Unit 10.2 – Working with Geometric Series BC ONLY

AP® Calculus BC | Mastering Geometric Series

Why This Matters: Geometric series are THE most important type of series in calculus! They're the only series where we have an exact formula for the sum, and they appear everywhere—from compound interest to probability to power series. This is the foundation you'll use throughout Unit 10. Master this and you're set!

🎯 The Geometric Series Formula

Geometric Series (THE MOST IMPORTANT FORMULA!)

STANDARD FORM:

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

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

Convergence and Sum:

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

\[ S = \frac{a}{1-r} \]

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

⚠️ MEMORIZE: The formula \(\frac{a}{1-r}\) with \(|r| < 1\) will be used constantly throughout this unit!

📋 Alternative Forms of Geometric Series

Form 1: Starting at n = 0

\[ \sum_{n=0}^{\infty} ar^n = a + ar + ar^2 + ar^3 + \cdots = \frac{a}{1-r} \]

(Same sum formula!)

Form 2: Starting at n = 1

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

Form 3: Without explicit a

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

(First term is \(r\))

Form 4: Starting at n = k

\[ \sum_{n=k}^{\infty} ar^n = ar^k + ar^{k+1} + ar^{k+2} + \cdots = \frac{ar^k}{1-r} \]

(First term is \(ar^k\))

📝 Key Insight: The sum formula is always \(\frac{\text{first term}}{1-r}\)!

🔍 Identifying the Common Ratio r

How to Find r:

Method 1: Divide Consecutive Terms

\[ r = \frac{a_{n+1}}{a_n} = \frac{\text{any term}}{\text{previous term}} \]

Method 2: Look at the Pattern

Each term = (previous term) × \(r\)

Example: \(2, 6, 18, 54, \ldots\) → \(r = 3\)

Method 3: From Formula

If \(a_n = a \cdot r^{n-1}\) or \(a_n = a \cdot r^n\), then \(r\) is the base

📖 Comprehensive Worked Examples

Example 1: Standard Geometric Series

Problem: Find the sum: \(\sum_{n=1}^{\infty} \frac{5}{2^n}\)

Solution:

Step 1: Rewrite in standard form

\[ \sum_{n=1}^{\infty} \frac{5}{2^n} = \sum_{n=1}^{\infty} 5 \cdot \left(\frac{1}{2}\right)^n \]

Step 2: Identify first term and r

When \(n = 1\): first term = \(5 \cdot \frac{1}{2} = \frac{5}{2}\)

Common ratio: \(r = \frac{1}{2}\)

Step 3: Check \(|r| < 1\)

\(|r| = \frac{1}{2} < 1\) ✓ Converges!

Step 4: Apply formula

\[ S = \frac{a}{1-r} = \frac{5/2}{1-1/2} = \frac{5/2}{1/2} = 5 \]

ANSWER: \(S = 5\)

Example 2: Negative Ratio

Problem: Does \(\sum_{n=0}^{\infty} 3(-0.5)^n\) converge? If so, find the sum.

Identify components:

First term (when \(n=0\)): \(a = 3 \cdot (-0.5)^0 = 3\)

Common ratio: \(r = -0.5\)

Check convergence:

\[ |r| = |-0.5| = 0.5 < 1 \]

Converges! (Note: absolute value!)

Calculate sum:

\[ S = \frac{3}{1-(-0.5)} = \frac{3}{1.5} = 2 \]

Example 3: Starting Index ≠ 0 or 1

Problem: Find \(\sum_{n=3}^{\infty} 2 \cdot 3^{-n}\)

Rewrite:

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

Find first term (n=3):

\[ a = 2 \cdot \left(\frac{1}{3}\right)^3 = \frac{2}{27} \]

Ratio: \(r = \frac{1}{3}\)

Apply formula:

\[ S = \frac{2/27}{1-1/3} = \frac{2/27}{2/3} = \frac{2}{27} \cdot \frac{3}{2} = \frac{1}{9} \]

Example 4: Expressing as Geometric Series

Problem: Express \(0.\overline{27}\) (repeating decimal) as a fraction.

Write as series:

\[ 0.\overline{27} = 0.272727\ldots = \frac{27}{100} + \frac{27}{10000} + \frac{27}{1000000} + \cdots \]

\[ = \sum_{n=1}^{\infty} \frac{27}{100^n} = \sum_{n=1}^{\infty} 27 \cdot \left(\frac{1}{100}\right)^n \]

Apply geometric series:

First term: \(a = \frac{27}{100}\), ratio: \(r = \frac{1}{100}\)

\[ S = \frac{27/100}{1-1/100} = \frac{27/100}{99/100} = \frac{27}{99} = \frac{3}{11} \]

Example 5: Divergent Geometric Series

Problem: Does \(\sum_{n=0}^{\infty} 5(1.2)^n\) converge?

Identify r:

\(r = 1.2\)

Check:

\[ |r| = 1.2 > 1 \]

DIVERGES because \(|r| \geq 1\)

⚠️ Special Cases and Tricks

Case 1: r = 1

\[ \sum_{n=1}^{\infty} a = a + a + a + \cdots = \infty \]

DIVERGES (constant series)

Case 2: r = -1

\[ \sum_{n=1}^{\infty} a(-1)^{n-1} = a - a + a - a + \cdots \]

DIVERGES (oscillates)

Case 3: Negative r with |r| < 1

Series still converges! Just use \(|r|\) to check.

Example: \(r = -\frac{1}{2}\) → converges to \(\frac{a}{1-(-1/2)} = \frac{a}{3/2}\)

📊 Common Geometric Series

Geometric Series Examples
Series	r	Converges?	Sum (if convergent)
\(\sum_{n=0}^{\infty} \left(\frac{1}{2}\right)^n\)	\(\frac{1}{2}\)	Yes	\(2\)
\(\sum_{n=1}^{\infty} \frac{1}{3^n}\)	\(\frac{1}{3}\)	Yes	\(\frac{1}{2}\)
\(\sum_{n=0}^{\infty} 2^n\)	\(2\)	No	—
\(\sum_{n=1}^{\infty} (-\frac{1}{2})^n\)	\(-\frac{1}{2}\)	Yes	\(-\frac{1}{3}\)

💡 Essential Tips & Strategies

✅ Success Strategies:

Always find the first term: Substitute starting value of n
Use \(|r|\) for convergence test: Even if r is negative
Formula is \(\frac{\text{first term}}{1-r}\): Not \(\frac{a}{1-r}\) unless a IS the first term
Check indices carefully: Starting at 0 vs 1 affects first term
Simplify before identifying r: Make the pattern clear
Factor out constants: To see geometric pattern
Repeating decimals = geometric series: Common application
Remember \(|r| < 1\) for convergence: Not \(r < 1\)!

🔥 Quick Recognition Patterns:

\(\sum a^n\): First term = \(a\), ratio = \(a\)
\(\sum \frac{1}{b^n}\): Ratio = \(\frac{1}{b}\)
\(\sum c \cdot r^n\): First term = \(c \cdot r^{\text{starting index}}\)
Each term ÷ previous = r: Constant ratio check

❌ Common Mistakes to Avoid

Mistake 1: Using \(a\) as first term when it's not (check starting index!)
Mistake 2: Forgetting absolute value: using \(r < 1\) instead of \(|r| < 1\)
Mistake 3: Wrong first term when series starts at \(n = k \neq 0, 1\)
Mistake 4: Saying series converges when \(|r| = 1\) (boundary case diverges!)
Mistake 5: Not simplifying \(1 - r\) correctly (especially with negative r)
Mistake 6: Confusing \(\sum_{n=0}\) vs \(\sum_{n=1}\) (different first terms!)
Mistake 7: Using formula when \(|r| \geq 1\) (series diverges, no sum!)
Mistake 8: Arithmetic errors in \(\frac{a}{1-r}\) calculation
Mistake 9: Not recognizing geometric series in disguise
Mistake 10: Forgetting to check convergence before finding sum

📝 Practice Problems

Determine convergence and find sum if convergent:

\(\sum_{n=0}^{\infty} \frac{4}{5^n}\)
\(\sum_{n=1}^{\infty} 3 \cdot (0.8)^{n-1}\)
\(\sum_{n=2}^{\infty} \frac{7}{2^n}\)
\(\sum_{n=0}^{\infty} (-\frac{2}{3})^n\)
Express \(0.\overline{6}\) as a fraction

Answers:

Converges to \(5\) (first term = 4, r = 1/5)
Converges to \(15\) (first term = 3, r = 0.8)
Converges to \(\frac{7}{4}\) (first term = 7/4, r = 1/2)
Converges to \(\frac{3}{5}\) (first term = 1, r = -2/3, use |r|)
\(\frac{2}{3}\) (series: \(\sum_{n=1}^{\infty} \frac{6}{10^n}\))

✏️ AP® Exam Success Tips

What AP® Graders Look For:

Explicitly identify as geometric: State "This is a geometric series"
Show first term calculation: Substitute starting index value
Show r identification: "Common ratio r = ..."
Check \(|r| < 1\): Show the inequality
Write the formula: \(S = \frac{a}{1-r}\)
Show substitution: Plug in values clearly
Simplify answer: Reduce fractions
State conclusion: "Converges to..." or "Diverges because..."

💯 Exam Strategy:

Identify series as geometric (look for constant ratio)
Find first term by substituting starting index
Find r (ratio of consecutive terms)
Check \(|r| < 1\) for convergence
If convergent: use \(S = \frac{\text{first term}}{1-r}\)
If divergent: state why (\(|r| \geq 1\))
Simplify and box final answer

⚡ Quick Reference Guide

GEOMETRIC SERIES ESSENTIALS

The Formula:

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

ONLY if \(|r| < 1\)

General Rule:

\[ S = \frac{\text{first term}}{1-r} \text{ when } |r| < 1 \]

Convergence Test:

\(|r| < 1\) → CONVERGES
\(|r| \geq 1\) → DIVERGES

Remember:

Use absolute value for convergence test!
First term depends on starting index!
Check convergence before using formula!

Master Geometric Series! The geometric series \(\sum ar^{n-1} = a + ar + ar^2 + \cdots\) converges to \(\frac{a}{1-r}\) when \(|r| < 1\) and diverges when \(|r| \geq 1\). Critical steps: (1) Identify first term (substitute starting index), (2) Find r (ratio of consecutive terms), (3) Check \(|r| < 1\) using ABSOLUTE VALUE, (4) Apply \(S = \frac{\text{first term}}{1-r}\). Common forms: starting at n=0 vs n=1 gives same sum formula but different first term; starting at n=k requires first term = \(ar^k\). Applications: repeating decimals, compound interest, probability. Remember: boundary cases \(|r| = 1\) DIVERGE (including r = 1 and r = -1). Negative r still converges if \(|r| < 1\). This is THE foundation for power series and Taylor series later—absolutely essential! Practice until recognition and calculation are automatic! 🎯✨