# Geometric Series: Sum Formula, Convergence and Examples

> Learn the geometric series sum formula, why it converges only when |r| < 1, how partial sums behave, and how to turn repeating decimals into fractions.

Source: https://practicecalculus.com/blog/geometric-series
Updated: 2026-10-06

## Key takeaways

- A geometric series multiplies each term by the same ratio $r$: $a + ar + ar^2 + \cdots$.
- It converges exactly when $|r| < 1$, and then its sum is $\frac{a}{1-r}$, where $a$ is the first term.
- The $n$th partial sum is $S_n = \frac{a(1-r^n)}{1-r}$. The infinite sum is what $S_n$ approaches as $r^n \to 0$.
- Every repeating decimal is a geometric series, so it always equals a fraction.

Add $1 + \frac12 + \frac14 + \frac18 + \cdots$ forever. Infinitely many positive terms, yet the total never passes 2. **Geometric series** are the one kind of infinite sum you can add up exactly with a single formula, and they are the foundation for everything else in the study of series, from convergence tests to power series.

## What is a geometric series?

A **geometric series** is a sum in which each term is the previous term multiplied by a fixed number $r$, called the **common ratio**:

$$\sum_{n=0}^{\infty} a r^n = a + ar + ar^2 + ar^3 + \cdots$$

Here $a$ is the **first term**. To find $r$, divide any term by the one before it. For $1 + \frac12 + \frac14 + \cdots$, the first term is $a = 1$ and the ratio is $r = \frac12$.

## The geometric series sum formula

The **$n$th partial sum** adds the first $n$ terms:

$$S_n = a + ar + \cdots + ar^{n-1} = \frac{a(1 - r^n)}{1 - r} \qquad (r \ne 1)$$

The infinite sum is the limit of $S_n$ as $n \to \infty$. If $|r| < 1$, then $r^n \to 0$ and the partial sums settle down:

$$\sum_{n=0}^\infty ar^n = \frac{a}{1-r} \qquad \text{when } |r| < 1$$

A handy way to remember it: **sum = first term over one minus ratio.**

### Where the formula comes from

Write out $S_n$, multiply by $r$ and subtract:

$$S_n = a + ar + ar^2 + \cdots + ar^{n-1}$$

$$rS_n = ar + ar^2 + \cdots + ar^{n-1} + ar^n$$

Every middle term cancels, leaving $S_n - rS_n = a - ar^n$. Factor and divide:

$$S_n(1 - r) = a(1 - r^n) \quad\Longrightarrow\quad S_n = \frac{a(1-r^n)}{1-r}$$

That subtraction trick is worth remembering. It works on finite and infinite geometric sums alike.

## When does a geometric series converge?

Everything depends on the size of $r$:

| Ratio            | What $r^n$ does            | The series                           |
| ---------------- | -------------------------- | ------------------------------------ |
| $\lvert r\rvert < 1$ | shrinks to 0           | converges to $\frac{a}{1-r}$         |
| $r = 1$          | stays 1                    | diverges ($a + a + a + \cdots$)      |
| $r = -1$         | flips between $1$ and $-1$ | diverges (partial sums bounce)       |
| $\lvert r\rvert > 1$ | grows without bound    | diverges                             |

(We assume $a \ne 0$ throughout.) So a geometric series converges **if and only if** $|r| < 1$. This clean rule is why geometric series are used as a benchmark in the [comparison and ratio tests](/blog/convergence-tests).

## Seeing the partial sums approach the limit

For $1 + \frac12 + \frac14 + \cdots$, the partial sum formula gives $S_n = \frac{1 - (1/2)^n}{1 - 1/2} = 2 - 2\left(\frac12\right)^n$. The first few values are $1, 1.5, 1.75, 1.875, 1.9375$. Each one closes half the remaining gap to 2.

```graph
title: 'Partial sums of 1 + 1/2 + 1/4 + ... approaching 2'
caption: 'The dots are the partial sums $S_n = 2 - 2\left(\frac12\right)^n$. They climb toward the dashed line $y = 2$, the sum of the series, but never pass it.'
x: [0, 7]
y: [0, 2.5]
functions:
  - expr: 2 - 2(0.5)^x
    label: partial sum formula
  - expr: '2'
    label: y = 2 (the sum)
    dashed: true
points:
  - [1, 1, '1']
  - [2, 1.5, '1.5']
  - [3, 1.75, '1.75']
  - [4, 1.875, '1.875']
  - [5, 1.9375, '1.94']
```

When $r$ is negative, the partial sums **alternate** around the limit instead of climbing to it. For $1 - \frac12 + \frac14 - \frac18 + \cdots$ (with $r = -\frac12$), the partial sums are $1, 0.5, 0.75, 0.625, 0.6875, \dots$, bouncing above and below the sum $\frac{1}{1 + 1/2} = \frac23$.

```graph
title: 'Partial sums of 1 − 1/2 + 1/4 − ... alternating around 2/3'
caption: 'With $r = -\frac12$, the partial sums overshoot and undershoot the sum $\frac23$, getting closer each time.'
x: [0, 7]
y: [0, 1.2]
functions:
  - expr: 2/3
    label: y = 2/3 (the sum)
    dashed: true
points:
  - [1, 1, '1']
  - [2, 0.5, '0.5']
  - [3, 0.75, '0.75']
  - [4, 0.625, '0.625']
  - [5, 0.6875, '0.6875']
```

## Worked example 1: a sum that starts at n = 1

Find $\sum_{n=1}^{\infty} 3\left(\frac25\right)^n$.

**Step 1. Find the first term.** Plug in $n = 1$: $a = 3\cdot\frac25 = \frac65$.

**Step 2. Find the ratio.** Each term is multiplied by $\frac25$, so $r = \frac25$.

**Step 3. Check convergence.** $|r| = \frac25 < 1$, so the series converges.

**Step 4. Apply the formula.**

$$\sum_{n=1}^\infty 3\left(\frac25\right)^n = \frac{6/5}{1 - 2/5} = \frac{6/5}{3/5} = 2$$

## Worked example 2: a negative ratio

Find $\sum_{n=0}^{\infty} \frac{(-2)^n}{3^{n+1}}$.

**Step 1. Rewrite in standard form.** $\frac{(-2)^n}{3^{n+1}} = \frac13\cdot\frac{(-2)^n}{3^n} = \frac13\left(-\frac23\right)^n$.

**Step 2. Read off $a$ and $r$.** At $n = 0$ the term is $\frac13$, so $a = \frac13$ and $r = -\frac23$.

**Step 3. Check.** $|r| = \frac23 < 1$. It converges.

**Step 4. Sum.**

$$\frac{1/3}{1 - (-2/3)} = \frac{1/3}{5/3} = \frac15$$

## Worked example 3: a divergent series

Does $\sum_{n=0}^\infty \frac{3^n}{2^n}$ converge?

This is geometric with $a = 1$ and $r = \frac32$. Since $|r| > 1$, the terms grow and the series **diverges**. Plugging into $\frac{a}{1-r}$ would give $\frac{1}{1 - 3/2} = -2$, a meaningless answer. Always check $|r| < 1$ before using the formula.

## Worked example: a finite geometric sum

The partial sum formula works for any finite geometric sum, even when $|r| > 1$. Find $2 + 6 + 18 + \cdots + 2\cdot 3^9$.

**Step 1. Identify $a$, $r$ and $n$.** The first term is $a = 2$, the ratio is $r = 3$, and the exponents run from $0$ to $9$, so there are $n = 10$ terms.

**Step 2. Apply $S_n = \frac{a(1-r^n)}{1-r}$.**

$$S_{10} = \frac{2(1 - 3^{10})}{1 - 3} = \frac{2(1 - 59{,}049)}{-2} = 59{,}048$$

Counting the terms correctly is the main trap here: from $3^0$ to $3^9$ is ten terms, not nine.

## Repeating decimals as geometric series

Every repeating decimal is secretly a geometric series, which is why every repeating decimal is a fraction.

### Worked example 4: 0.272727...

$$0.\overline{27} = \frac{27}{100} + \frac{27}{10{,}000} + \frac{27}{1{,}000{,}000} + \cdots$$

Here $a = \frac{27}{100}$ and $r = \frac{1}{100}$, so

$$0.\overline{27} = \frac{27/100}{1 - 1/100} = \frac{27/100}{99/100} = \frac{27}{99} = \frac{3}{11}$$

### Worked example 5: 0.999... equals 1

$$0.\overline{9} = \frac{9}{10} + \frac{9}{100} + \cdots = \frac{9/10}{1 - 1/10} = \frac{9/10}{9/10} = 1$$

This is not an approximation. The infinite decimal and 1 are the same number.

### Worked example 6: a decimal that repeats later

Write $1.2\overline{3} = 1.2333\ldots$ as a fraction.

Separate the non-repeating part: $1.2333\ldots = 1.2 + 0.0333\ldots$. The repeating part is geometric with $a = \frac{3}{100}$ and $r = \frac{1}{10}$:

$$0.0\overline{3} = \frac{3/100}{1 - 1/10} = \frac{3/100}{9/10} = \frac{3}{90} = \frac{1}{30}$$

So $1.2\overline{3} = \frac{6}{5} + \frac{1}{30} = \frac{36}{30} + \frac{1}{30} = \frac{37}{30}$.

## A real-world example: the bouncing ball

A ball is dropped from 10 meters, and each bounce reaches $\frac34$ of the previous height. How far does it travel in total?

**Step 1. The first drop** is 10 meters.

**Step 2. The rebounds.** The ball rises to $7.5$ m, then $5.625$ m, and so on, a geometric series with $a = 7.5$ and $r = \frac34$. Their sum is $\frac{7.5}{1 - 3/4} = 30$ meters.

**Step 3. Count each rebound twice**, once up and once down: $2 \cdot 30 = 60$ meters.

**Step 4. Total:** $10 + 60 = 70$ meters, even though the ball bounces infinitely many times.

## From geometric series to power series

Replace the ratio with a variable and the sum formula becomes a statement about functions:

$$\frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots \qquad \text{for } |x| < 1$$

This single identity is the seed of [power series](/blog/power-series). Substitute $-x$ for $x$ and you get $\frac{1}{1+x} = 1 - x + x^2 - \cdots$. Substitute $-x^2$ and you get $\frac{1}{1+x^2} = 1 - x^2 + x^4 - \cdots$, which integrates term by term to the series for $\arctan x$. The condition $|x| < 1$ comes straight from $|r| < 1$, and it is the first example of an interval of convergence.

## Common geometric series mistakes

1. **Using the wrong first term.** When the sum starts at $n = 1$ or higher, $a$ is the first term you actually add, not the coefficient in front.
2. **Applying the formula when $|r| \ge 1$.** The formula only works for convergent series. A negative "sum" of positive terms is a red flag.
3. **Misreading the ratio.** In $\frac{2^n}{3^{n+1}}$, the ratio is $\frac23$, not $\frac{2}{3^{n+1}}$. Rewrite terms as $a\cdot r^n$ first.
4. **Forgetting the sign in $1 - r$.** With $r = -\frac23$, the denominator is $1 + \frac23$.
5. **Confusing the sequence and the series.** The terms $\left(\frac12\right)^n$ approach 0. Their sum approaches 2.

## Quick reference

| Quantity                     | Formula                                |
| ---------------------------- | -------------------------------------- |
| General term                 | $ar^n$ (starting at $n = 0$)           |
| Common ratio                 | $r = \frac{\text{term}_{n+1}}{\text{term}_n}$ |
| $n$th partial sum            | $S_n = \frac{a(1-r^n)}{1-r}$           |
| Infinite sum, $\lvert r\rvert < 1$ | $\frac{a}{1-r}$                  |
| Diverges when                | $\lvert r\rvert \ge 1$                 |

## Practice until the formula is automatic

Geometric series show up again and again: in the ratio test, in comparison arguments, and as the starting point for [power series](/blog/power-series). Getting quick at finding $a$ and $r$ pays off for the rest of the course. The interactive lesson below gives you instant feedback on each step, so a wrong first term or a sign slip gets caught right away.

## Frequently asked questions

### What is the formula for the sum of a geometric series?

If the first term is $a$ and the common ratio is $r$ with $|r| < 1$, the infinite sum is $\frac{a}{1-r}$. For a finite number of terms, the sum of the first $n$ terms is $\frac{a(1-r^n)}{1-r}$.

### How do you know if a geometric series converges?

Find the common ratio $r$ by dividing any term by the one before it. The series converges if $|r| < 1$ and diverges if $|r| \ge 1$. There are no other cases, which makes geometric series the easiest series to classify.

### Why does 0.999... equal 1?

Write $0.999\ldots$ as $\frac{9}{10} + \frac{9}{100} + \frac{9}{1000} + \cdots$, a geometric series with $a = \frac{9}{10}$ and $r = \frac{1}{10}$. Its sum is $\frac{9/10}{1 - 1/10} = \frac{9/10}{9/10} = 1$. The infinite decimal is exactly 1, not just close to it.

### What is the difference between a geometric sequence and a geometric series?

A geometric sequence is a list of numbers like $1, \frac12, \frac14, \frac18, \ldots$, where each term is the previous one times $r$. A geometric series is the sum of those terms, $1 + \frac12 + \frac14 + \cdots$, which here equals 2.

### How do you find the first term when the sum starts at n = 1?

Plug the starting index into the formula for the terms. For $\sum_{n=1}^\infty 3\left(\frac25\right)^n$, the first term is $3\cdot\frac25 = \frac65$, not 3. Using the wrong first term is the most common error with the sum formula.