AP®MathsCollege BoardCalculus BCStudy GuidesUnit 10: Infinite Sequences and SeriesIntroduction to Infinite SeriesGeometric Series

Geometric Series (College Board AP® Calculus BC): Study Guide

Written by: Roger B

Reviewed by: Mark Curtis

Updated on 19 May 2026

Geometric series

What is a geometric series?

A geometric series is a series of the form $\sum_{n = 0}^{\infty} a r^{n} = a + a r + a r^{2} + . . .$
- where $a \neq 0$ is a real number constant
  - $a$ is also the first term of the series
- and $r$ is a real number known as the common ratio
- Note that the series starts with $n = 0$
Note that there is a constant ratio between one term and the next
- For example, $\sum_{n = 0}^{\infty} 9 \cdot {(\frac{1}{3})}^{n} = 9 + 3 + 1 + \frac{1}{3} + . . .$
  - Each term is the preceding term multiplied by the common ratio $\frac{1}{3}$

When does a geometric series converge?

The sum of the terms up to and including $a_{k}$ is given by
- $\sum_{n = 0}^{k} a \cdot r^{n} = \frac{a (1 - r^{k + 1})}{1 - r}$
- If $r = 1$ , then the series is $a + a + a + . . .$ so $a (k + 1)$
If $|r| < 1$ then the geometric series $\sum_{n = 0}^{\infty} a r^{n}$ converges
- and $\sum_{n = 0}^{\infty} a r^{n} = \frac{a}{1 - r}$
  - This comes from $\sum_{n = 0}^{\infty} a r^{n} = \lim_{n \to \infty} (\frac{a (1 - r^{n + 1})}{1 - r}) = \frac{a (1 - 0)}{1 - r} = \frac{a}{1 - r}$
If $|r| \geq 1$ then the geometric series diverges
- For example, if $r = 1$ , then the series is $\sum_{n = 0}^{\infty} a \cdot 1^{n} = a + a + a + a + . . .$
  - The sequence of partial sums is $a, 2 a, 3 a, 4 a, . . .$ , which diverges to $\pm \infty$ (depending on whether $a$ is positive or negative)
  - So the series diverges
- Or if $a = 5$ and $r = 2$ , then the series is $\sum_{n = 0}^{\infty} 5 \cdot 2^{n} = 5 + 10 + 20 + 40 + . . .$
  - Each term added on is bigger than the one before
  - The sequence of partial sums is $5, 15, 35, 75, . . .$ , which diverges to $+ \infty$
  - So the series diverges

Examiner Tips and Tricks

Take note of the lower limit of a summation.

If it starts with 1 rather than 0, then you can rewrite it so that it starts at 0.

For example, $\sum_{n = 1}^{\infty} (\frac{3}{5^{n}}) = \sum_{n = 0}^{\infty} \frac{3}{5} \cdot (\frac{1}{5^{n}})$ .

However, you only need to know the first term and the common ratio in order to use the formula. In the above example, $a = \frac{3}{5}$ and $r = \frac{1}{5}$ .

Worked Example

Consider the infinite series $\sum_{n = 1}^{\infty} \frac{1}{2^{n}}$ . Show that the series converges and determine the value of the sum.

Answer:

Rewrite the series so that it has the standard form of a geometric series

$\begin{array}{rcl} \sum_{n = 1}^{\infty} \frac{1}{2^{n}} & = & \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + . . . \\ = & \frac{1}{2} \cdot {(\frac{1}{2})}^{0} + \frac{1}{2} \cdot {(\frac{1}{2})}^{1} + \frac{1}{2} \cdot {(\frac{1}{2})}^{2} + . . \\ = & \sum_{n = 0}^{\infty} \frac{1}{2} \cdot {(\frac{1}{2})}^{n} \end{array}$

Note that we've changed the sum to start with $n = 0$

That is a geometric series with $a = \frac{1}{2}$ and $r = \frac{1}{2}$

Show that it satisfies the convergence condition

That is a geometric series with $|r| = |\frac{1}{2}| = \frac{1}{2} < 1$ so it converges

Use the formula $\sum_{n = 0}^{\infty} a r^{n} = \frac{a}{1 - r}$

So $\sum_{n = 1}^{\infty} \frac{1}{2^{n}} = \sum_{n = 0}^{\infty} \frac{1}{2} \cdot {(\frac{1}{2})}^{n} = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1$

$\sum_{n = 1}^{\infty} \frac{1}{2^{n}} = 1$

Worked Example

Consider the number $0 . \dot{6} 2 \dot{1} = 0.621621621621 . . .$ .

(a) Write the number in the form of a geometric series.

(b) Use the result from part (a) to find the value of $0 . \dot{6} 2 \dot{1}$ as a fraction in lowest terms.

Answer:

(a)

Use the fact that $0 . \dot{6} 2 \dot{1} = 0.621 + 0.000621 + 0.000000621 + . . .$

$\begin{array}{rcl} 0 . \dot{6} 2 \dot{1} & = & \frac{621}{1000} + \frac{621}{1000000} + \frac{621}{1000000000} + . . . \\ = & \frac{621}{1000} + \frac{621}{{(1000)}^{2}} + \frac{621}{{(1000)}^{3}} + . . . \\ = & \frac{621}{1000} \cdot {(\frac{1}{1000})}^{0} + \frac{621}{1000} \cdot {(\frac{1}{1000})}^{1} + \frac{621}{1000} \cdot {(\frac{1}{1000})}^{2} + . . . \\ = & \sum_{n = 0}^{\infty} \frac{621}{1000} \cdot {(\frac{1}{1000})}^{n} \end{array}$

$\begin{array}{rcl} 0 . \dot{6} 2 \dot{1} & = & \sum_{n = 0}^{\infty} \frac{621}{1000} \cdot {(\frac{1}{1000})}^{n} \end{array}$

(b)

$\begin{array}{rcl} \sum_{n = 0}^{\infty} \frac{621}{1000} \cdot {(\frac{1}{1000})}^{n} \end{array}$ is a geometric series with $a = \frac{621}{1000}$ and $r = \frac{1}{1000}$

First show that the series converges

For that geometric series, $|r| = |\frac{1}{1000}| = \frac{1}{1000} < 1$ so it converges

Now you can use the formula $\sum_{n = 0}^{\infty} a r^{n} = \frac{a}{1 - r}$

Therefore

$\begin{array}{rcl} \sum_{n = 0}^{\infty} \frac{621}{1000} \cdot {(\frac{1}{1000})}^{n} & = & \frac{\frac{621}{1000}}{1 - \frac{1}{1000}} \\ = & \frac{\frac{621}{1000}}{\frac{999}{1000}} \\ = & \frac{621}{999} \\ = & \frac{69}{111} \\ = & \frac{23}{37} \end{array}$

$\begin{array}{rcl} 0 . \dot{6} 2 \dot{1} & = & \frac{23}{37} \end{array}$

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Geometric Series (College Board AP® Calculus BC): Study Guide

Geometric series

What is a geometric series?

When does a geometric series converge?

Examiner Tips and Tricks

Worked Example

Worked Example

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Author: Roger B

Reviewer: Mark Curtis