Infinite geometric sums and convergence | Year 11

01 / LEARN

Understand the idea.

A sum approached by finite sums

An infinite series converges when its partial sums approach a finite limit. For a geometric series with $∣ r ∣ < 1$ , $r^{n}$ approaches zero, so the finite-sum formula approaches $a / (1 - r)$ .

For a nonzero first term, $∣ r ∣ \geq 1$ prevents geometric convergence. Terms approaching zero are necessary for convergence of any series, but this alone is not enough for a general series; the geometric formula supplies the justification here.

S_{\infty} = \frac{a}{1 - r} when ∣ r ∣ < 1

There is no final term

There is no last term in an infinite series. After taking $n$ terms beginning at $a$ , the omitted tail starts at $a r^{n}$ . Its signed sum is $R_{n} = a r^{n} / (1 - r)$ ; use $∣ R_{n} ∣$ when measuring approximation error.

To obtain an error below a stated tolerance, solve the tail inequality and find the smallest permitted integer count. Check the count just before it as well. An infinite-process model, such as idealised bouncing, relies on assumptions that real behaviour may eventually stop following.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

An alternating infinite sum

Find $9 - 3 + 1 - 1 / 3 + \dots$ .

Worked solution

1
Identify and test the ratio.
$a = 9, r = - \frac{1}{3}, ∣ r ∣ = \frac{1}{3} < 1$
A ratio of magnitude below one makes the geometric partial sums converge.
2
Apply the limiting sum formula.
$S_{\infty} = \frac{9}{1 - (- 1 / 3)} = \frac{9}{4 / 3}$
Subtracting a negative ratio increases the denominator.
3
Simplify the fraction.
$S_{\infty} = 9 \cdot \frac{3}{4} = \frac{27}{4}$
Dividing by a fraction means multiplying by its reciprocal.

Answer

$27 / 4$ .

EXAMPLE 02

Controlling a remaining tail

For $1 + 1 / 2 + 1 / 4 + \dots$ , find how many terms give a remaining tail below 0.01.

Worked solution

1
Express the tail after the first n terms.
$R_{n} = \frac{{(1 / 2)}^{n}}{1 - 1 / 2} = 2^{1 - n}$
The first omitted term is the term with exponent n.
2
Check the boundary counts.
$\begin{aligned} R_{7} & = \frac{1}{64} = 0.015625 > 0.01 \\ R_{8} & = \frac{1}{128} = 0.0078125 < 0.01 \end{aligned}$
Seven terms fail the strict tolerance and eight terms meet it.
3
Establish the minimum count.
$R_{n + 1} = \frac{1}{2} R_{n}, n \geq 8$
The positive tail halves with each extra term, so every count of at least eight works.

Answer

At least 8 terms.

03 / YOUR TURN

Try it yourself.

Check your understanding

Does $4 + 6 + 9 + \dots$ converge?

Show answer & reasoning

No.

The ratio is $6 / 4 = 9 / 6 = 3 / 2$ . Its magnitude exceeds one, so the terms grow instead of tending to zero and the series does not converge.

Keep practising

6 worksheet questions · 13 marks

Open worksheet Worked solutions

Read the questions online

Find $12 + 3 + \frac{3}{4} + \dots$ to infinity.
2 marks
Find the infinite sum for $a = 5, r = - 0.6$ .
2 marks
A geometric series starts at 8 and sums to 20. Find its ratio.
2 marks
Use partial sums to explain why $1 - 1 + 1 - 1 + \dots$ does not converge.
2 marks
A ball falls 2 m and rebounds to half its preceding height each time. Find the ideal total vertical distance.
3 marks
For $10 + 5 + 2.5 + \dots$ , find the tail after four terms.
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Find $12 + 3 + \frac{3}{4} + \dots$ to infinity.
2 marks
$\frac{12}{1 - \frac{1}{4}} = 16$ .
$16$ .
Find the infinite sum for $a = 5, r = - 0.6$ .
2 marks
$\frac{5}{1 + 0.6} = 3.125$ .
$\frac{25}{8} = 3.125$ .
A geometric series starts at 8 and sums to 20. Find its ratio.
2 marks
$\frac{8}{1 - r} = 20$ gives $r = \frac{3}{5}$ .
$\frac{3}{5}$ .
Use partial sums to explain why $1 - 1 + 1 - 1 + \dots$ does not converge.
2 marks
They do not approach one value.
Partial sums alternate between 1 and 0.
A ball falls 2 m and rebounds to half its preceding height each time. Find the ideal total vertical distance.
3 marks
The first fall is 2 m; later paired journeys total $2 (1 + \frac{1}{2} + \dots) = 4$ m.
$6 m$ .
For $10 + 5 + 2.5 + \dots$ , find the tail after four terms.
2 marks
The fifth term is $10 {(\frac{1}{2})}^{4} = \frac{5}{8}$ ; its tail sums to $\frac{\frac{5}{8}}{\frac{1}{2}} = \frac{5}{4}$ .
$1.25$ .

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics sequence and series content with introductory induction, regrouped into an original school-sized chapter.

Official curriculum source ↗

In this lesson

Understand the idea.

A sum approached by finite sums

There is no final term

See the method in action.

An alternating infinite sum

Identify and test the ratio.

Apply the limiting sum formula.

Simplify the fraction.

Controlling a remaining tail

Express the tail after the first n terms.

Check the boundary counts.

Establish the minimum count.

Try it yourself.

Check your understanding

Keep practising