Sum to infinity of a geometric progression

AS · P1 · 11 min

Add $1 + \tfrac{1}{2} + \tfrac{1}{4} + \tfrac{1}{8} + \cdots$ for ever and the total never passes $2$ : each new term covers half the remaining gap. A geometric progression whose terms shrink fast enough like this has a finite sum to infinity. Paper 1 asks you to state when a GP converges, to use $S_\infty = \dfrac{a}{1 - r}$ , and to combine it with the $n$ th term and sum formulas to find unknowns, often in a context such as a bouncing ball or a recurring decimal.

Partial sums that settle down

For the GP $8 + 4 + 2 + 1 + \cdots$ (first term $a = 8$ , common ratio $r = \tfrac{1}{2}$ ), the sums of the first $n$ terms are

S_1 = 8, \quad S_2 = 12, \quad S_3 = 14, \quad S_4 = 15, \quad S_5 = 15.5, \quad S_6 = 15.75, \quad \ldots

The gap to $16$ halves each time: $8, 4, 2, 1, 0.5, 0.25, \ldots$ So the sums get as close to $16$ as you like, without ever reaching it. We say the series converges and its sum to infinity is $16$ .

y = 16(1 - 0.5^x) y = 16 (1, 8) (2, 12) (3, 14) (4, 15) (5, 15.5)

The plotted points are the partial sums $S_n$ ; the curve through them approaches the line $y = 16$ but never reaches it.

When does a GP converge?

The sum of the first $n$ terms is

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

The first part does not depend on $n$ . Everything depends on what $r^n$ does as $n$ grows:

Ratio	What $r^n$ does	The series
$-1 < r < 1$	gets closer and closer to $0$	converges to $\dfrac{a}{1 - r}$
$r > 1$	grows without limit	diverges (the sum grows without limit)
$r = 1$	stays $1$	diverges: $S_n = na$
$r = -1$	alternates $1, -1, \ldots$	diverges: $S_n$ alternates between $a$ and $0$
$r < -1$	alternates in sign and grows in size	diverges

Key result

A geometric progression converges if and only if

-1 < r < 1 \qquad (\text{that is, } |r| < 1)

and then its sum to infinity is

S_\infty = \frac{a}{1 - r}

This formula is in the list of formulae (MF19), but the condition $|r| < 1$ is not: you must state it.

A convergent GP with a negative ratio has terms that alternate in sign, and its partial sums swing above and below $S_\infty$ , closing in from both sides.

The remaining sum

The difference between the sum to infinity and the sum of the first $n$ terms is the sum of all the terms after the $n$ th:

S_\infty - S_n = \frac{a r^n}{1 - r}

This is itself the sum to infinity of a GP, with first term $ar^n$ (the $(n + 1)$ th term) and the same ratio. It answers questions like "how many terms are needed for the sum to be within $0.001$ of the sum to infinity?"

Sum to infinity problems

Identify $a$ and $r$ (find $r$ by dividing a term by the one before).
Check, and state, that $|r| < 1$ , so the sum to infinity exists.
Use $S_\infty = \dfrac{a}{1 - r}$ , together with $u_n = ar^{n-1}$ or $S_n = \dfrac{a(1 - r^n)}{1 - r}$ for any other facts given.
With two unknowns, write two equations and eliminate $a$ (often by substituting $a = S_\infty(1 - r)$ ).
Check every value of $r$ you find satisfies $|r| < 1$ ; reject any that do not.

Ratios that contain a variable

If the ratio is an expression such as $2x - 1$ , the condition $-1 < r < 1$ becomes an inequality to solve for $x$ . This gives the set of values of $x$ for which the series converges. For any $x$ in that set, $S_\infty$ is a function of $x$ .

Recurring decimals

A recurring decimal is a convergent GP in disguise:

0.\dot{4}\dot{5} = 0.454545\ldots = 0.45 + 0.0045 + 0.000045 + \cdots

Here $a = 0.45$ and $r = 0.01$ , so the value is $\dfrac{0.45}{1 - 0.01} = \dfrac{0.45}{0.99} = \dfrac{45}{99} = \dfrac{5}{11}$ . This explains the familiar rule that a two-digit repeating block over $99$ gives the fraction.

Worked examples

A negative ratio

Find the sum to infinity of the geometric progression $24, -12, 6, -3, \ldots$

Solution

$r = \dfrac{-12}{24} = -\dfrac{1}{2}$ . Since $|r| < 1$ , the sum to infinity exists:

S_\infty = \frac{24}{1 - \left(-\tfrac{1}{2}\right)} = \frac{24}{\tfrac{3}{2}} = 16

Convergence with a variable ratio

The series $1 + (1 - 2x) + (1 - 2x)^2 + \cdots$ is geometric.

(a) Find the set of values of $x$ for which the series converges.

(b) Find an expression for the sum to infinity in terms of $x$ , and find $x$ when the sum to infinity is $5$ .

Solution

(a) $r = 1 - 2x$ . The series converges when

-1 < 1 - 2x < 1 \quad\Rightarrow\quad -2 < -2x < 0 \quad\Rightarrow\quad 0 < x < 1

(Dividing by $-2$ reverses both inequality signs.)

(b)

S_\infty = \frac{1}{1 - (1 - 2x)} = \frac{1}{2x}

$\dfrac{1}{2x} = 5$ gives $x = \dfrac{1}{10}$ , which is in the interval $0 < x < 1$ , so it is valid.

A recurring decimal

Express $0.2\dot{3} = 0.2333\ldots$ as a fraction in its lowest terms.

Solution

Separate the non-repeating part:

0.2333\ldots = 0.2 + (0.03 + 0.003 + 0.0003 + \cdots)

The bracket is a GP with $a = 0.03$ and $r = 0.1$ :

0.2 + \frac{0.03}{1 - 0.1} = \frac{1}{5} + \frac{0.03}{0.9} = \frac{1}{5} + \frac{1}{30} = \frac{7}{30}

Sum to infinity and the first three terms

The sum to infinity of a geometric progression is three times its first term, and the sum of the first three terms is $38$ . Find the first term and the common ratio.

Solution

\frac{a}{1 - r} = 3a \quad\Rightarrow\quad 1 = 3(1 - r) \quad\Rightarrow\quad r = \frac{2}{3}

(Dividing by $a$ is fine, since $a \neq 0$ .) Then

a\left(1 + \tfrac{2}{3} + \tfrac{4}{9}\right) = 38 \quad\Rightarrow\quad \frac{19a}{9} = 38 \quad\Rightarrow\quad a = 18

Two possible progressions

The second term of a geometric progression is $12$ and its sum to infinity is $64$ . Find the two possible values of the first term and the corresponding common ratios.

Solution

$ar = 12$ and $\dfrac{a}{1 - r} = 64$ , so $a = 64(1 - r)$ . Substituting:

\begin{aligned} 64r(1 - r) &= 12 \\ 64r^2 - 64r + 12 &= 0 \\ 16r^2 - 16r + 3 &= 0 \\ (4r - 1)(4r - 3) &= 0 \end{aligned}

$r = \tfrac{1}{4}$ gives $a = 64 \times \tfrac{3}{4} = 48$ . $r = \tfrac{3}{4}$ gives $a = 64 \times \tfrac{1}{4} = 16$ .

Both ratios satisfy $|r| < 1$ , so both progressions are valid: $48, 12, 3, \ldots$ and $16, 12, 9, \ldots$

How many terms are needed

A GP has first term $8$ and common ratio $\tfrac{1}{2}$ . Find the least value of $n$ for which the sum of the first $n$ terms is within $0.001$ of the sum to infinity.

Solution

$S_\infty = \dfrac{8}{1 - \frac{1}{2}} = 16$ , and

S_\infty - S_n = \frac{8\left(\tfrac{1}{2}\right)^n}{\tfrac{1}{2}} = 16\left(\tfrac{1}{2}\right)^n

We need $16\left(\tfrac{1}{2}\right)^n < 0.001$ , i.e. $\left(\tfrac{1}{2}\right)^n < 0.0000625$ . By trial:

16\left(\tfrac{1}{2}\right)^{13} = 0.00195 > 0.001, \qquad 16\left(\tfrac{1}{2}\right)^{14} = 0.000977 < 0.001

So $n = 14$ .

A bouncing ball

A ball is dropped from a height of $10$ m. Each time it hits the ground it rebounds to $\tfrac{3}{4}$ of the height from which it fell. Find the total distance the ball travels before it comes to rest.

Solution

The ball falls $10$ m. After that, each bounce goes up and comes back down the same height: $7.5$ m up and $7.5$ m down, then $5.625$ m up and down, and so on.

The rebound heights $7.5, 5.625, \ldots$ form a GP with $a = 7.5$ and $r = \tfrac{3}{4}$ , with sum to infinity $\dfrac{7.5}{1 - 0.75} = 30$ .

\text{Total distance} = 10 + 2 \times 30 = 70 \text{ m}

The model assumes infinitely many bounces in a finite time, so in reality the ball stops sooner, but the total distance is a good estimate.

The odd-numbered terms

A geometric progression has sum to infinity $15$ . The sum to infinity of its odd-numbered terms (the $1$ st, $3$ rd, $5$ th, $\ldots$ ) is $10$ . Find the first term and the common ratio.

Solution

The odd-numbered terms $a, ar^2, ar^4, \ldots$ form a GP with ratio $r^2$ . So

\frac{a}{1 - r} = 15, \qquad \frac{a}{1 - r^2} = 10

Divide the first equation by the second:

\frac{1 - r^2}{1 - r} = \frac{15}{10} \quad\Rightarrow\quad \frac{(1 - r)(1 + r)}{1 - r} = 1.5 \quad\Rightarrow\quad 1 + r = 1.5

So $r = \tfrac{1}{2}$ and $a = 15\left(1 - \tfrac{1}{2}\right) = 7.5$ .

Watch out

Not checking $|r| < 1$ . Using $\dfrac{a}{1 - r}$ with $r = 2$ gives a negative "sum" for a series of positive terms. The formula is meaningless unless $|r| < 1$ .

Sign errors with a negative ratio. $1 - (-0.4) = 1.4$ , not $0.6$ .

Inequality direction. Solving $-1 < 1 - 2x < 1$ involves dividing by $-2$ , which reverses both signs.

Bouncing balls. The first drop is counted once; every rebound height is counted twice (up and down).

Writing $S_\infty$ for a finite sum. $S_\infty$ means the limit of $S_n$ ; use $S_n$ for the sum of a fixed number of terms.

Exam tip

State the condition. "The sum to infinity exists because $|r| = \tfrac{1}{2} < 1$ " is often worth a mark, and examiners report it being omitted.
Reject invalid ratios. If solving gives $r = 2$ and $r = \tfrac{1}{3}$ , write " $r = 2$ is rejected since $|r| < 1$ is needed for a sum to infinity".
Convergence sets. "Find the set of values of $x$ for which the progression is convergent" wants an inequality such as $0 < x < 1$ , from $-1 < r < 1$ .
Least $n$ . As with other GP questions on Paper 1, find $n$ by trial and show the values either side of the target.
Exact fractions. Recurring decimals should be given as fractions in lowest terms.

Summary

A GP converges exactly when $-1 < r < 1$ ; otherwise it diverges.
$S_\infty = \dfrac{a}{1 - r}$ for $|r| < 1$ .
$S_\infty - S_n = \dfrac{ar^n}{1 - r}$ : the sum of the terms after the $n$ th.
A ratio containing $x$ gives a convergence condition $-1 < r(x) < 1$ to solve.
Recurring decimals are convergent GPs: $0.\dot{4}\dot{5} = \dfrac{0.45}{0.99} = \dfrac{5}{11}$ .
Alternate terms form a GP with ratio $r^2$ .
Combine $S_\infty$ with $u_n$ or $S_n$ and eliminate $a$ ; reject any $r$ with $|r| \ge 1$ .

Practice questions

Question

Find the sum to infinity of the geometric progression $81, 54, 36, \ldots$
Find the sum to infinity of $5 - 2 + 0.8 - 0.32 + \cdots$
Explain why the geometric progression $1 + 1.1 + 1.21 + \cdots$ does not have a sum to infinity.
Express (a) $0.\dot{3}\dot{6}$ and (b) $0.1\dot{2}$ as fractions in their lowest terms.
A geometric progression has first term $20$ and sum to infinity $80$ . Find the common ratio and the exact value of the fifth term.
(a) Find the set of values of $x$ for which the series $1 + 3x + 9x^2 + \cdots$ converges. (b) Find the value of $x$ for which the sum to infinity is $4$ .
For $0 < \theta < \tfrac{\pi}{2}$ , show that the sum to infinity of $1 + \cos^2\theta + \cos^4\theta + \cdots$ is $\dfrac{1}{\sin^2\theta}$ , and find $\theta$ when this sum is $4$ .
The second term of a geometric progression is $6$ and the sum to infinity is $27$ . Find the two possible values of the common ratio and the corresponding first terms.
A geometric progression has first term $6$ and common ratio $0.4$ . Find the least value of $n$ for which the difference between the sum to infinity and the sum of the first $n$ terms is less than $0.01$ .
A ball is dropped from a height of $2$ m. After each bounce it rises to $60\%$ of the height from which it last fell. (a) Find the total distance travelled by the ball before it comes to rest. (b) Find after which bounce the greatest height reached first falls below $1$ cm.

Answers

$r = \tfrac{54}{81} = \tfrac{2}{3}$ , so $S_\infty = \dfrac{81}{1 - \frac{2}{3}} = 243$ .
$r = -\tfrac{2}{5} = -0.4$ , $|r| < 1$ . $S_\infty = \dfrac{5}{1 + 0.4} = \dfrac{5}{1.4} = \dfrac{25}{7}$ .
The common ratio is $1.1$ , and $|r| > 1$ , so the terms grow and the sums increase without limit: the progression is divergent.
(a) $\dfrac{0.36}{1 - 0.01} = \dfrac{36}{99} = \dfrac{4}{11}$ . (b) $0.1 + \dfrac{0.02}{1 - 0.1} = \dfrac{1}{10} + \dfrac{1}{45} = \dfrac{11}{90}$ .
$\dfrac{20}{1 - r} = 80$ , so $1 - r = \tfrac{1}{4}$ and $r = \tfrac{3}{4}$ . $u_5 = 20\left(\tfrac{3}{4}\right)^4 = 20 \times \tfrac{81}{256} = \tfrac{405}{64}$ .
(a) $r = 3x$ , so $-1 < 3x < 1$ : $-\tfrac{1}{3} < x < \tfrac{1}{3}$ . (b) $\dfrac{1}{1 - 3x} = 4$ , so $1 - 3x = \tfrac{1}{4}$ and $x = \tfrac{1}{4}$ , which is in the interval.
$r = \cos^2\theta$ , and $0 < \cos^2\theta < 1$ for $0 < \theta < \tfrac{\pi}{2}$ , so the series converges. $S_\infty = \dfrac{1}{1 - \cos^2\theta} = \dfrac{1}{\sin^2\theta}$ . If this is $4$ , $\sin^2\theta = \tfrac{1}{4}$ , so $\sin\theta = \tfrac{1}{2}$ (positive in this interval) and $\theta = \tfrac{\pi}{6}$ .
$ar = 6$ and $a = 27(1 - r)$ , so $27r(1 - r) = 6$ , giving $27r^2 - 27r + 6 = 0$ , i.e. $9r^2 - 9r + 2 = 0$ , $(3r - 1)(3r - 2) = 0$ . $r = \tfrac{1}{3}$ with $a = 18$ , or $r = \tfrac{2}{3}$ with $a = 9$ .
$S_\infty - S_n = \dfrac{6(0.4)^n}{0.6} = 10(0.4)^n < 0.01$ , so $0.4^n < 0.001$ . $0.4^7 = 0.00164$ and $0.4^8 = 0.000655$ , so $n = 8$ .
(a) Rebound heights $1.2, 0.72, \ldots$ : a GP with $a = 1.2$ , $r = 0.6$ and sum to infinity $\tfrac{1.2}{0.4} = 3$ . Total distance $= 2 + 2 \times 3 = 8$ m. (b) The height after the $n$ th bounce is $2(0.6)^n$ m. Need $2(0.6)^n < 0.01$ . $2(0.6)^{10} = 0.0121$ and $2(0.6)^{11} = 0.00726$ , so after the $11$ th bounce.

Partial sums that settle down

When does a GP converge?

The remaining sum

Ratios that contain a variable

Recurring decimals

Worked examples

Practice questions

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