum equals a FINITE number!

🌟 Geometric Series (MOST IMPORTANT!)

Geometric Series Formula

GEOMETRIC SERIES:

\[ \sum_{n=1}^{\infty} ar^{n-1} = a + ar + ar^2 + ar^3 + \cdots \]

where \(a\) = first term, \(r\) = common ratio

Convergence Rule:

If \(|r| < 1\): Series CONVERGES to

\[ \sum_{n=1}^{\infty} ar^{n-1} = \frac{a}{1-r} \]

If \(|r| \geq 1\): Series DIVERGES

⚠️ MEMORIZE THIS: The geometric series formula is THE most important series formula. You'll use it constantly!

🚫 The Divergence Test (n-th Term Test)

Test for Divergence

THE DIVERGENCE TEST:

If \(\lim_{n \to \infty} a_n \neq 0\), then \(\sum a_n\) DIVERGES

Contrapositive: If \(\sum a_n\) converges, then \(\lim_{n \to \infty} a_n = 0\)

⚠️ CRITICAL WARNING: This test can ONLY prove DIVERGENCE! If \(\lim_{n \to \infty} a_n = 0\), the test is INCONCLUSIVE (series might converge OR diverge).

📝 Remember: "Terms going to zero" is NECESSARY but NOT SUFFICIENT for convergence!

⚙️ Properties of Convergent Series

Series Properties

1. Constant Multiple:

If \(\sum a_n\) converges, then for any constant \(c\):

\[ \sum ca_n = c \sum a_n \]

2. Sum/Difference:

If \(\sum a_n\) and \(\sum b_n\) both converge:

\[ \sum (a_n \pm b_n) = \sum a_n \pm \sum b_n \]

3. Adding/Removing Terms:

Adding or removing a finite number of terms does NOT affect convergence (but changes the sum)

4. Reindexing:

Starting index doesn't affect convergence:

\[ \sum_{n=1}^{\infty} a_n \text{ and } \sum_{n=0}^{\infty} a_n \text{ have same convergence behavior} \]

📖 Comprehensive Worked Examples

Example 1: Geometric Series

Problem: Does \(\sum_{n=1}^{\infty} \frac{3}{2^n}\) converge? If so, find its sum.

Solution:

Step 1: Identify as geometric

\[ \sum_{n=1}^{\infty} \frac{3}{2^n} = \sum_{n=1}^{\infty} 3 \cdot \left(\frac{1}{2}\right)^n \]

This is geometric with \(a = \frac{3}{2}\) (first term) and \(r = \frac{1}{2}\)

Step 2: Check \(|r| < 1\)

\(|r| = \frac{1}{2} < 1\), so it converges!

Step 3: Find sum

\[ \sum_{n=1}^{\infty} \frac{3}{2^n} = \frac{a}{1-r} = \frac{3/2}{1-1/2} = \frac{3/2}{1/2} = 3 \]

ANSWER: Converges to 3

Example 2: Divergence Test

Problem: Does \(\sum_{n=1}^{\infty} \frac{n}{n+1}\) converge?

Apply divergence test:

\[ \lim_{n \to \infty} \frac{n}{n+1} = \lim_{n \to \infty} \frac{1}{1+1/n} = 1 \neq 0 \]

Since the limit is not zero, the series DIVERGES by the divergence test.

Example 3: Divergence Test is Inconclusive

Problem: What about \(\sum_{n=1}^{\infty} \frac{1}{n}\)?

Check the limit:

\[ \lim_{n \to \infty} \frac{1}{n} = 0 \]

The divergence test is INCONCLUSIVE! (Later we'll learn this is the harmonic series, which DIVERGES despite terms going to zero)

Example 4: Telescoping Series

Problem: Find the sum: \(\sum_{n=1}^{\infty} \left(\frac{1}{n} - \frac{1}{n+1}\right)\)

Write out partial sums:

\[ S_n = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \cdots + \left(\frac{1}{n} - \frac{1}{n+1}\right) \]

Most terms cancel (telescope):

\[ S_n = 1 - \frac{1}{n+1} \]

Take limit:

\[ \lim_{n \to \infty} S_n = \lim_{n \to \infty} \left(1 - \frac{1}{n+1}\right) = 1 \]

📊 Quick Reference: Series Behavior

Common Series
Series	Convergence	Notes
\(\sum ar^{n-1}\)	Conv. if \(\|r\| < 1\)	Geometric series
\(\sum \frac{1}{n}\)	Diverges	Harmonic series
\(\sum \frac{1}{n^p}\)	Conv. if \(p > 1\)	p-series
\(\sum c\)	Diverges (if \(c \neq 0\))	Constant series

💡 Essential Tips & Strategies

✅ Success Strategies:

Recognize geometric series: Look for \(ar^n\) pattern
ALWAYS check divergence test first: Easiest test!
Remember \(|r| < 1\): For geometric convergence
Memorize geometric sum formula: \(\frac{a}{1-r}\)
Terms → 0 doesn't guarantee convergence: Common trap!
Look for telescoping: Terms that cancel
Rewrite in standard form: Makes patterns clearer
Partial sums are key: Convergence = limit of \(S_n\)

🔥 Common Series Patterns:

Geometric: Constant ratio between consecutive terms
Telescoping: Terms cancel in partial sums
p-series: Form \(\sum \frac{1}{n^p}\)
Harmonic: \(\sum \frac{1}{n}\) (diverges!)

❌ Common Mistakes to Avoid

Mistake 1: Thinking \(\lim a_n = 0\) means the series converges
Mistake 2: Forgetting absolute value in \(|r| < 1\) for geometric series
Mistake 3: Using wrong formula for geometric series (wrong first term)
Mistake 4: Confusing sequence convergence with series convergence
Mistake 5: Not checking if series is geometric before using formula
Mistake 6: Thinking divergence test can prove convergence
Mistake 7: Adding series when one or both diverge
Mistake 8: Wrong common ratio in geometric series
Mistake 9: Not simplifying before identifying series type
Mistake 10: Confusing \(\sum_{n=0}^{\infty}\) vs \(\sum_{n=1}^{\infty}\) (affects first term)

📝 Practice Problems

Determine convergence and find sum if possible:

\(\sum_{n=1}^{\infty} \frac{5}{3^n}\)
\(\sum_{n=1}^{\infty} \frac{n^2}{n^2+1}\)
\(\sum_{n=0}^{\infty} \left(\frac{2}{3}\right)^n\)
\(\sum_{n=1}^{\infty} \frac{1}{n(n+1)}\)

Answers:

Converges to \(\frac{5}{2}\) (geometric with \(r = 1/3\))
Diverges (divergence test: limit = 1)
Converges to 3 (geometric with \(a=1, r=2/3\))
Converges to 1 (telescoping)

✏️ AP® Exam Success Tips

What AP® Graders Look For:

Identify series type: State "geometric series" or "use divergence test"
Show limit calculations: Don't skip steps
For geometric: Identify \(a\) and \(r\) explicitly
Check \(|r| < 1\): Show the inequality
Write formula: \(\sum ar^{n-1} = \frac{a}{1-r}\)
State conclusion: "Converges to..." or "Diverges because..."
For divergence test: Show \(\lim a_n \neq 0\)
Be clear: Distinguish between sequence and series

💯 Exam Strategy:

Check divergence test first (quickest)
Look for geometric series pattern
If geometric: find \(a\) and \(r\), check \(|r| < 1\)
Look for telescoping if not geometric
Show all work clearly
State conclusions explicitly
Remember: later units have more tests!

⚡ Quick Reference Guide

SERIES CONVERGENCE ESSENTIALS

Geometric Series:

\(\sum ar^{n-1}\) converges if \(|r| < 1\)
Sum: \(\frac{a}{1-r}\) when \(|r| < 1\)

Divergence Test:

If \(\lim_{n \to \infty} a_n \neq 0\) → DIVERGES
If \(\lim_{n \to \infty} a_n = 0\) → INCONCLUSIVE

Convergence Definition:

\(\sum a_n\) converges ⟺ \(\lim_{n \to \infty} S_n = L\) (finite)

Remember:

Geometric series formula is #1 priority!
Divergence test can only prove divergence
Terms → 0 is necessary but not sufficient

Master Series Fundamentals! A series is the sum \(\sum_{n=1}^{\infty} a_n\) while a sequence is the list \(\{a_n\}\). Series converges if \(\lim_{n\to\infty} S_n = L\) (finite). Geometric series \(\sum ar^{n-1}\) converges to \(\frac{a}{1-r}\) if \(|r| < 1\), diverges if \(|r| \geq 1\)—MEMORIZE THIS! Divergence test: if \(\lim a_n \neq 0\), series diverges; but if \(\lim a_n = 0\), test is INCONCLUSIVE (can't conclude convergence). Necessary condition for convergence: terms must go to zero. Properties: can add convergent series, multiply by constants, add/remove finite terms. This is the foundation for all series work—Unit 10 is entirely BC content and heavily tested! Coming up: convergence tests, power series, Taylor series. Practice geometric series until automatic! 🎯✨