# Geometric Series Formula: Sum a/(1 − r), Convergence and Examples

> The geometric series formula: a + ar + ar² + ... = a/(1 − r) when |r| < 1. Partial sums, the convergence condition and worked examples.

Source: https://practicecalculus.com/rules/geometric-series-formula
Updated: 2026-11-03

$$ \sum_{n=0}^{\infty} a\,r^n = \frac{a}{1 - r}, \quad \lvert r\rvert < 1 $$

## At a glance

- **First term:** $a$, the term you get when the exponent is $0$
- **Common ratio:** $r$, the factor that turns each term into the next
- **Converges when:** $\lvert r\rvert < 1$
- **Diverges when:** $\lvert r\rvert \ge 1$ (and $a \ne 0$)
- **Partial sum:** $S_N = a\,\dfrac{1 - r^N}{1 - r}$ for the first $N$ terms, $r \ne 1$

## Key takeaways

- An infinite geometric series converges to $\frac{a}{1-r}$ exactly when $\lvert r\rvert < 1$.
- $a$ is the first term of the series as written, not necessarily the coefficient in front of $r^n$.
- The finite sum of the first $N$ terms is $a\frac{1 - r^N}{1 - r}$.

A **geometric series** adds terms that each come from the one before by multiplying by the same number $r$. It is one of the few infinite series whose exact sum you can write down, which makes it the benchmark that every [convergence test](/blog/convergence-tests) is compared against.

## The geometric series formula

With first term $a$ and common ratio $r$:

$$a + ar + ar^2 + ar^3 + \cdots = \sum_{n=0}^{\infty} a\,r^n = \frac{a}{1-r}, \qquad \lvert r\rvert < 1$$

If $\lvert r\rvert \ge 1$ and $a \ne 0$, the series **diverges**.

The sum of the first $N$ terms is finite for every $r \ne 1$:

$$S_N = a + ar + \cdots + ar^{N-1} = a\,\frac{1 - r^N}{1 - r}$$

## Why the formula works

Multiply $S_N$ by $r$ and subtract. Almost every term cancels:

$$S_N - rS_N = a - ar^N \quad\Longrightarrow\quad S_N = a\,\frac{1 - r^N}{1 - r}$$

When $\lvert r\rvert < 1$, $r^N \to 0$ as $N \to \infty$, so $S_N \to \frac{a}{1-r}$. When $\lvert r\rvert \ge 1$, $r^N$ does not go to zero and the partial sums have no finite limit.

```graph
title: 'Partial sums of 1 + 1/2 + 1/4 + ... approaching 2'
caption: 'Each partial sum $S_N = 2 - 2^{1-N}$ closes half of the remaining gap to the limit $\frac{1}{1 - 1/2} = 2$.'
x: [0, 7]
y: [0, 2.5]
functions:
  - expr: '2'
    label: y = 2 (the sum)
    dashed: true
points:
  - [1, 1, 'S1 = 1']
  - [2, 1.5, 'S2 = 1.5']
  - [3, 1.75, 'S3 = 1.75']
  - [4, 1.875]
  - [5, 1.9375]
  - [6, 1.96875]
```

## How to use it

1. **Find $a$**: the first term of the series as written. Plug in the starting index.
2. **Find $r$**: divide any term by the previous one.
3. **Check $\lvert r\rvert < 1$**. If not, stop: the series diverges.
4. **Compute $\frac{a}{1-r}$.**

## Worked examples

**Example 1: a classic.** $1 + \frac{1}{2} + \frac{1}{4} + \cdots$ has $a = 1$, $r = \frac{1}{2}$:

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

**Example 2: the index does not start at 0.** In $\sum_{n=1}^{\infty} 3\left(\frac{2}{3}\right)^n$ the first term is $3\cdot\frac{2}{3} = 2$, not $3$:

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

**Example 3: a negative ratio.** $\sum_{n=0}^{\infty} \left(-\frac{1}{3}\right)^n$ has $a = 1$, $r = -\frac{1}{3}$:

$$\frac{1}{1 - \left(-\frac{1}{3}\right)} = \frac{3}{4}$$

**Example 4: a repeating decimal.** $0.272727\ldots = \frac{27}{100} + \frac{27}{10000} + \cdots$ has $a = \frac{27}{100}$ and $r = \frac{1}{100}$:

$$\frac{27/100}{99/100} = \frac{27}{99} = \frac{3}{11}$$

**Example 5: divergence.** $\sum_{n=0}^{\infty} \left(\frac{5}{4}\right)^n$ has $r = \frac{5}{4} > 1$, so it diverges. The terms grow instead of shrinking.

## Finite vs infinite sums

The finite formula works for every ratio except $r = 1$, so even a series with $r = 3$ has a perfectly good sum of the first ten terms: $\sum_{n=0}^{9} 3^n = \frac{1 - 3^{10}}{1 - 3} = 29524$. Only the infinite sum needs $\lvert r\rvert < 1$. When $r = 1$ every term equals $a$, so the first $N$ terms add to $Na$, and the infinite series diverges.

## Common mistakes

1. **Using the coefficient instead of the first term.** When the sum starts at $n = 1$ or later, compute the first term directly.
2. **Ignoring the condition.** $\frac{a}{1-r}$ with $r = 2$ gives $-a$, a meaningless number. Always check $\lvert r\rvert < 1$ first.
3. **Mistaking other series for geometric.** $\sum \frac{1}{n^2}$ and $\sum \frac{1}{2n}$ have no constant ratio, so the formula does not apply.
4. **Dropping a sign in $r$.** For an alternating geometric series, $r$ is negative, and $1 - r$ becomes $1 + \lvert r\rvert$.

## Where it leads

Geometric series are the model for the [ratio test](/rules/ratio-test), which asks whether a series eventually behaves like a geometric one with $\lvert r\rvert < 1$. They also give the first power series, $\frac{1}{1-x} = \sum x^n$ for $\lvert x\rvert < 1$, the starting point for [power series](/blog/power-series). For more examples, read the full guide to [geometric series](/blog/geometric-series).

## Frequently asked questions

### What is the formula for the sum of an infinite geometric series?

If the first term is $a$ and the common ratio $r$ satisfies $\lvert r\rvert < 1$, the sum is $\frac{a}{1-r}$. If $\lvert r\rvert \ge 1$ the series diverges.

### How do you find r in a geometric series?

Divide any term by the term before it. For $6 + 2 + \frac{2}{3} + \cdots$, the ratio is $\frac{2}{6} = \frac{1}{3}$.

### Why does a geometric series diverge when |r| is at least 1?

If $\lvert r\rvert \ge 1$ the terms $ar^n$ do not shrink to zero, so by the nth-term test the series cannot converge.

### How do you write 0.999... as a geometric series?

$0.999\ldots = \frac{9}{10} + \frac{9}{100} + \cdots$, a geometric series with $a = \frac{9}{10}$ and $r = \frac{1}{10}$. Its sum is $\frac{9/10}{9/10} = 1$.