6.4: Infinite geometric series

Elementary mathematics is predominantly about equations and identities. But it is often impossible to capture important mathematical relations in the form of exact equations. This is one reason why inequalities become more central as we progress; another reason is because inequalities allow us to make precise statements about certain infinite processes.

One of the simplest infinite process arises in the formula for the “sum” of an infinite geometric series:

$1+\mathit{r}+{\mathit{r}}^2+{\mathit{r}}^3+\cdots +{\mathit{r}}^{\mathit{n}}+\cdots$ (for ever).

Despite the use of the familiar-looking “+” signs, this can be no ordinary addition. Ordinary addition is defined for two summands; and by repeating the process, we can add three summands (thanks in part to the associative law of addition). We can then add four, or any finite number of summands. But this does not allow us to “add” infinitely many terms as in the above sum. To get round this we combine ordinary addition (of finitely many terms) and simple inequalities to find a new way of giving a meaning to the above “endless sum”. In Problem 116 you used the factorisation

${\mathit{r}}^{\mathit{n}+1}-1=(\mathit{r}-1)\left(1+\mathit{r}+{\mathit{r}}^2+{\mathit{r}}^3+\cdots +{\mathit{r}}^{\mathit{n}}\right)$

to derive the closed formula:

$1+\mathit{r}+{\mathit{r}}^2+{\mathit{r}}^3+\cdots +{\mathit{r}}^{\mathit{n}}=\frac{1-{\mathit{r}}^{\mathit{n}+1}}{1-\mathit{r}}$.

This formula for the sum of a finite geometric series can be rewritten in the form

$$1+\mathit{r}+{\mathit{r}}^2+{\mathit{r}}^3+\cdots +{\mathit{r}}^{\mathit{n}}=\frac{1}{1-\mathit{r}}-\frac{{\mathit{r}}^{\mathit{n}+1}}{1-\mathit{r}}$$

At first sight, this may not look like a clever move! However, it separates the part that is independent of n from the part on the RHS that depends on n; and it allows us to see how the second part behaves as n gets large:

when $|\mathit{r}|<1$, successive powers of r get smaller and smaller and converge rapidly towards 0,

so the above form of the identity may be interpreted as having the form:

$1+\mathit{r}+{\mathit{r}}^2+{\mathit{r}}^3+\cdots +{\mathit{r}}^{\mathit{n}}=\frac{1}{1-\mathit{r}}$ — (an “error term”).

Moreover if $|\mathit{r}|<1$, then the “error term” converges towards 0 as $\mathit{n}\to \infty$. In particular, if $1>\mathit{r}>0$, the error term is always positive, so we have proved, for all $$n⩾1:

$1+\mathit{r}+{\mathit{r}}^2+{\mathit{r}}^3+\cdots +{\mathit{r}}^{\mathit{n}}<\frac{1}{1-\mathit{r}}$

and

the difference between the two sides tends rapidly to 0 as $\mathit{n}\to \infty$.

We then make the natural (but bold) step to interpret this, when $|\mathit{r}|<1$, as offering a new definition which explains precisely what is meant by the endless sum

$1+\mathit{r}+{\mathit{r}}^2+{\mathit{r}}^3+\cdots$ (for ever),

declaring that, when $|\mathit{r}|<1$,

$1+\mathit{r}+{\mathit{r}}^2+{\mathit{r}}^3+\cdots$ (for ever) = $\frac{1}{1-\mathit{r}}$.

More generally, if we multiply every term by a, we see that

$\mathit{a}+\mathit{ar}+{}^{}$ar2+ar3+⋯ (for ever) = $\frac{\mathit{a}}{1-\mathit{r}}$.

Problem 243 Interpret the recurring decimal 0.037037037 ··· (for ever) as an infinite geometric series, and hence find its value as a fraction.

Problem 244 Interpret the following endless processes as infinite geometric series.

(a) A square cake is cut into four quarters, with two perpendicular cuts through the centre, parallel to the sides. Three people receive one quarter each - leaving a smaller square piece of cake. This smaller piece is then cut in the same way into four quarters, and each person receives one (even smaller) piece - leaving an even smaller residual square piece, which is then cut in the same way. And so on for ever. What fraction of the original cake does each person receive as a result of this endless process?

(b) I give you a whole cake. Half a minute later, you give me half the cake back. One quarter of a minute later, I return one quarter of the cake to you. One eighth of a minute later you return one eighth of the cake to me. And so on. Adding the successive time intervals, we see that

$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots$ (for ever) = 1,

so the whole process is completed in exactly 1 minute. How much of the cake do I have at the end, and how much do you have?

Problem 245 When John von Neumann (1903-1957) was seriously ill in hospital, a visitor tried (rather insensitively) to distract him with the following elementary mathematics problem.

Have you heard the one about the two trains and the fly? Two trains are on a collision course on the same track, each travelling at 30 km/h. A super-fly starts on Train A when the trains are 120 km apart, and flies at a constant speed of 50 km/h - from Train A to Train B, then back to Train A, and so on. Eventually the two trains collide and the fly is squashed. How far did the fly travel before this sad outcome?