Math Core

Can you add infinitely many numbers and get a finite answer? Surprisingly, yes, as long as the numbers shrink fast enough. Infinite geometric series explain why $0.999\ldots = 1$ , let you turn any repeating decimal into a fraction, and tell you the total distance a bouncing ball travels.

Partial sums

Walk halfway across a room, then half of the remaining distance, then half of what's left, and so on. If the room is $2$ meters wide, the distances you walk are

1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots

You can't add infinitely many terms directly, so look at the partial sums $S_n$ , the sum of the first $n$ terms:

$n$	$1$	$2$	$3$	$4$	$5$	$6$
$S_n$	$1$	$1.5$	$1.75$	$1.875$	$1.9375$	$1.96875$

Each partial sum closes half the gap to $2$ . The partial sums get as close to $2$ as you like, but never pass it. We say the series converges to $2$ and write $1 + \frac12 + \frac14 + \cdots = 2$ .

Partial sums of 1 + 1/2 + 1/4 + … approach the dashed line y = 2.Open in grapher →

Why the formula works

The finite sum formula is

S_n = \frac{a_1\left(1 - r^n\right)}{1 - r}.

What happens to $r^n$ as $n$ grows? It depends on the size of $r$ :

If $|r| < 1$ , then $r^n$ shrinks toward $0$ . For example, $\left(\tfrac12\right)^{20} \approx 0.00000095$ . So $S_n$ approaches $\dfrac{a_1(1 - 0)}{1 - r}$ .
If $|r| \ge 1$ , then $r^n$ doesn't shrink. The terms stay large (or grow), and the partial sums never settle on one value. The series diverges and has no sum.

Sum of an infinite geometric series

If $|r| < 1$ , the infinite geometric series $a_1 + a_1 r + a_1 r^2 + \cdots$ converges, and

S = \sum_{k=1}^{\infty} a_1 r^{k-1} = \frac{a_1}{1 - r}.

If $|r| \ge 1$ (and $a_1 \ne 0$ ), the series diverges.

Check it on the room example: $a_1 = 1$ , $r = \tfrac12$ , so $S = \dfrac{1}{1 - \frac12} = 2$ . ✓

Common mistake

Always check $|r| < 1$ before using $\dfrac{a_1}{1 - r}$ . The formula will happily give a number for $2 + 4 + 8 + \cdots$ (namely $\dfrac{2}{1 - 2} = -2$ ), but that's nonsense: the terms grow without bound, so the series has no sum.

Worked example: A converging series

Find $12 + 4 + \dfrac{4}{3} + \dfrac{4}{9} + \cdots$ .

The ratio is $r = \dfrac{4}{12} = \dfrac{1}{3}$ , and $\left|\tfrac13\right| < 1$ , so the series converges:

S = \frac{12}{1 - \frac{1}{3}} = \frac{12}{\frac{2}{3}} = 18.

Worked example: Converge or diverge?

Decide whether each series converges. If it does, find its sum.

(a) $3 - 6 + 12 - 24 + \cdots$

Here $r = -2$ and $|-2| = 2 \ge 1$ . The series diverges.

(b) $20 - 5 + \dfrac{5}{4} - \dfrac{5}{16} + \cdots$

Here $r = \dfrac{-5}{20} = -\dfrac{1}{4}$ and $\left|-\tfrac14\right| < 1$ . The series converges:

S = \frac{20}{1 - \left(-\frac{1}{4}\right)} = \frac{20}{\frac{5}{4}} = 16.

The partial sums bounce above and below $16$ ( $20, 15, 16.25, 15.9375, \dots$ ) while closing in on it.

Repeating decimals

Every repeating decimal is an infinite geometric series in disguise, which gives a clean way to write it as a fraction.

Worked example: A repeating decimal as a fraction

Write $0.\overline{72} = 0.727272\ldots$ as a fraction in lowest terms.

Split the decimal into blocks of the repeating part:

0.727272\ldots = 0.72 + 0.0072 + 0.000072 + \cdots

This is geometric with $a_1 = 0.72 = \dfrac{72}{100}$ and $r = \dfrac{1}{100}$ . So

S = \frac{\frac{72}{100}}{1 - \frac{1}{100}} = \frac{\frac{72}{100}}{\frac{99}{100}} = \frac{72}{99} = \frac{8}{11}.

The same method shows $0.\overline{9} = \dfrac{0.9}{1 - 0.1} = \dfrac{0.9}{0.9} = 1$ . So $0.999\ldots$ and $1$ are the same number.

The bouncing ball

Worked example: Total distance traveled

A ball is dropped from a height of $10$ feet. After each bounce it rises to $60\%$ of the height it fell from. Assuming it keeps bouncing forever, what total vertical distance does it travel?

The ball falls $10$ feet. Then it rises and falls the same distance on each bounce: up $6$ and down $6$ , then up $3.6$ and down $3.6$ , and so on.

\text{Total} = 10 + 2(6 + 3.6 + 2.16 + \cdots).

The series in parentheses has $a_1 = 6$ and $r = 0.6$ , so its sum is $\dfrac{6}{1 - 0.6} = \dfrac{6}{0.4} = 15$ . The total distance is

10 + 2(15) = 40 \text{ feet}.

Tip

In bouncing-ball problems, the first drop happens only once, but every height after that is traveled twice (up, then down). Handle the first drop separately, then double the rest.