The sum of a geometric progression

The terms of a G.P. can grow aston­ish­ingly fast, and their sums even faster. There is a story of a king who offered a reward of one grain of rice on the first square of a chess­board, two on the sec­ond, four on the third, and so on: the total for 64 squares is more rice than the world has ever grown. In this les­son we find a for­mula for the sum of the first $n$ terms of a G.P., and then a sur­pris­ing fact: when the terms shrink, the sum of infi­nitely many of them can be a finite num­ber.

The sum of n terms of a G.P.#

Let $S_n = a + ar + ar^2 + \cdots + ar^{n-1}$. Mul­ti­ply every term by $r$:

$$rS_n = ar + ar^2 + ar^3 + \cdots + ar^n$$

Now sub­tract the first line from the sec­ond. All the mid­dle terms can­cel, leav­ing only the last term of $rS_n$ and the first term of $S_n$:

$$rS_n - S_n = ar^n - a \implies S_n(r - 1) = a(r^n - 1)$$

So, pro­vided $r \neq 1$,

$$\displaystyle S_n = \frac{a(r^n - 1)}{r - 1} = \frac{a(1 - r^n)}{1 - r}$$

The two forms are the same (mul­ti­ply top and bot­tom by $-1$). Use the first when $r > 1$ and the sec­ond when $r < 1$, so that the num­bers stay pos­i­tive.

If $r = 1$, every term equals $a$ and the for­mula can­not be used; sim­ply $S_n = na$.

Exam­ple 1#

Find the sum of the first 6 terms of $2 + 6 + 18 + \cdots$

$a = 2$, $r = 3$, $n = 6$:

$$\displaystyle S_6 = \frac{2(3^6 - 1)}{3 - 1} = \frac{2 \times 728}{2} = 728$$

Exam­ple 2#

Find the sum of the first 8 terms of $\displaystyle 1 + \frac{1}{2} + \frac{1}{4} + \cdots$

$a = 1$, $\displaystyle r = \frac{1}{2}$:

$$\displaystyle S_8 = \frac{1 - \left(\frac{1}{2}\right)^8}{1 - \frac{1}{2}} = \frac{1 - \frac{1}{256}}{\frac{1}{2}} = 2 \times \frac{255}{256} = \frac{255}{128}$$

Exam­ple 3#

How many terms of $3 + 6 + 12 + \cdots$ must be added to give 765?

$$\displaystyle \frac{3(2^n - 1)}{2 - 1} = 765 \implies 2^n - 1 = 255 \implies 2^n = 256 = 2^8 \implies n = 8$$

Exam­ple 4#

The rice on the chess­board: find $1 + 2 + 4 + \cdots$ to 64 terms.

$a = 1$, $r = 2$, $n = 64$: $S_{64} = 2^{64} - 1$, a num­ber with 20 dig­its, more than 18 mil­lion mil­lion mil­lion grains.

The sum to infin­ity#

Now look at $\displaystyle 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots$. The sums of the first few terms are

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

They creep closer and closer to 2 with­out ever pass­ing it. Each new term cov­ers half the remain­ing gap. We say that the sum to infin­ity is 2.

In gen­eral, look at $\displaystyle S_n = \frac{a(1 - r^n)}{1 - r}$. If $-1 < r < 1$, the pow­ers $r^n$ get closer and closer to 0 as $n$ grows: for $\displaystyle r = \frac{1}{2}$, $r^{10}$ is already less than $0.001$. So $S_n$ gets closer and closer to $\displaystyle \frac{a}{1 - r}$. This value is the sum to infin­ity:

$$\displaystyle S_\infty = \frac{a}{1 - r}, \qquad \text{valid only when } -1 < r < 1$$

If $r \ge 1$ or $r \le -1$, the terms do not shrink and the series has no finite sum. For exam­ple $1 + 2 + 4 + \cdots$ just keeps grow­ing.

Exam­ple 5#

Find the sum to infin­ity of $\displaystyle 9 + 3 + 1 + \frac{1}{3} + \cdots$

$a = 9$, $\displaystyle r = \frac{1}{3}$, which lies between $-1$ and 1:

$$\displaystyle S_\infty = \frac{9}{1 - \frac{1}{3}} = \frac{9}{\frac{2}{3}} = \frac{27}{2}$$

Exam­ple 6#

Find the sum to infin­ity of $8 - 4 + 2 - 1 + \cdots$

$a = 8$, $\displaystyle r = -\frac{1}{2}$: $\displaystyle S_\infty = \frac{8}{1 + \frac{1}{2}} = \frac{8}{\frac{3}{2}} = \frac{16}{3}$.

Recur­ring dec­i­mals as frac­tions#

A recur­ring dec­i­mal is an infi­nite G.P. in dis­guise.

Exam­ple 7#

Write $0.\overline{7} = 0.777\ldots$ as a frac­tion.

$$\displaystyle 0.777\ldots = \frac{7}{10} + \frac{7}{100} + \frac{7}{1000} + \cdots$$

This is a G.P. with $\displaystyle a = \frac{7}{10}$ and $\displaystyle r = \frac{1}{10}$. So the value is $\displaystyle \frac{7/10}{1 - 1/10} = \frac{7/10}{9/10} = \frac{7}{9}$.

Exam­ple 8#

Write $0.\overline{45} = 0.4545\ldots$ as a frac­tion.

$$\displaystyle 0.4545\ldots = \frac{45}{100} + \frac{45}{10000} + \cdots = \frac{45/100}{1 - 1/100} = \frac{45}{99} = \frac{5}{11}$$

A bounc­ing ball#

Exam­ple 9#

A ball is dropped from a height of 12 m. Each time it hits the ground it rises to $\displaystyle \frac{2}{3}$ of the height it fell from. Find the total dis­tance it trav­els before com­ing to rest.

The ball first falls 12 m. Then it rises 8 m and falls 8 m, rises $\displaystyle \frac{16}{3}$ m and falls $\displaystyle \frac{16}{3}$ m, and so on. The rises form a G.P. $\displaystyle 8 + \frac{16}{3} + \cdots$ with $a = 8$ and $\displaystyle r = \frac{2}{3}$, whose sum is $\displaystyle \frac{8}{1 - 2/3} = 24$. The falls after the first are the same. So the total dis­tance is

$$12 + 2 \times 24 = 60 \text{ m}$$

Prac­tice#

1. Find the sum of the first 7 terms of $1 + 3 + 9 + \cdots$
2. Find the sum of the first 6 terms of $5 + 10 + 20 + \cdots$
3. Find the sum of the first 5 terms of $16 + 8 + 4 + \cdots$
4. Find the sum of the first 6 terms of $1 - 2 + 4 - \cdots$
5. How many terms of $2 + 4 + 8 + \cdots$ must be added to give 510?
6. Find the sum to infin­ity of $\displaystyle 12 + 4 + \frac{4}{3} + \cdots$
7. Find the sum to infin­ity of $\displaystyle 5 - \frac{5}{2} + \frac{5}{4} - \cdots$
8. Write $0.\overline{3}$ and $0.\overline{36}$ as frac­tions.
9. The sum to infin­ity of a G.P. is 20 and its first term is 5. Find the com­mon ratio.
10. A ball is dropped from 20 m and each time rises to $\displaystyle \frac{3}{5}$ of the height it fell. Find the total dis­tance trav­elled.

Answers#