Sequences and Series

What you'll learn

How to recognise and use arithmetic and geometric sequences.
How to find sums of finite series and sums to infinity.
How to handle recurrence relations, periodic sequences, and sigma notation.
How to approach worded modelling questions involving salaries, savings, production, and training plans.

1. The basic language

A sequence is a list of terms in order: for example, 4, 7, 10, 13, ...

A series is what you get when you add the terms of a sequence.

Definition

Sequences, terms and series

A sequence is an ordered list of numbers.
A term is one number in the sequence.
The notation $u_n$ means the $n$ th term of a sequence.
A series is the sum of terms in a sequence.
The notation $∑\sum$ means “add up”.

For example,

∑r=15r2\sum_{r=1}^{5} r^2

means

1^2+2^2+3^2+4^2+5^2

The lower number tells you where to start; the upper number tells you where to stop.

The sigma notation is labelled to show the lower limit, upper limit, index variable, and expression being summed.

Example

Reading a sigma series

A series is given by $∑r=418(5r−1)\sum_{r=4}^{18}(5r-1)$ . Find the first term, the common difference, and the number of terms.

The first term comes from the lower limit, so substitute $r = 4$ : $5 (4) - 1 = 19$ .
The next term comes from $r = 5$ : $5 (5) - 1 = 24$ , so the common difference is 5.
Count the terms inclusively: $18 - 4 + 1 = 15$ .

Common Mistake

Starting at the wrong value

In $∑r=418(5r−1)\sum_{r=4}^{18}(5r-1)$ , the first term uses $r = 4$ , not $r = 1$ . Always check the lower limit.

2. Arithmetic sequences and series

An arithmetic sequence is one where you add the same amount each time.

Definition

Arithmetic sequence

The first term is usually called $a$ .
The common difference is called $d$ .
The $n$ th term is
$u_n=a+(n-1)d$

So if the first term is 6 and the common difference is 4, the sequence is 6, 10, 14, 18, ...

The arithmetic sum formula

For an arithmetic series,

Sn=a+(a+d)+(a+2d)+⋯+(a+(n−1)d)S_n=a+(a+d)+(a+2d)+\cdots+\bigl(a+(n-1)d\bigr)

where $S_n$ means the sum of the first $n$ terms.

Write the sum forwards and backwards:

Sn=a+(a+d)+⋯+(a+(n−1)d)Sn=(a+(n−1)d)+⋯+(a+d)+a\begin{aligned} S_n&=a+(a+d)+\cdots+\bigl(a+(n-1)d\bigr)\\ S_n&=\bigl(a+(n-1)d\bigr)+\cdots+(a+d)+a \end{aligned}

Adding these gives $n$ identical pairs:

Writing an arithmetic series forwards and backwards creates identical pairs, leading to the arithmetic sum formula.

2Sn=n(2a+(n−1)d)2S_n=n\bigl(2a+(n-1)d\bigr)

Therefore,

Sn=n2(2a+(n−1)d)S_n=\frac{n}{2}\bigl(2a+(n-1)d\bigr)

Key Idea

Arithmetic formulae

Use $u_n=a+(n-1)d$ for a single term.
Use $Sn=n2(2a+(n−1)d)S_n=\frac{n}{2}\bigl(2a+(n-1)d\bigr)$ for a total.

Example

Finding an arithmetic sequence and a limit

The third term of an arithmetic sequence is -2. The sum of the first eight terms is 44. Find $a$ and $d$ , then find the greatest value of $n$ for which $S_n<250$ .

Use the third term condition:
$a + 2 d = - 2$
Use the sum condition with $n = 8$ :
$S8=82(2a+7d)44=4(2a+7d)2a+7d=11\begin{aligned} S_8&=\frac{8}{2}(2a+7d)\\ 44&=4(2a+7d)\\ 2a+7d&=11 \end{aligned}$
Solve the simultaneous equations:
$a+2d=−22a+7d=11d=5,a=−12\begin{aligned} a+2d&=-2\\ 2a+7d&=11\\ d&=5,\quad a=-12 \end{aligned}$
Write a formula for $S_n$ :
$Sn=n2(2(−12)+(n−1)5)Sn=n2(5n−29)\begin{aligned} S_n&=\frac{n}{2}\bigl(2(-12)+(n-1)5\bigr)\\ S_n&=\frac{n}{2}(5n-29) \end{aligned}$
Test the boundary: $S_{13}=234$ and $S_{14}=287$ , so the greatest value is 13.

Tip

Single term or total?

If the question asks “amount in week 100”, use $u_{100}$ . If it asks “total over 100 weeks”, use $S_{100}$ .

Maximum arithmetic sums

If $d < 0$ , the terms eventually become negative. The running total increases while you are adding positive terms, then decreases once you start adding negative terms.

For an arithmetic series with negative common difference, the running total is largest just before negative terms start being added.

Example

Greatest value of an arithmetic sum

An arithmetic series has first term 500 and sixth term 250. Find the greatest possible value of $S_n$ .

Use the sixth term to find $d$ :
$500 + 5 d = 250$
Solve for the common difference:
$d = - 50$
The $n$ th term is therefore $u_n=500+(n-1)(-50)=550-50n$ .
Find when the terms stop being positive: $u_{10}=50$ and $u_{11}=0$ , so adding term 11 does not change the sum.
The greatest sum is $S_{10}$ or $S_{11}$ :
$S10=102(500+50)=2750S_{10}=\frac{10}{2}(500+50)=2750$

3. Geometric sequences and series

A geometric sequence is one where you multiply by the same amount each time.

Definition

Geometric sequence

The first term is $a$ .
The common ratio is $r$ .
The $n$ th term is
$u_n=ar^{n-1}$

For example, 3, 6, 12, 24, ... has common ratio 2.

The finite geometric sum is

Sn=a(1−rn)1−rS_n=\frac{a(1-r^n)}{1-r}

provided $r≠1r\neq1$ .

Example

Finding a geometric sequence from two terms

The fifth term of a geometric sequence is 18 and the eighth term is 144. Find the common ratio, the first term, and the sum of the first 12 terms to the nearest whole number.

Write the two pieces of information using $u_n=ar^{n-1}$ : $ar^4=18$ and $ar^7=144$ .
Divide the second equation by the first: $r^3=8$ , so $r = 2$ .
Substitute into $ar^4=18$ : $a⋅24=18a\cdot2^4=18$ , so $a=98a=\frac{9}{8}$ .
Use the geometric sum formula:
$S12=98(212−1)2−1=368558≈4607S_{12}=\frac{\frac{9}{8}(2^{12}-1)}{2-1}=\frac{36855}{8}\approx4607$

Sum to infinity

Some geometric series get closer and closer to a fixed total.

A convergent geometric series has partial sums that approach a horizontal limiting value when |r|<1.

Definition

Sum to infinity

For a geometric series with $∣ r ∣ < 1$ , the sum to infinity is

S∞=a1−rS_\infty=\frac{a}{1-r}

Common Mistake

When $S_{\infty}$ exists

The formula for $S∞S_\infty$ only works when $∣ r ∣ < 1$ . If the ratio has magnitude at least 1, the terms do not settle to zero.

Example

Using sum to infinity

A convergent geometric series has second term 6 and sum to infinity 27. Find the possible values of $r$ and $a$ . For the larger value of $r$ , find the smallest $n$ such that $S_n>26$ .

The second term gives $a r = 6$ , and the sum to infinity gives $a1−r=27\frac{a}{1-r}=27$ .
Substitute $a=6ra=\frac{6}{r}$ :
$6r(1−r)=279r2−9r+2=0(3r−1)(3r−2)=0\begin{aligned} \frac{6}{r(1-r)}&=27\\ 9r^2-9r+2&=0\\ (3r-1)(3r-2)&=0 \end{aligned}$
Hence $r=13r=\frac{1}{3}$ or $r=23r=\frac{2}{3}$ . The corresponding values of $a$ are 18 and 9.
For the larger ratio, $a = 9$ and $r=23r=\frac{2}{3}$ . We need $27(1−(23)n)>2627\left(1-\left(\frac{2}{3}\right)^n\right)>26$ , so $(23)n<127\left(\frac{2}{3}\right)^n<\frac{1}{27}$ .
Use logarithms: $n>log⁡(1/27)log⁡(2/3)≈8.13n>\frac{\log(1/27)}{\log(2/3)}\approx8.13$ , so the smallest integer is 9.

Common Mistake

Percentage multipliers

A 4% increase means multiply by 1.04 each time. A 15% decrease means multiply by 0.85 each time.

4. Recurrence relations and periodic sequences

Some sequences are defined by a rule that uses previous terms.

A recurrence relation generates each term by feeding the previous term into the same rule.

Definition

Recurrence relation

A recurrence relation defines a term using earlier terms, such as $u_{n+1}=2u_n-3$ . You also need starting value information, such as $u_1=5$ .

Example

Working with a recurrence relation

A sequence is defined by $x_1=3$ and $x_{n+1}=bx_n-4$ . Find $x_2$ , show that $x_3=3b^2-4b-4$ , and find $b$ if $x_3=8$ .

Substitute $x_1=3$ into the rule: $x_2=3b-4$ .
Substitute $x_2$ into the rule again:
$x_3=b(3b-4)-4=3b^2-4b-4$
Set this equal to 8 and solve:
$3b2−4b−4=83b2−4b−12=0b=2±2103\begin{aligned} 3b^2-4b-4&=8\\ 3b^2-4b-12&=0\\ b&=\frac{2\pm2\sqrt{10}}{3} \end{aligned}$

Definition

Periodic sequence

A periodic sequence repeats its values in a cycle. The period is the length of the shortest repeating block.

A periodic sequence is shown as a repeating block, with the period equal to the length of the shortest cycle.

Example

Spotting a periodic sequence

A sequence is defined by $u1=35u_1=\frac{3}{5}$ and $un+1=1unu_{n+1}=\frac{1}{u_n}$ . Find $∑r=1100ur\sum_{r=1}^{100}u_r$ .

Work out the first few terms: $u1=35u_1=\frac{3}{5}$ , $u2=53u_2=\frac{5}{3}$ , and $u3=35u_3=\frac{3}{5}$ .
The sequence repeats every 2 terms, so the first 100 terms form 50 identical pairs.
Each pair sums to $35+53=3415\frac{3}{5}+\frac{5}{3}=\frac{34}{15}$ .
Therefore the total is $50⋅3415=340350\cdot\frac{34}{15}=\frac{340}{3}$ .

5. Telescoping sums

Some sums collapse because most middle terms cancel.

In a telescoping product or sum, consecutive middle factors cancel so that only the start and end remain.

Definition

Telescoping sum

A telescoping sum is a sum where consecutive terms cancel, leaving only a few terms at the start and end.

Example

A logarithmic telescoping sum

Show that $∑r=115log⁡2(r+1r)=4\sum_{r=1}^{15}\log_2\left(\frac{r+1}{r}\right)=4$ .

Use the log rule that a sum of logs is the log of a product:
$∑r=115log⁡2(r+1r)=log⁡2(21⋅32⋅43⋯1615)\sum_{r=1}^{15}\log_2\left(\frac{r+1}{r}\right) = \log_2\left(\frac{2}{1}\cdot\frac{3}{2}\cdot\frac{4}{3}\cdots\frac{16}{15}\right)$
Most factors cancel, leaving $log_2(16)$ .
Since $2^4=16$ , the sum is 4.

Tip

Nearly identical sigma sums

If you see $∑k=0nf(k)−∑k=0n−1f(k)\sum_{k=0}^{n}f(k)-\sum_{k=0}^{n-1}f(k)$ , everything cancels except the final term, so the result is $f (n)$ .

Exam technique

In the exam

Decide first whether the question is asking for a term, $u_n$ , or a total, $S_n$ .
For worded questions, define the first term carefully and count how many terms are included.
For inequalities involving $n$ , solve algebraically if possible, then check the nearest integers.

Self review

Check yourself

Can you explain the difference between an arithmetic sequence and a geometric sequence?
When does a geometric series have a sum to infinity?
If a sigma sum starts at $r = 5$ , what value of $r$ gives the first term?

What you'll learn

How to recognise and use arithmetic and geometric sequences.
How to find sums of finite series and sums to infinity.
How to handle recurrence relations, periodic sequences, and sigma notation.
How to approach worded modelling questions involving salaries, savings, production, and training plans.

1. The basic language

A sequence is a list of terms in order: for example, 4, 7, 10, 13, ...

A series is what you get when you add the terms of a sequence.

Definition

Sequences, terms and series

A sequence is an ordered list of numbers.
A term is one number in the sequence.
The notation $u_n$ means the $n$ th term of a sequence.
A series is the sum of terms in a sequence.
The notation $∑\sum$ means “add up”.

For example,

∑r=15r2\sum_{r=1}^{5} r^2

means

1^2+2^2+3^2+4^2+5^2

The lower number tells you where to start; the upper number tells you where to stop.

The sigma notation is labelled to show the lower limit, upper limit, index variable, and expression being summed.

Example

Reading a sigma series

A series is given by $∑r=418(5r−1)\sum_{r=4}^{18}(5r-1)$ . Find the first term, the common difference, and the number of terms.

The first term comes from the lower limit, so substitute $r = 4$ : $5 (4) - 1 = 19$ .
The next term comes from $r = 5$ : $5 (5) - 1 = 24$ , so the common difference is 5.
Count the terms inclusively: $18 - 4 + 1 = 15$ .

Common Mistake

Starting at the wrong value

In $∑r=418(5r−1)\sum_{r=4}^{18}(5r-1)$ , the first term uses $r = 4$ , not $r = 1$ . Always check the lower limit.