Mathematics 9709 · Pure Mathematics 1

Infinite geometric series

The surprise of the chapter: if a geometric series gets small fast enough, then endless terms still add up to a number that is not endless.
$8+4+2+1+\dots$ never ends, but its total is exactly $16$ .

When $|r|<1$ , the terms get smaller and smaller. The series then has a sum to infinity that is a real number: $S_\infty=\dfrac{a}{1-r}$ . If $|r|\ge 1$ , there is no such sum. So check the ratio first.

When |r| < 1, the sum has a limit

In the normal sum $S_n=\dfrac{a(1-r^{\,n})}{1-r}$ , the only part that changes is $r^{\,n}$ .
When $|r|<1$ , multiplying by $r$ again and again makes the terms smaller and smaller. So $r^{\,n}\to 0$ .
Drop it, and the formula becomes one simple fraction:

$S_\infty=\frac{a}{1-r},\qquad |r|<1$

8 + 4 + 2 + 1 + ...: each piece is half the one before, so the running total fills the bar but never reaches the gold line at 16.

If $|r|\ge 1$ the terms do not get smaller. So there is no sum to infinity. The series diverges (the total has no limit).
So every sum-to-infinity question has two parts. First check $|r|<1$ . Then use the formula.

Drag the ratio below. The bars are the terms $a,ar,ar^2,\dots$ , and the green bar is their running sum. Inside $-1<r<1$ the green bar stops at the gold line $\dfrac{a}{1-r}$ . Go past $\pm1$ and the bars grow with no limit.

ratio r

r = +0.60·converges S∞ = 2.50

Why r ⁿ goes to zero (the full reason)

If you multiply by a number whose size is below $1$ , the answer gets smaller. Do it again and again, and the answer gets very close to zero: $(\tfrac12)^{10}\approx 0.001$ , $(\tfrac12)^{20}\approx 0.000001$ , and so on with no limit. So as $n$ grows, $r^{\,n}\to 0$ .

Put that back into $S_n=\dfrac{a(1-r^{\,n})}{1-r}$ . The $1-r^{\,n}$ on top becomes $1-0=1$ , which leaves $\dfrac{a}{1-r}$ . That is the value the running total gets close to but never passes. It is the gold line at $16$ in the picture.

Worked examples

Worked example

A quick sum to infinity

The first four terms of a GP are $5,\,4,\,3.2,\,2.56$ . The ratio is one term divided by the one before, $r=\tfrac{4}{5}$ . Since $|r|<1$ , a sum to infinity exists:

$S_\infty=\frac{a}{1-r}=\frac{5}{1-\tfrac45}=\frac{5}{\tfrac15}$

Answer

S_\infty=25

Worked example

A recurring decimal is really a sum to infinity

You have used this since primary school without knowing it. $0.3333\dots$ is the never-ending sum $\tfrac{3}{10}+\tfrac{3}{100}+\tfrac{3}{1000}+\dots$ . It is a GP with $a=\tfrac{3}{10}$ and $r=\tfrac{1}{10}$ (each piece is a tenth of the last). Since $|r|<1$ , just use the formula:

$0.3333\dots=\frac{a}{1-r}=\frac{3/10}{1-1/10}=\frac{3/10}{9/10}=\tfrac13$

Answer

0.3333\dots=\tfrac13

Worked example

Real exam: first term $216$ , fourth term $64$

The first term of a GP is $216$ and the fourth term is $64$ . Find the sum to infinity.

The fourth term is $ar^3$ , so $216\,r^3=64$ , giving $r^3=\tfrac{64}{216}=\tfrac{8}{27}$ .
Cube-root: $r=\sqrt[3]{\tfrac{8}{27}}=\tfrac23$ . Check that $|r|<1$ : yes, so a sum to infinity exists.
Quote the formula: $S_\infty=\dfrac{a}{1-r}=\dfrac{216}{1-\tfrac23}=\dfrac{216}{\tfrac13}$ .

Answer

S_\infty=648

(9709/13 Nov 2022 Q9a) The mark scheme only gives the final $S_\infty$ mark if you show $|r|<1$ . So write that check down. Do not just assume it.

Worked example

Real exam: the sum to infinity gives a quadratic in $a$

A geometric progression has first term $a$ and second term $\dfrac{a^2}{a+2}$ , with $a>0$ . The sum to infinity is $264$ . Find $a$ .

First find the ratio, the second term over the first. Then write $1-r$ as a single fraction:

$r=\dfrac{a^2/(a+2)}{a}=\dfrac{a}{a+2},\qquad 1-r=\dfrac{2}{a+2}$

Now put that into $S_\infty=\dfrac{a}{1-r}$ . Dividing by $\tfrac{2}{a+2}$ turns it over and moves it to the top:

$S_\infty=\frac{a}{\,2/(a+2)\,}=\frac{a(a+2)}{2}=264$

Multiply up: $a(a+2)=528$ . This rearranges to the quadratic $a^2+2a-528=0$ . Solve it on the calculator first. Then write the bracket form to show the method (the same way as in the Quadratics chapter). The roots are $22$ and $-24$ , so

$a^2+2a-528=(a-22)(a+24)=0$

Reject $a=-24$ because $a>0$ . A quick check: at $a=22$ the ratio is $r=\tfrac{22}{24}=\tfrac{11}{12}$ , which is below $1$ . So a sum to infinity really does exist. (9709/13 Jun 2023 Q8)

Answer

a=22

Where the marks go

Common mistake

Always check $|r|<1$ before you use

S_\infty

. If

|r|\ge 1

the series diverges and has no finite sum. So writing one down loses the mark.

Common mistake

Do not confuse

S_n

and

S_\infty

. The sum to infinity is just

\dfrac{a}{1-r}

. It has no

r^{\,n}

term and no value of

n

. The finite

S_n

has both. (9709/13 Jun 2023 Q8)

Now you try

The fourth and sixth terms of a GP are $36$ and $6$ respectively, and the common ratio is positive. Find the sum to infinity, giving your answer in exact form. (9709/12 Jun 2025 Q10b)

Your turn— tap to reveal the worked answer (9709/12 Jun 2025 Q10b)

The two terms are two steps apart. Divide to cancel $a$ : $\dfrac{ar^5}{ar^3}=\dfrac{6}{36}\Rightarrow r^2=\tfrac16$ .

The ratio is positive, so take the positive root: $r=\dfrac{1}{\sqrt6}$ . It is below $1$ , so a sum to infinity exists.

Find $a$ from $ar^3=36$ : $a=36\,r^{-3}=36\,(\sqrt6)^3=216\sqrt6$ . Then $S_\infty=\dfrac{a}{1-r}=\dfrac{216\sqrt6}{1-\tfrac{1}{\sqrt6}}$ .

Multiply the top and the bottom by $\sqrt6$ to remove the surd from the bottom, then rationalise:

$S_\infty=\frac{216\sqrt6\cdot\sqrt6}{\sqrt6-1}=\frac{1296}{\sqrt6-1}$

Answer

S_\infty=\dfrac{1296}{\sqrt6-1}=\dfrac{1296(\sqrt6+1)}{5}\approx 894

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Infinite geometric series

The surprise of the chapter: if a geometric series gets small fast enough, then endless terms still add up to a number that is not endless.
$8+4+2+1+\dots$ never ends, but its total is exactly $16$ .

When |r| < 1, the sum has a limit

In the normal sum $S_n=\dfrac{a(1-r^{\,n})}{1-r}$ , the only part that changes is $r^{\,n}$ .
When $|r|<1$ , multiplying by $r$ again and again makes the terms smaller and smaller. So $r^{\,n}\to 0$ .
Drop it, and the formula becomes one simple fraction:

$S_\infty=\frac{a}{1-r},\qquad |r|<1$

8 + 4 + 2 + 1 + ...: each piece is half the one before, so the running total fills the bar but never reaches the gold line at 16.

If $|r|\ge 1$ the terms do not get smaller. So there is no sum to infinity. The series diverges (the total has no limit).
So every sum-to-infinity question has two parts. First check $|r|<1$ . Then use the formula.

ratio r

r = +0.60·converges S∞ = 2.50

Why r ⁿ goes to zero (the full reason)

Worked examples

Worked example

A quick sum to infinity

The first four terms of a GP are $5,\,4,\,3.2,\,2.56$ . The ratio is one term divided by the one before, $r=\tfrac{4}{5}$ . Since $|r|<1$ , a sum to infinity exists:

$S_\infty=\frac{a}{1-r}=\frac{5}{1-\tfrac45}=\frac{5}{\tfrac15}$

Answer

S_\infty=25

Worked example

A recurring decimal is really a sum to infinity

$0.3333\dots=\frac{a}{1-r}=\frac{3/10}{1-1/10}=\frac{3/10}{9/10}=\tfrac13$

Answer

0.3333\dots=\tfrac13

Worked example

Real exam: first term $216$ , fourth term $64$

The first term of a GP is $216$ and the fourth term is $64$ . Find the sum to infinity.

The fourth term is $ar^3$ , so $216\,r^3=64$ , giving $r^3=\tfrac{64}{216}=\tfrac{8}{27}$ .
Cube-root: $r=\sqrt[3]{\tfrac{8}{27}}=\tfrac23$ . Check that $|r|<1$ : yes, so a sum to infinity exists.
Quote the formula: $S_\infty=\dfrac{a}{1-r}=\dfrac{216}{1-\tfrac23}=\dfrac{216}{\tfrac13}$ .

Answer

S_\infty=648

(9709/13 Nov 2022 Q9a) The mark scheme only gives the final $S_\infty$ mark if you show $|r|<1$ . So write that check down. Do not just assume it.

Worked example

Real exam: the sum to infinity gives a quadratic in $a$

A geometric progression has first term $a$ and second term $\dfrac{a^2}{a+2}$ , with $a>0$ . The sum to infinity is $264$ . Find $a$ .

First find the ratio, the second term over the first. Then write $1-r$ as a single fraction:

$r=\dfrac{a^2/(a+2)}{a}=\dfrac{a}{a+2},\qquad 1-r=\dfrac{2}{a+2}$

Now put that into $S_\infty=\dfrac{a}{1-r}$ . Dividing by $\tfrac{2}{a+2}$ turns it over and moves it to the top:

$S_\infty=\frac{a}{\,2/(a+2)\,}=\frac{a(a+2)}{2}=264$

$a^2+2a-528=(a-22)(a+24)=0$

Reject $a=-24$ because $a>0$ . A quick check: at $a=22$ the ratio is $r=\tfrac{22}{24}=\tfrac{11}{12}$ , which is below $1$ . So a sum to infinity really does exist. (9709/13 Jun 2023 Q8)

Answer

a=22

Where the marks go

Common mistake

Always check $|r|<1$ before you use

S_\infty

. If

|r|\ge 1

the series diverges and has no finite sum. So writing one down loses the mark.

Common mistake

Do not confuse

S_n

and

S_\infty

. The sum to infinity is just

\dfrac{a}{1-r}

. It has no

r^{\,n}

term and no value of

n

. The finite

S_n

has both. (9709/13 Jun 2023 Q8)

Now you try

The fourth and sixth terms of a GP are $36$ and $6$ respectively, and the common ratio is positive. Find the sum to infinity, giving your answer in exact form. (9709/12 Jun 2025 Q10b)

Your turn— tap to reveal the worked answer (9709/12 Jun 2025 Q10b)

The two terms are two steps apart. Divide to cancel $a$ : $\dfrac{ar^5}{ar^3}=\dfrac{6}{36}\Rightarrow r^2=\tfrac16$ .

The ratio is positive, so take the positive root: $r=\dfrac{1}{\sqrt6}$ . It is below $1$ , so a sum to infinity exists.

Find $a$ from $ar^3=36$ : $a=36\,r^{-3}=36\,(\sqrt6)^3=216\sqrt6$ . Then $S_\infty=\dfrac{a}{1-r}=\dfrac{216\sqrt6}{1-\tfrac{1}{\sqrt6}}$ .

Multiply the top and the bottom by $\sqrt6$ to remove the surd from the bottom, then rationalise:

$S_\infty=\frac{216\sqrt6\cdot\sqrt6}{\sqrt6-1}=\frac{1296}{\sqrt6-1}$

Answer

S_\infty=\dfrac{1296}{\sqrt6-1}=\dfrac{1296(\sqrt6+1)}{5}\approx 894