# Working with Geometric Series

> AP Calculus BC · Unit 10: Infinite Sequences and Series
> Source: https://www.owlsprep.com/study/ap-calculus-bc-u10-working-with-geometric-series/

This module covers core properties of finite and infinite geometric series, convergence rules, sum calculations, repeating decimal conversion, and real-world applications, a key foundational topic for AP Calculus BC Unit 10.

**Prerequisites:** Limits of sequences; Basic exponent rules; Algebraic factoring

## Learning objectives

- Identify geometric sequences and series and their common ratio
- Determine convergence of infinite geometric series using the $|r|<1$ rule
- Calculate exact sums of convergent geometric series in standard and non-standard forms
- Convert repeating decimals to exact fractions using geometric series
- Solve real-world application problems involving infinite geometric series

## Definition and Convergence of Geometric Series

A geometric series is the sum of terms of a geometric sequence, where each term after the first is the previous term multiplied by a constant non-zero common ratio $r$. Unlike most other infinite series, convergent geometric series have an exact closed-form sum, making them a foundational tool for more advanced series topics.

**Geometric Sequence and Series** — A geometric sequence has a constant common ratio between consecutive terms. A geometric series is the sum of the terms of a geometric sequence.

*Notation:* $a_n = a_1 r^{n-1}, \sum_{n=0}^\infty a_1 r^n$

*Example:* $\sum_{n=0}^\infty 3\left(\frac{1}{2}\right)^n$ has $a_1=3, r=\frac{1}{2}$

$$S_n = \sum_{k=0}^{n-1} a_1 r^k = a_1 \frac{1 - r^n}{1 - r}, \quad r \neq 1$$

To find the sum of an infinite geometric series, we take the limit of the partial sum as $n \to \infty$. If $|r| < 1$, $r^n \to 0$, so we get a finite convergent sum:

$$S = \sum_{n=0}^{\infty} a_1 r^n = \frac{a_1}{1 - r}, \quad |r| < 1$$

An infinite geometric series converges **if and only if** $|r| < 1$. If $|r| \geq 1$, the partial sums do not approach a finite limit, so the series diverges.

**Worked example:** Determine whether the infinite series $\sum_{n=1}^{\infty} 4 \left(\frac{3}{4}\right)^{n}$ converges. If it converges, find its exact sum.

1. Confirm the series is geometric by calculating the ratio of consecutive terms:
2. $$n=1: 4 \cdot \left(\frac{3}{4}\right) = 3; \quad n=2: 4 \cdot \left(\frac{3}{4}\right)^2 = \frac{9}{4}; \quad r = \frac{9/4}{3} = \frac{3}{4}$$
3. Check the convergence condition: $|r| = \frac{3}{4} < 1$, so the series converges.
4. Calculate the first term at the starting index $n=1$:
5. $$a_1 = 4 \cdot \left(\frac{3}{4}\right)^1 = 3$$
6. Apply the infinite sum formula:
7. $$S = \frac{a_1}{1 - r} = \frac{3}{1 - 3/4} = 12$$

> **Exam tip:** Always confirm $|r| < 1$ before writing a finite sum. AP exam readers will deduct points if you state a finite sum for a divergent geometric series, even if you correctly plug values into the formula.

## Rewriting Non-Standard Geometric Series

Most AP exam questions do not present geometric series in the neat $\sum ar^n$ standard form. You will often need to simplify exponents, factor constants, or reindex the series to correctly identify $a_1$ and $r$. The core strategy is to isolate the power of the index variable $n$ in the exponent, so that you can write every term as $c \cdot r^n$.

**Worked example:** Find the exact sum of the convergent series $\sum_{n=0}^{\infty} \frac{2^{n+2}}{3^{n-1}}$.

1. Use exponent rules to split constant terms away from terms with $n$:
2. $$\frac{2^{n+2}}{3^{n-1}} = 2^n \cdot 2^2 \cdot 3^{-n} \cdot 3^{1} = 12 \left(\frac{2}{3}\right)^n$$
3. Check convergence: $|r| = \frac{2}{3} < 1$, so the series converges.
4. Identify the first term for $n=0$:
5. $$a_1 = 12 \left(\frac{2}{3}\right)^0 = 12$$
6. Apply the infinite sum formula:
7. $$S = \frac{12}{1 - 2/3} = 36$$

> **Exam tip:** When rewriting exponents, always separate all constants from the index term before identifying $r$ to avoid mixing up the common ratio.

## Converting Repeating Decimals to Fractions

One of the most common concrete applications of convergent geometric series is converting repeating decimals to their exact fractional form. A repeating decimal can be split into a finite non-repeating part and an infinite geometric repeating series, with a common ratio of $10^{-k}$ where $k$ is the number of digits in the repeating block.

**Worked example:** Convert the repeating decimal $0.3\overline{14}$ (equal to $0.3141414...$) to an exact fraction in lowest terms.

1. Split the decimal into non-repeating and infinite repeating parts:
2. $$0.3\overline{14} = 0.3 + 0.014 + 0.00014 + 0.0000014 + ...$$
3. Identify $a_1$ and $r$: the repeating block has 2 digits, so $r = \frac{1}{10^2} = \frac{1}{100}$, and the first term of the repeating series is $a_1 = 0.014 = \frac{14}{1000}$.
4. Calculate the sum of the repeating series:
5. $$S_{\text{repeat}} = \frac{a_1}{1 - r} = \frac{14/1000}{99/100} = \frac{14}{990}$$
6. Add the non-repeating part with a common denominator:
7. $$0.3 = \frac{3}{10} = \frac{297}{990}, \quad \text{Total} = \frac{297 + 14}{990} = \frac{311}{990}$$
8. 311 is prime, so this fraction is already in lowest terms.

> **Exam tip:** Count the number of digits in the repeating block correctly. A 2-digit repeating block always gives $r = 1/100$, count again to confirm.

## Applications and Practice

**Worked example:** A ball is dropped from a height of 10 meters. Each time it bounces, it reaches 75% of the height of the previous bounce. What is the total vertical distance the ball travels before coming to rest? Include units in your answer.

1. The ball first travels 10 meters downward before the first bounce. After each bounce, it travels up to the new height then falls the same distance, so every bounce after the first contributes twice the height to the total distance.
2. Write the total distance $D$ as an infinite geometric series:
3. $$D = 10 + 2\left(10 \cdot 0.75\right) + 2\left(10 \cdot 0.75^2\right) + 2\left(10 \cdot 0.75^3\right) + ...$$
4. Factor out constants to get a standard geometric series:
5. $$D = 10 + 15 \sum_{n=0}^{\infty} (0.75)^n$$
6. The infinite series has $a=1, r=0.75$, so its sum is:
7. $$\frac{1}{1 - 0.75} = 4$$
8. Calculate the total distance:
9. $$D = 10 + 15(4) = 70 \text{ meters}$$

**Check your understanding**

Test your understanding of non-standard starting indexes:

1. Which of the following is the sum of $\sum_{n=3}^{\infty} 2 \left(-\frac{1}{2}\right)^n$?

   - $\frac{1}{12}$
   - $-\frac{1}{12}$
   - $\frac{1}{6}$
   - $-\frac{1}{6}$

   *Answer:* $-\frac{1}{6}$

   *Why:* First check convergence: $|r| = 1/2 < 1$, so the series converges. The first term at $n=3$ is $2(-1/2)^3 = -1/4$. Then $S = \frac{-1/4}{1 - (-1/2)} = -\frac{1}{6}$, which matches option D.

## Common pitfalls

- **Wrong:** For $\sum_{n=1}^{\infty} 5 (1/2)^n$, you use $a_1 = 5$ to get a sum of $10$.
  - Why it fails: You confuse starting indexes, assuming the constant coefficient is always the first term even when the exponent is non-zero at the starting index.
  - Correct: Always plug in the starting value of $n$ to calculate the first term explicitly instead of assuming the constant is the first term.
- **Wrong:** You conclude a geometric series with $r=-0.9$ diverges because $-0.9 < 1$.
  - Why it fails: You forget the convergence condition uses the absolute value of $r$, not $r$ itself.
  - Correct: Always compute $|r|$ first when checking convergence, regardless of the sign of $r$.
- **Wrong:** You write the sum of $\sum_{n=0}^{\infty} (5/4)^n$ as $1/(1 - 5/4) = -4$ and accept that as the final answer.
  - Why it fails: You remember the formula but forget it is only valid when the series converges.
  - Correct: Always check $|r| < 1$ first; if $|r| \geq 1$, state the series diverges and does not have a finite sum, do not apply the formula.
- **Wrong:** For $\sum_{n=2}^{\infty} 3 (1/2)^n$, you take the full sum from $n=0$ and subtract twice the first term to get the sum from $n=2$.
  - Why it fails: You incorrectly assume skipping the first two terms just means subtracting twice the first term, instead of subtracting the actual values of the first two terms.
  - Correct: When finding the sum of a geometric series starting at $n = k > 0$, either reindex explicitly to get the new first term, or factor out $r^k$ to get the correct starting constant.
- **Wrong:** When converting $0.\overline{12}$, you set $a_1 = 12$ and $r=1/10$, and end up with an answer greater than 1.
  - Why it fails: You forget to shift the decimal correctly for the first term of the repeating series.
  - Correct: Write out the first term of the repeating part as a decimal explicitly, or use $a_1 = \frac{\text{repeating block}}{10^k}$ where $k$ is the number of digits after the decimal point.

## Cheatsheet

| Category | Formula | Notes |
| --- | --- | --- |
| Finite geometric sum ($n$ terms starting at $n=0$) | $S_n = a_1 \frac{1 - r^n}{1 - r}, r \neq 1$ | Valid for any $r \neq 1$, whether the infinite series converges or not |
| Infinite geometric convergence | Converges $\iff \|r\| < 1$ | Diverges if $\|r\| \geq 1$, regardless of the sign of $r$ |
| Sum of convergent infinite geometric (starts at $n=0$) | $S = \frac{a_1}{1 - r}$ | $a_1$ is the first term of the series |
| Sum starting at $n = k$ | $S = \frac{a_0 r^k}{1 - r}$ | $a_0$ is the first term for $n=0$; factor out $r^k$ to get this |
| Repeating decimal $0.\overline{d}$ ($d$ is $k$-digits) | $S = \frac{d}{10^k - 1}$ | Only applies when the repeating block starts after the decimal |
| Repeating decimal $0.a\overline{b}$ ($a$: $m$ digits, $b$: $n$ digits) | $S = \frac{10^m a + b - a}{10^m(10^n - 1)}$ | $a, b$ are integer values of the non-repeating and repeating blocks |
| Exponent rewrite rule | $\frac{c^{n + k}}{d^{n + m}} = c^k d^{-m} \left(\frac{c}{d}\right)^n$ | Isolate constants to identify $r$ correctly |
| Constant multiple rule | $\sum c a r^n = c \cdot \frac{a}{1 - r}$ | Constants factor out of convergent infinite sums, same as finite sums |

## What's next

Mastery of geometric series is a foundational prerequisite for almost all other topics in AP Calculus BC Unit 10. You will use geometric series to find the interval of convergence of power series, derive closed-form expressions for common power series, and as a comparison in convergence tests like the ratio test. Without solid proficiency in geometric series, more advanced topics like Taylor series and power series integration will be much harder to master on exam day.

- [The nth term test for divergence](https://www.owlsprep.com/study/ap-calculus-bc-u10-the-nth-term-test-for/)
- [Integral Test for Convergence](https://www.owlsprep.com/study/ap-calculus-bc-u10-integral-test-for-convergence/)
- [Harmonic series and p-series](https://www.owlsprep.com/study/ap-calculus-bc-u10-harmonic-series-and-p-series/)

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