10.2 Working with Geometric Series

A geometric series has a constant ratio $r$ between consecutive terms, written as $\sum_{n=0}^{\infty} ar^n$. It converges when $|r|<1$ and diverges when $|r|\geq 1$, and when it converges its sum equals $\frac{a}{1-r}$. For AP Calculus BC, identify $a$ and $r$ before applying the convergence rule.

How Do Geometric Series Work?

A geometric series adds terms that keep getting multiplied by the same common ratio. For AP Calculus BC, identify the first term $a$ and common ratio $r$, check whether $|r|<1$, and only use $\frac{a}{1-r}$ when the series converges.

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Why This Matters for the AP Calculus Exam

Geometric series are the first convergence test you learn in AP Calculus BC, and they show up across Unit 10. Recognizing a series as geometric lets you decide convergence quickly and find an exact sum, which other tests often cannot do. The skill also carries forward into power series and Taylor series, where geometric series logic helps you find intervals of convergence and rewrite functions as series. On both multiple-choice and free-response questions, you may need to identify the values of $a$ and $r$, justify convergence or divergence, and compute a sum with clear supporting work.

Key Takeaways

• A geometric series has a constant ratio $r$ between successive terms and an initial term $a$.
• The series $\sum_{n=0}^{\infty} a r^n$ converges only when $|r| < 1$; it diverges when $|r| \geq 1$.
• When it converges, the sum is $\frac{a}{1-r}$.
• An index shift between $\sum_{n=0}^{\infty} a r^n$ and $\sum_{n=1}^{\infty} a r^{n-1}$ does not change $a$, $r$, or the answer.
• The ratio $r$ can be negative; convergence depends on $|r|$, not the sign.
• To justify your answer, state that the result follows from the geometric series test and show the values of $a$ and $r$.

What Is a Geometric Series?

A geometric series is a series with a constant ratio between successive terms. You will see it written in two common forms. Watch where the index starts: the first begins at 0 and the second begins at 1.

Here $a$ is the initial term and $r$ is the ratio between any two consecutive terms. The values of $a$ and $r$ stay the same even though the index shifts, so both forms give the same answer when you check convergence or find a sum.

If you want a refresher on convergent versus divergent series first, see the 10.1 guide: Defining Convergent and Divergent Infinite Series.

This is an AP Calculus BC topic only. If you are taking Calculus AB, you can skip it.

Building a Geometric Series from a Sequence

Given a sequence, you can write it as a geometric series by finding $a$ and $r$.

Example 1

Write the geometric series for:

Step 1: Find $a$ and $r$. The initial term is $a = 27$. To find $r$, see what factor turns one term into the next. Since $9 \cdot \frac{1}{3} = 3$, you get $r = \frac{1}{3}$.

Step 2: Plug into the geometric form.

Example 2

Write the geometric series for:

Step 1: Find $a$ and $r$.

Step 2: Plug into the geometric form.

Notice that $r$ is negative here. The ratio can be any real number, because convergence is decided by the absolute value of $r$, not its sign.

The Geometric Series Test

A geometric series converges when $|r| < 1$ and diverges when $|r| \geq 1$, where $a$ is the first term and $r$ is the common ratio.

When the series converges, its sum is:

Example 1 continued

Since $r = \frac{1}{3}$ and $|r| < 1$, the series converges. Find the sum:

A clear answer would state: the sum of the series is $40.5$ by the geometric series test. Your supporting work just needs the series in geometric form and the sum computation.

Example 2 continued

Since $r = -3$, we have $|r| \geq 1$, so the series diverges. State that it diverges by the geometric series test.

How to Use This on the AP Calculus Exam

MCQ

• Spot a geometric series by checking for a constant ratio between terms.
• Read off $a$ and $r$ carefully, then test $|r|$ against 1 to decide convergence.
• If it converges, apply $\frac{a}{1-r}$. Watch out for answer choices that use the wrong starting term.

Free Response

• When a problem gives a series, show that it is geometric by identifying $a$ and $r$.
• For convergence, state the condition you checked: $|r| < 1$ converges, $|r| \geq 1$ diverges.
• For a sum, write $\frac{a}{1-r}$ with your values plugged in, then simplify. Clean notation makes your reasoning easy to follow.

Common Trap

• If a series starts at an index other than 0 or 1, re-identify the true first term $a$ before using $\frac{a}{1-r}$. The formula uses the actual first term of the series you are summing, not just the coefficient out front.

Common Misconceptions

• Thinking the sum formula works for any geometric series. The formula $\frac{a}{1-r}$ only applies when $|r| < 1$. If $|r| \geq 1$, the series diverges and has no finite sum.
• Letting a negative $r$ trick you. A negative ratio is fine; convergence depends on $|r|$. For example, $r = -\frac{1}{2}$ converges because $|r| < 1$.
• Confusing $a$ with the coefficient when the index is shifted. The value of $a$ is the actual first term being added. Switching between the $n=0$ and $n=1$ forms does not change $a$ or $r$.
• Mixing up the convergence value with a partial sum. $\frac{a}{1-r}$ gives the full infinite sum when the series converges, not just the sum of the first few terms.
• Assuming geometric and arithmetic series behave the same. Geometric series have a constant ratio between terms, while arithmetic series have a constant difference. The convergence rules here apply only to geometric series.

Related AP Calculus Guides

• Unit 10 Overview: Infinite Series and Sequences
• 10.1 Defining Convergent and Divergent Infinite Series
• 10.3 The nth Term Test for Divergence
• 10.5 Harmonic Series and p-Series
• 10.4 Integral Test for Convergence
• 10.6 Comparison Tests for Convergence