Geometric Sequences and Infinite Geometric Series

Q: How can an infinite series have a finite sum?

If each term is a fixed fraction of the one before (with |r| < 1), the terms shrink so quickly that the running total approaches a fixed value, which we call the sum.

Q: Is 0.999 recurring really equal to 1?

Yes. As a geometric series with first term 0.9 and ratio 0.1, its sum is exactly 0.9/0.9 = 1.

How geometric sequences grow and shrink, the $n$ th term formula, how an infinite sum can have a finite answer, and why $0.999\ldots = 1$ : your first real taste of a limit.

4 min read Supports 4 lessons

An arithmetic sequence adds the same amount each step. A geometric sequence multiplies by the same amount, the common ratio $r$ .

3,\; 6,\; 12,\; 24,\; 48,\; \dots \qquad r = 2

a_{1}

a_{2}

a_{3}

a_{4}

a_{5}

a_{6}

Each bar is double the last.

Find $r$ by dividing any term by the one before it. If $r$ is between $-1$ and $1$ , the terms shrink towards zero; a negative $r$ makes the signs alternate.

The nth term

nth term

a_n = a_1 \cdot r^{\,n-1}

Reaching the $n$ th term takes $n - 1$ multiplications by $r$ .

Key idea

A geometric sequence is exponential growth (or decay) observed at whole-number steps. $a_n = 3 \cdot 2^{n-1}$ is the exponential function $3 \cdot 2^{x-1}$ sampled at $x = 1, 2, 3, \dots$

Worked example

Find the 8th term of $5, 15, 45, \dots$

1
$a_1 = 5$ and $r = 3$ .
$a_8 = 5 \cdot 3^{7}$

Adding infinitely many terms

Walk halfway to a wall. Then half of the remaining distance. Then half of what remains, and so on for ever. You take infinitely many steps, yet you never travel further than the wall.

\frac12 + \frac14 + \frac18 + \frac{1}{16} + \cdots = 1

0.5

a_{1}

0.75

a_{2}

0.875

a_{3}

0.938

a_{4}

0.969

a_{5}

0.984

a_{6}

0.992

a_{7}

The running totals creep up towards $1$ but never pass it.

When the common ratio is between $-1$ and $1$ , the terms shrink fast enough that the running total settles on a finite value.

Infinite sum

S_\infty = \frac{a_1}{1 - r}, \qquad |r| < 1

$1 + \frac12 + \frac14 + \cdots = \frac{1}{1 - \frac12} = 2$ .

Why the formula works

$S = a + ar + ar^2 + ar^3 + \cdots$
Call the infinite sum $S$ .

1/4

A surprising consequence

Worked example

Show that $0.999\ldots = 1$ .

1
Write it as a geometric series.
$0.9 + 0.09 + 0.009 + \cdots$

Watch out

The formula only works when $|r| < 1$ . For $1 + 2 + 4 + 8 + \cdots$ the total grows without limit, and plugging into the formula gives nonsense.

Key idea

An infinite sum with a finite answer is a limit: the value the running totals approach. Calculus is built on exactly this idea.

Tip

For a finite number of terms, $S_n = \frac{a_1(1 - r^n)}{1 - r}$ . As $n$ grows and $|r| < 1$ , $r^n$ shrinks to $0$ , giving the infinite formula.

Common questions

How can an infinite series have a finite sum?: If each term is a fixed fraction of the one before (with $|r| < 1$ ), the terms shrink so quickly that the running total approaches a fixed value, which we call the sum.
Is 0.999 recurring really equal to 1?: Yes. As a geometric series with first term $0.9$ and ratio $0.1$ , its sum is exactly $\frac{0.9}{0.9} = 1$ .

Practise geometric sequences and series

The Geometric Sequences lesson has worked examples and unlimited practice that adapts to you.

Open the lesson

Common questions

How can an infinite series have a finite sum?

If each term is a fixed fraction of the one before (with $|r| < 1$ ), the terms shrink so quickly that the running total approaches a fixed value, which we call the sum.

Is 0.999 recurring really equal to 1?

Yes. As a geometric series with first term $0.9$ and ratio $0.1$ , its sum is exactly $\frac{0.9}{0.9} = 1$ .