Infinite Geometric Series: When an Endless Sum Has a Total

Q: What is the formula for an infinite geometric series?

S = a/(1 − r), where a is the first term and r is the common ratio. It works only when −1 < r < 1.

Q: When does an infinite geometric series have a sum?

When the ratio is strictly between −1 and 1. Then the terms shrink toward zero and the partial sums close in on a single number.

Q: How can infinitely many numbers add up to a finite total?

Each new term covers a fixed fraction of the gap that remains, so the gap shrinks toward zero. The partial sums approach the total as closely as you like.

Q: Does 0.999… really equal 1?

Yes. 0.999… = 0.9 + 0.09 + 0.009 + ⋯, a geometric series with a = 0.9 and r = 0.1, and its sum is 0.9/(1 − 0.1) = 1.

Q: What happens when r = 1 or r = −1?

There is no sum. With r = 1 the partial sums grow without limit; with r = −1 they bounce between two values forever.

Quick answer

An infinite geometric series a + ar + ar² + ⋯ never stops, yet it can still have a sum. When −1 < r < 1 the terms shrink toward zero, the partial sums settle down, and the series converges to S = a/(1 − r). When r ≤ −1 or r ≥ 1 the partial sums never settle, and the series has no sum. The formula turns repeating decimals into fractions and totals the distance a bouncing ball travels.

What you'll learn

Decide whether an infinite geometric series has a sum
Find the sum of a convergent geometric series
Write a repeating decimal as a fraction
Model a bouncing ball or a repeated process with an infinite series

Partial sums that settle

Keep adding halves: $1 + \tfrac{1}{2} + \tfrac{1}{4} + \tfrac{1}{8} + \cdots$ . The running totals, called partial sums, are

1, \quad 1.5, \quad 1.75, \quad 1.875, \quad 1.9375, \quad \ldots

They climb toward $2$ , and each one closes half of the gap that is left.

S = 2

Partial sums of 1 + 1/2 + 1/4 + ⋯

The geometric series formula shows where they are headed. The sum of the first $n$ terms is

S_n = \frac{a\left(1 - r^n\right)}{1 - r}

When $-1 < r < 1$ , the power $r^n$ shrinks toward $0$ as $n$ grows, so $S_n$ closes in on a single number. That number is the sum of the infinite geometric series:

a + ar + ar^2 + \cdots = \frac{a}{1 - r} \qquad -1 < r < 1

For the halves, $a = 1$ and $r = \tfrac{1}{2}$ , and the sum is $\tfrac{1}{1 - 1/2} = 2$ .

Why the sum is a/(1 − r)

Split the partial sum into the limit and a leftover:

S_n = \frac{a}{1 - r} - \frac{a r^n}{1 - r}

The first part never changes. The leftover is a fixed number times $r^n$ . When $-1 < r < 1$ each extra term multiplies $r^n$ by a number smaller than $1$ in size, so the leftover shrinks toward $0$ . The partial sums differ from $\tfrac{a}{1 - r}$ by a multiple of $r^n$ , so they close in on it exactly when $r^n$ shrinks to zero: when $-1 < r < 1$ . For $r \ge 1$ or $r \le -1$ , $r^n$ does not shrink, and there is no sum.

The halves fit together in a picture: half of a square, then half of what is left, and so on, fill the whole square.

Halves of a square

Worked examples

Common mistakes

Practice problems

Find $6 + 3 + 1.5 + \cdots$ .

Answer

$12$

Full solution

$a = 6$ and $r = \tfrac{1}{2}$ , so $S = \tfrac{6}{1/2} = 12$ .
Find $1 - \tfrac{1}{3} + \tfrac{1}{9} - \cdots$ .

Answer

$\tfrac{3}{4}$

Full solution

$a = 1$ and $r = -\tfrac{1}{3}$ , so $S = \tfrac{1}{4/3} = \tfrac{3}{4}$ .
Does $3 + 4.5 + 6.75 + \cdots$ have a sum?

Answer

No

Full solution

The ratio is $\tfrac{4.5}{3} = 1.5$ , so the terms grow and the partial sums increase without limit.
Write $0.999\ldots$ as a fraction.

Answer

$1$

Full solution

$0.9 + 0.09 + 0.009 + \cdots$ has $a = 0.9$ and $r = 0.1$ , so the sum is $\tfrac{0.9}{0.9} = 1$ . The partial sums $0.9, 0.99, 0.999, \ldots$ come as close to $1$ as you like, and $1$ is the only number they approach.
Write $0.121212\ldots$ as a fraction.

Answer

$\tfrac{4}{33}$

Full solution

$a = \tfrac{12}{100}$ and $r = \tfrac{1}{100}$ , so the sum is $\tfrac{12}{99} = \tfrac{4}{33}$ .
Find $\displaystyle\sum_{n=1}^{\infty} 10(0.8)^{n-1}$ .

Answer

$50$

Full solution

$a = 10$ and $r = 0.8$ , so $S = \tfrac{10}{0.2} = 50$ .
Find $\displaystyle\sum_{n=1}^{\infty} 3\left(\tfrac{1}{4}\right)^{n}$ .

Answer

$1$

Full solution

The first term, at $n = 1$ , is $\tfrac{3}{4}$ , and $r = \tfrac{1}{4}$ . So $S = \tfrac{3/4}{3/4} = 1$ .
An infinite geometric series has first term $9$ and sum $12$ . Find its ratio.

Answer

$r = \tfrac{1}{4}$

Full solution

$\tfrac{9}{1 - r} = 12$ gives $1 - r = \tfrac{3}{4}$ .
A ball is dropped from $12$ m, and each bounce rises to $\tfrac{3}{4}$ of the previous height. What total distance does it travel?

Answer

$84$ m

Full solution

The first fall is $12$ . The rebound heights $9 + 6.75 + \cdots$ sum to $\tfrac{9}{1/4} = 36$ , and each is traveled up and down: $12 + 2(36) = 84$ .
A student writes $1 + 3 + 9 + \cdots = \tfrac{1}{1 - 3} = -\tfrac{1}{2}$ . What went wrong?

Hint

What is the ratio, and what does the formula require?

Answer

The ratio is $3$ , so the series has no sum. The formula needs $-1 < r < 1$ .

Full solution

The partial sums $1, 4, 13, 40, \ldots$ grow without limit. The formula comes from the leftover term $r^n$ shrinking to zero, and $3^n$ grows instead.

Frequently asked questions

What is the formula for an infinite geometric series?

S = a/(1 − r), where a is the first term and r is the common ratio. It works only when −1 < r < 1.

When does an infinite geometric series have a sum?