Math Core

Geometric series are the one family of infinite series whose sums you can always write down exactly. They show up everywhere: repeating decimals, bouncing balls, drug doses, and (at the end of this unit) as the model for every power series.

What makes a series geometric

A series is geometric if each term is a fixed multiple of the previous one. That fixed multiplier is the common ratio $r$ :

\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 example, $6 + 2 + \dfrac{2}{3} + \dfrac{2}{9} + \cdots$ has $a = 6$ and $r = \dfrac{2}{6} = \dfrac{1}{3}$ .

The partial sums

There is a neat trick for the $n$ th partial sum. Let

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

Multiply by $r$ : $\; rS_n = ar + ar^2 + \cdots + ar^{n-1} + ar^n.$ Subtracting, almost everything cancels:

S_n - rS_n = a - ar^n \quad\Longrightarrow\quad S_n = \frac{a(1 - r^n)}{1 - r} \quad (r \ne 1).

Now let $n \to \infty$ . If $|r| < 1$ , then $r^n \to 0$ and $S_n \to \dfrac{a}{1 - r}$ . If $|r| \ge 1$ , the terms $ar^n$ do not go to $0$ (for $a \ne 0$ ), so the series diverges by the nth term test.

Geometric series

The geometric series $\displaystyle\sum_{n=0}^{\infty} a r^n$ (with $a \ne 0$ )

converges when $|r| < 1$ , and its sum is $\dfrac{a}{1 - r} = \dfrac{\text{first term}}{1 - \text{ratio}}$ ;
diverges when $|r| \ge 1$ .

Partial sums of 1 + 1/2 + 1/4 + ⋯ close half the remaining gap to 2 at every step.Open in grapher →

Watch the starting index

The formula $\dfrac{a}{1-r}$ uses the first term actually in the series, whatever the index is. The safest habit is to write out the first term and the ratio instead of trusting the formula's letters. For instance,

\sum_{n=1}^{\infty} 3\left(\frac{2}{5}\right)^n = \frac{6}{5} + \frac{12}{25} + \cdots

has first term $\dfrac{6}{5}$ (not $3$ ) and ratio $\dfrac{2}{5}$ , so its sum is $\dfrac{6/5}{1 - 2/5} = \dfrac{6/5}{3/5} = 2$ .

Common mistake

The most common error is using the wrong first term. Plug in the starting value of $n$ to get the first term. Also check the ratio carefully when the expression has powers in both the numerator and the denominator: $\dfrac{2^{n}}{3^{n+1}} = \dfrac{1}{3}\left(\dfrac{2}{3}\right)^n$ , so $r = \dfrac{2}{3}$ .

Worked examples

Worked example: A negative ratio

Find the sum of $\displaystyle\sum_{n=0}^{\infty} 8\left(-\frac{3}{4}\right)^n$ .

Solution. The first term ( $n = 0$ ) is $8$ , and $r = -\dfrac{3}{4}$ . Since $|r| = \dfrac{3}{4} < 1$ , the series converges:

\frac{8}{1 - \left(-\frac{3}{4}\right)} = \frac{8}{\frac{7}{4}} = \frac{32}{7}.

Worked example: A repeating decimal

Write $0.\overline{45} = 0.454545\ldots$ as a fraction.

Solution. Split it into blocks:

0.\overline{45} = \frac{45}{100} + \frac{45}{100^2} + \frac{45}{100^3} + \cdots.

This is geometric with first term $\dfrac{45}{100}$ and $r = \dfrac{1}{100}$ :

\frac{45/100}{1 - 1/100} = \frac{45/100}{99/100} = \frac{45}{99} = \frac{5}{11}.

Worked example: Rewriting to find the ratio

Determine whether $\displaystyle\sum_{n=1}^{\infty} \frac{5^n}{4^{n+1}}$ converges.

Solution. Rewrite: $\dfrac{5^n}{4^{n+1}} = \dfrac{1}{4}\left(\dfrac{5}{4}\right)^n$ . The ratio is $r = \dfrac{5}{4}$ , and $|r| \ge 1$ , so the series diverges.

Worked example: A bouncing ball

A ball is dropped from a height of 6 feet. Each time it hits the ground, it rebounds to $\dfrac{2}{3}$ of its previous height. Find the total vertical distance the ball travels.

Solution. The ball falls 6 feet. After that, each bounce contributes an up-and-down trip. The rebound heights are $4, \frac{8}{3}, \frac{16}{9}, \dots$ , a geometric sequence with first term $4$ and ratio $\frac{2}{3}$ . So

\text{distance} = 6 + 2\left(4 + \tfrac{8}{3} + \tfrac{16}{9} + \cdots\right) = 6 + 2\cdot\frac{4}{1 - \frac{2}{3}} = 6 + 2(12) = 30 \text{ feet}.

Geometric series with a variable

When the ratio contains $x$ , the series converges only for some values of $x$ . For example, $\displaystyle\sum_{n=0}^{\infty} x^n = \frac{1}{1-x}$ exactly when $|x| < 1$ , that is, $-1 < x < 1$ . This single fact is the seed of power series, which you will study in the last lessons of the unit.

The same idea works with any ratio that involves $x$ . The series $\displaystyle\sum_{n=0}^{\infty} \left(\frac{x}{3}\right)^n$ has ratio $\frac{x}{3}$ , so it converges when $\left|\frac{x}{3}\right| < 1$ , that is, $-3 < x < 3$ , and on that interval its sum is $\dfrac{1}{1 - x/3} = \dfrac{3}{3 - x}$ . Outside the interval, including at the endpoints $x = \pm 3$ , the ratio has absolute value at least $1$ and the series diverges.

Recognizing geometric series in disguise

Geometric series don't always come labeled. Signs that a series is geometric:

The variable $n$ appears only in exponents, such as $\dfrac{2^{n+1}}{5^{n}}$ or $3 \cdot 4^{-n}$ .
Dividing a term by the one before it gives the same number every time.

If $n$ appears anywhere else, for instance $\dfrac{n}{2^n}$ or $\dfrac{1}{2^n + 1}$ , the series is not geometric. It may still converge, but you will need one of the tests from later in this unit (such as the ratio or comparison tests) to decide, and you usually won't get an exact sum.

Tip

Before using $\dfrac{a}{1-r}$ , always state that $|r| < 1$ . On free-response questions, that one sentence is often worth a point.