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Working with
Geometric Series

Master convergence, partial sums, and infinite series for sequences with a constant ratio between consecutive terms.

AP Calculus BC

What Is a Geometric Series?

A geometric series is the sum of the terms of a geometric sequence: a sequence where each term is obtained by multiplying the previous term by a fixed number called the common ratio \( r \).

The key question in calculus: does this infinite sum converge to a finite value, or does it diverge to infinity?

Visualizing Shrinking Terms

When \( |r| < 1 \), each term gets smaller and smaller. The partial sums approach a finite limit. Below, the bars show terms of \( \sum 1 \cdot (0.5)^n \):

0 0.5 1.0 1 0.5 0.25 0.125 0.0625 0.031 0.016 n=0 n=1 n=2 n=3 n=4 n=5 n=6

Why Geometric Series Matter

Geometric series appear throughout calculus, physics, economics, and computer science. They are one of the few infinite series with a closed-form sum.

Power Series
Many Taylor series are built from geometric series as a foundation: \( \frac{1}{1-x} = \sum x^n \).
Physics
Bouncing ball distances, electrical circuit feedback, and radioactive decay all follow geometric patterns.
Finance
Present value of annuities and compound interest calculations rely on geometric sums.

The Convergence Theorem

-1 0 1
Converges: \(|r| < 1\)
Diverges Diverges

The open circles at \( r = -1 \) and \( r = 1 \) indicate that the series diverges at those endpoints as well.

A bouncing ball leaves a shrinking ladder of travel

An orange basketball rises above a gym floor while two faint silhouettes suggest earlier and later stages of diminishing rebounds; their heights are illustrative.A ball falls 10 m and rebounds to 60% of its preceding height. Count the initial fall once and each rebound twice: the geometric series gives 40 m of total travel.

A bouncing ball leaves a shrinking ladder of travel

A ball falls 10 m and rebounds to 60% of its preceding height. Count the initial fall once and each rebound twice: the geometric series gives 40 m of total travel.A ball falls 10 m and rebounds to 60% of its preceding height. Count the initial fall once and each rebound twice: the geometric series gives 40 m of total travel.

The Partial Sum Formula

Before tackling infinite sums, we need the finite partial sum of the first \( n \) terms:

1
Write \( S_n = a + ar + ar^2 + \cdots + ar^{n-1} \).
2
Multiply both sides by \( r \): \( rS_n = ar + ar^2 + \cdots + ar^n \).
3
Subtract: \( S_n - rS_n = a - ar^n \), so \( S_n(1-r) = a(1-r^n) \).
4
Divide by \( (1-r) \) to get the formula. As \( n \to \infty \) with \( |r| < 1 \), \( r^n \to 0 \), giving \( S = \frac{a}{1-r} \).

Finite vs. Infinite Geometric Sums

Finite Sum \( S_n \) Infinite Sum \( S \)
Formula \( a \cdot \frac{1-r^n}{1-r} \) \( \frac{a}{1-r} \)
Restriction on \( r \) \( r \neq 1 \) (any other \( r \)) \( |r| < 1 \) strictly
Number of terms Exactly \( n \) terms Infinitely many terms
Always finite? Yes Only if \( |r| < 1 \)
AP Exam usage Less common Very common

Watch the Starting Index

Not every geometric series starts at \( n=0 \). The formula \( \frac{a}{1-r} \) applies when the first term is \( a \). If the series starts at \( n=k \), adjust accordingly:

Starts at \( n=0 \)
\( \sum_{n=0}^{\infty} ar^n = \frac{a}{1-r} \)
First term is \( a \).
Starts at \( n=1 \)
\( \sum_{n=1}^{\infty} ar^n = \frac{ar}{1-r} \)
First term is \( ar \).

Step-by-Step Strategy

1
Identify the form. Rewrite the series as \( \sum ar^n \). Find the first term \( a \) and common ratio \( r \).
2
Check convergence. Is \( |r| < 1 \)? If not, state that the series diverges.
3
Apply the formula. Use \( S = \frac{a}{1-r} \) where \( a \) is the actual first term (plug in the starting index).
4
Simplify. Reduce the fraction. Watch for negative \( r \) values: \( 1-(-r) = 1+r \).
5
Verify. Check by computing the first few partial sums: do they approach your answer?

Example 1: Basic Convergent Series

Find the sum of \( \displaystyle\sum_{n=0}^{\infty} 3\left(\frac{1}{4}\right)^n \).

1
Identify: \( a = 3 \), \( r = \frac{1}{4} \).
2
Check: \( |r| = \frac{1}{4} < 1 \) ✓ Series converges.
3
Apply: \( S = \frac{3}{1 - \frac{1}{4}} = \frac{3}{\frac{3}{4}} = 3 \cdot \frac{4}{3} = 4 \).

Example 2: Negative Common Ratio

Find the sum of \( \displaystyle\sum_{n=0}^{\infty} 5\left(-\frac{2}{3}\right)^n \).

1
Identify: \( a = 5 \), \( r = -\frac{2}{3} \).
2
Check: \( |r| = \frac{2}{3} < 1 \) ✓ Series converges (alternating terms).
3
Apply: \( S = \frac{5}{1 - \left(-\frac{2}{3}\right)} = \frac{5}{1 + \frac{2}{3}} = \frac{5}{\frac{5}{3}} = 5 \cdot \frac{3}{5} = 3 \).

Example 3: Non-Zero Starting Index

Evaluate \( \displaystyle\sum_{n=2}^{\infty} \frac{7}{5^n} \).

1
Rewrite: \( \sum_{n=2}^{\infty} 7 \cdot \left(\frac{1}{5}\right)^n \). Here \( r = \frac{1}{5} \).
2
First term: Plug in \( n = 2 \): first term = \( \frac{7}{25} \). So \( a = \frac{7}{25} \).
3
Apply: \( S = \frac{7/25}{1 - 1/5} = \frac{7/25}{4/5} = \frac{7}{25} \cdot \frac{5}{4} = \frac{7}{20} \).

Partial Sums Approaching the Limit

For \( \sum_{n=0}^{\infty} 3(1/4)^n \), the partial sums \( S_n \) converge to 4:

0 1 2 3 4 S=4 n (number of terms) 3 3.75 3.94 1 2 3 4 5 6 7

Rewriting as Geometric Form

Many AP problems disguise geometric series. The key is rewriting into \( \sum ar^n \) form:

1
Exponential in numerator: \( \sum \frac{2^n}{3^n} = \sum \left(\frac{2}{3}\right)^n \). Here \( r = \frac{2}{3} \).
2
Constant in numerator: \( \sum \frac{4}{7^n} = 4 \sum \left(\frac{1}{7}\right)^n \). Here \( a = 4, r = \frac{1}{7} \).
3
Split exponents: \( \sum \frac{(-1)^n}{4^n} = \sum \left(\frac{-1}{4}\right)^n \). Here \( r = -\frac{1}{4} \).
4
Power functions: \( \sum \frac{3^{n+1}}{5^n} = 3 \sum \left(\frac{3}{5}\right)^n \). Factor out the extra 3.

Common Mistakes to Avoid

✕ Wrong First Term
Using the coefficient \( a \) instead of the actual first term when the index starts at \( n \neq 0 \). Always plug in the starting index.
✕ Forgetting \( |r| \) Check
Applying \( \frac{a}{1-r} \) without verifying \( |r| < 1 \). If \( |r| \geq 1 \), the series diverges: there is no sum.
✕ Sign Error with Negative \( r \)
Writing \( 1 - (-r) \) as \( 1 - r \) instead of \( 1 + r \). The double negative is the most common algebra slip.

Summary Reference Table

Scenario Formula Result
Finite sum (\( n \) terms) \( S_n = a \cdot \frac{1-r^n}{1-r} \) Always a number
Infinite, \( |r| < 1 \) \( S = \frac{a}{1-r} \) Converges
Infinite, \( |r| \geq 1 \) No formula applies Diverges
Starts at \( n = k \) \( S = \frac{a \cdot r^k}{1-r} \) Use first term = \( ar^k \)
Negative \( r \) \( S = \frac{a}{1+|r|} \) Alternating signs

Question 1 of 5

What is the sum of the geometric series \( \displaystyle\sum_{n=0}^{\infty} 6\left(\frac{1}{3}\right)^n \)?

Question 2 of 5

Which of the following geometric series diverges?

Question 3 of 5

Find the sum: \( \displaystyle\sum_{n=1}^{\infty} \frac{8}{3^n} \).

Question 4 of 5

Evaluate \( \displaystyle\sum_{n=0}^{\infty} \frac{(-2)^n}{5^n} \).

Question 5 of 5

The series \( \displaystyle\sum_{n=0}^{\infty} ar^n \) converges to \( 10 \). If \( a = 4 \), what is \( r \)?