1.4.6 Increasing, decreasing and periodic sequences (A-level only)

What you'll learn

How to recognise and prove that a sequence is increasing or decreasing.
How the difference $u_{n+1}-u_n$ determines the behaviour of consecutive terms.
How to identify periodic sequences and find their least period.
How to work with sequences whose behaviour depends on their starting value.

Sequence notation

A sequence is an ordered list of terms. We usually write it as

…u_1,\ u_2,\ u_3,\ \ldots

Here, $u_n$ is the term in position $n$ , and $u_{n+1}$ is the term immediately after it.

A sequence may be defined:

explicitly, by giving $u_n$ directly in terms of $n$ ;
recursively, by giving each term in terms of one or more earlier terms.

For example, $u_n=n^2+1$ is explicit, while $u_{n+1}=u_n+3$ with $u_1=2$ is recursive.

Three discrete sequence plots showing increasing, decreasing and periodic behaviour

Notice that a sequence is defined only at positive integer values of $n$ . Its graph therefore consists of separate points, not a continuous curve.

Increasing sequences

Definition

Increasing sequence

A sequence is increasing if every term is greater than the term before it:

u_{n+1}>u_n

for every relevant positive integer $n$ .

To investigate whether a sequence is increasing, subtract one term from the next:

u_{n+1}-u_n.

If this difference is positive for every relevant $n$ , then $u_{n+1}>u_n$ , so the sequence is increasing.

Key Idea

Test consecutive terms

To prove that a sequence is increasing, show that

u_{n+1}-u_n>0

for every positive integer $n$ in its domain.

Example

Proving that a quadratic sequence is increasing

The sequence is defined by $u_n=n^2+2n$ . Show that it is increasing for $n≥1n\geq1$ .

Find the next term by replacing $n$ with $n + 1$ :
$un+1=(n+1)2+2(n+1)=n2+4n+3.\begin{aligned} u_{n+1} &=(n+1)^2+2(n+1)\\ &=n^2+4n+3. \end{aligned}$
Subtract the original term:
$un+1−un=(n2+4n+3)−(n2+2n)=2n+3.\begin{aligned} u_{n+1}-u_n &=(n^2+4n+3)-(n^2+2n)\\ &=2n+3. \end{aligned}$
Since $n≥1n\geq1$ , we have $2 n + 3 > 0$ . Therefore $u_{n+1}>u_n$ for every positive integer $n$ , so the sequence is increasing.

Common Mistake

Checking only the first few terms

Calculating several increasing terms suggests a pattern, but it does not prove that the sequence continues to increase. Use an inequality valid for every relevant $n$ .

Decreasing sequences

Definition

Decreasing sequence

A sequence is decreasing if every term is less than the term before it:

u_{n+1}<u_n

for every relevant positive integer $n$ .

The same consecutive-term test applies. If

u_{n+1}-u_n<0,

then the sequence is decreasing.

Example

Proving that a reciprocal sequence is decreasing

Show that the sequence defined by

un=1n+2u_n=\frac{1}{n+2}

is decreasing for $n≥1n\geq1$ .

Write the next term:
$un+1=1n+3.u_{n+1}=\frac{1}{n+3}.$
Find the difference using a common denominator:
$un+1−un=1n+3−1n+2=n+2−(n+3)(n+3)(n+2)=−1(n+3)(n+2).\begin{aligned} u_{n+1}-u_n &=\frac{1}{n+3}-\frac{1}{n+2}\\ &=\frac{n+2-(n+3)}{(n+3)(n+2)}\\ &=-\frac{1}{(n+3)(n+2)}. \end{aligned}$
For $n≥1n\geq1$ , both $n + 2$ and $n + 3$ are positive. The denominator is therefore positive, so
$−1(n+3)(n+2)<0.-\frac{1}{(n+3)(n+2)}<0.$
Hence $u_{n+1}<u_n$ , and the sequence is decreasing.

Tip

Choose a useful form

For an explicitly defined sequence, $u_{n+1}-u_n$ is usually the clearest test. For a recurrence, you may already have an expression for $u_{n+1}-u_n$ after rearranging its defining rule.

Strict and non-strict behaviour

The definitions above use strict inequalities. This means consecutive equal terms are not allowed.

Some questions or textbooks also use the following terms:

non-decreasing means $un+1≥unu_{n+1}\geq u_n$ ;
non-increasing means $un+1≤unu_{n+1}\leq u_n$ .

A constant sequence is both non-decreasing and non-increasing, but it is neither strictly increasing nor strictly decreasing.

Common Mistake

Check the inequality convention

If a question defines “increasing” using $≥\geq$ rather than $>$ , follow the definition given in that question. Otherwise, distinguish carefully between increasing and non-decreasing.

When the behaviour depends on the term number

A sequence might increase for some values of $n$ and decrease for others. In that case, solve the appropriate inequality to locate the change.

Example

Finding where a sequence begins to increase

Let $u_n=n^2-10n$ . Determine when the sequence is increasing.

Calculate the difference between consecutive terms:
$un+1−un=((n+1)2−10(n+1))−(n2−10n)=2n−9.\begin{aligned} u_{n+1}-u_n &=\left((n+1)^2-10(n+1)\right)-(n^2-10n)\\ &=2n-9. \end{aligned}$
For the sequence to increase from $u_n$ to $u_{n+1}$ , require
$2 n - 9 > 0,$
which gives $n>92n>\frac{9}{2}$ .
Since $n$ is an integer, this means $n≥5n\geq5$ . Therefore $u_{n+1}>u_n$ for $n≥5n\geq5$ . The sequence starts increasing from $u_5$ to $u_6$ .

Common Mistake

Ignoring integer values of n

After solving an inequality, convert the result to the appropriate integer values. Here, $n>92n>\frac{9}{2}$ means $n≥5n\geq5$ , not $n≥92n\geq\frac{9}{2}$ .

Using a recurrence

For a recursively defined sequence, whether it increases can depend on the current term rather than directly on $n$ .

Suppose

u_{n+1}=f(u_n).

Then

u_{n+1}-u_n=f(u_n)-u_n.

The sign of this expression tells you whether the next term is above or below the current one.

Example

Determining behaviour from a recurrence

A sequence satisfies $u_{n+1}=3u_n-4$ .

Compare consecutive terms by subtracting $u_n$ :
$u_{n+1}-u_n=2u_n-4=2(u_n-2).$
If $u_n>2$ , then $2(u_n-2)>0$ , so $u_{n+1}>u_n$ . The sequence increases at that step.
If $u_n<2$ , then $2(u_n-2)<0$ , so $u_{n+1}<u_n$ . If $u_n=2$ , then $u_{n+1}=2$ , so the next term is unchanged.

This example identifies the possible behaviour at a particular step. To prove that the entire sequence is increasing, you must also establish that $u_n>2$ for every relevant $n$ , often using induction or a property of the recurrence.

Periodic sequences

Definition

Periodic sequence

A sequence is periodic if its terms repeat after a fixed positive integer number of positions. In symbols, there is a positive integer $p$ such that

u_{n+p}=u_n

for every relevant $n$ . The number $p$ is called a period.

The least period, sometimes called the fundamental period, is the smallest positive value of $p$ for which the repetition occurs.

For example, the sequence

…1,\ 3,\ 2,\ 1,\ 3,\ 2,\ \ldots

has least period 3. It also repeats after 6 or 9 positions, but these are not its least period.

Example

Finding the least period

Consider the sequence defined by $un=cos⁡(πn2)u_n=\cos\left(\frac{\pi n}{2}\right)$ .

Calculate enough terms to reveal a possible repeating block:
$u1=0,u2=−1,u3=0,u4=1,u5=0.u_1=0,\quad u_2=-1,\quad u_3=0,\quad u_4=1,\quad u_5=0.$
This suggests the repeating block $0, - 1, 0, 1$ .
Use the periodicity of cosine:
$un+4=cos⁡(π(n+4)2)=cos⁡(πn2+2π)=cos⁡(πn2)=un.\begin{aligned} u_{n+4} &=\cos\left(\frac{\pi(n+4)}{2}\right)\\ &=\cos\left(\frac{\pi n}{2}+2\pi\right)\\ &=\cos\left(\frac{\pi n}{2}\right)=u_n. \end{aligned}$
Therefore 4 is a period.
Smaller positive integers do not reproduce every term: for instance, $u_1=0$ but $u_2=-1$ , and $u_2=-1$ but $u_4=1$ . Hence the least period is 4.

Common Mistake

Giving any period instead of the least period

If a sequence has least period 3, then 6 is also a period. However, when asked to identify “the period”, questions usually expect the smallest positive period.

Periodic does not mean constant

A constant sequence satisfies $u_{n+1}=u_n$ , so it is periodic with least period 1. However, most periodic sequences move repeatedly through a block of different values.

A non-constant periodic sequence cannot be strictly increasing or strictly decreasing forever. After one complete period, it returns to an earlier value, contradicting permanent strict movement in one direction.

Key Idea

Classifying the behaviour

Increasing and decreasing describe how consecutive terms compare.
Periodic describes repetition: $u_{n+p}=u_n$ .
A sequence may be neither increasing, decreasing nor periodic.
A non-constant periodic sequence cannot be strictly increasing or strictly decreasing.

Exam technique

In the exam

Write down $u_{n+1}$ carefully by replacing every occurrence of $n$ with $n + 1$ .
Simplify $u_{n+1}-u_n$ and determine its sign over the stated integer domain.
For a periodic sequence, identify a candidate repeating block and then verify $u_{n+p}=u_n$ ; check that no smaller positive period works.
State the conclusion explicitly, including any restriction such as “increasing for $n≥5n\geq5$ ”.

Self review

Check yourself

How would you prove that $u_n=3n^2-n$ is increasing for positive integer values of $n$ ?
For which values of $n$ is the sequence $u_n=n^2-14n$ increasing?
What is the least period of the sequence $2,−1,4,2,−1,4,…2,-1,4,2,-1,4,\ldots$ ?

What you'll learn

How to recognise and prove that a sequence is increasing or decreasing.
How the difference $u_{n+1}-u_n$ determines the behaviour of consecutive terms.
How to identify periodic sequences and find their least period.
How to work with sequences whose behaviour depends on their starting value.

Sequence notation

A sequence is an ordered list of terms. We usually write it as

…u_1,\ u_2,\ u_3,\ \ldots

Here, $u_n$ is the term in position $n$ , and $u_{n+1}$ is the term immediately after it.

A sequence may be defined:

explicitly, by giving $u_n$ directly in terms of $n$ ;
recursively, by giving each term in terms of one or more earlier terms.

For example, $u_n=n^2+1$ is explicit, while $u_{n+1}=u_n+3$ with $u_1=2$ is recursive.

Three discrete sequence plots showing increasing, decreasing and periodic behaviour

Notice that a sequence is defined only at positive integer values of $n$ . Its graph therefore consists of separate points, not a continuous curve.

Increasing sequences

Definition

Increasing sequence

A sequence is increasing if every term is greater than the term before it:

u_{n+1}>u_n

for every relevant positive integer $n$ .

To investigate whether a sequence is increasing, subtract one term from the next:

u_{n+1}-u_n.

If this difference is positive for every relevant $n$ , then $u_{n+1}>u_n$ , so the sequence is increasing.

Key Idea

Test consecutive terms

To prove that a sequence is increasing, show that

u_{n+1}-u_n>0

for every positive integer $n$ in its domain.

Example

Proving that a quadratic sequence is increasing

The sequence is defined by $u_n=n^2+2n$ . Show that it is increasing for $n≥1n\geq1$ .

Find the next term by replacing $n$ with $n + 1$ :
$un+1=(n+1)2+2(n+1)=n2+4n+3.\begin{aligned} u_{n+1} &=(n+1)^2+2(n+1)\\ &=n^2+4n+3. \end{aligned}$
Subtract the original term:
$un+1−un=(n2+4n+3)−(n2+2n)=2n+3.\begin{aligned} u_{n+1}-u_n &=(n^2+4n+3)-(n^2+2n)\\ &=2n+3. \end{aligned}$
Since $n≥1n\geq1$ , we have $2 n + 3 > 0$ . Therefore $u_{n+1}>u_n$ for every positive integer $n$ , so the sequence is increasing.

Common Mistake

Checking only the first few terms

Calculating several increasing terms suggests a pattern, but it does not prove that the sequence continues to increase. Use an inequality valid for every relevant $n$ .

Decreasing sequences

Definition

Decreasing sequence

A sequence is decreasing if every term is less than the term before it:

u_{n+1}<u_n

for every relevant positive integer $n$ .

The same consecutive-term test applies. If

u_{n+1}-u_n<0,

then the sequence is decreasing.

Example

Proving that a reciprocal sequence is decreasing

Show that the sequence defined by

un=1n+2u_n=\frac{1}{n+2}

is decreasing for $n≥1n\geq1$ .

Write the next term:
$un+1=1n+3.u_{n+1}=\frac{1}{n+3}.$
Find the difference using a common denominator:
$un+1−un=1n+3−1n+2=n+2−(n+3)(n+3)(n+2)=−1(n+3)(n+2).\begin{aligned} u_{n+1}-u_n &=\frac{1}{n+3}-\frac{1}{n+2}\\ &=\frac{n+2-(n+3)}{(n+3)(n+2)}\\ &=-\frac{1}{(n+3)(n+2)}. \end{aligned}$
For $n≥1n\geq1$ , both $n + 2$ and $n + 3$ are positive. The denominator is therefore positive, so
$−1(n+3)(n+2)<0.-\frac{1}{(n+3)(n+2)}<0.$
Hence $u_{n+1}<u_n$ , and the sequence is decreasing.

Tip

Choose a useful form

For an explicitly defined sequence, $u_{n+1}-u_n$ is usually the clearest test. For a recurrence, you may already have an expression for $u_{n+1}-u_n$ after rearranging its defining rule.

Strict and non-strict behaviour

The definitions above use strict inequalities. This means consecutive equal terms are not allowed.

Some questions or textbooks also use the following terms:

non-decreasing means $un+1≥unu_{n+1}\geq u_n$ ;
non-increasing means $un+1≤unu_{n+1}\leq u_n$ .

A constant sequence is both non-decreasing and non-increasing, but it is neither strictly increasing nor strictly decreasing.

Common Mistake

Check the inequality convention

If a question defines “increasing” using $≥\geq$ rather than $>$ , follow the definition given in that question. Otherwise, distinguish carefully between increasing and non-decreasing.

When the behaviour depends on the term number

A sequence might increase for some values of $n$ and decrease for others. In that case, solve the appropriate inequality to locate the change.

Example

Finding where a sequence begins to increase

Let $u_n=n^2-10n$ . Determine when the sequence is increasing.

Calculate the difference between consecutive terms:
$un+1−un=((n+1)2−10(n+1))−(n2−10n)=2n−9.\begin{aligned} u_{n+1}-u_n &=\left((n+1)^2-10(n+1)\right)-(n^2-10n)\\ &=2n-9. \end{aligned}$
For the sequence to increase from $u_n$ to $u_{n+1}$ , require
$2 n - 9 > 0,$
which gives $n>92n>\frac{9}{2}$ .
Since $n$ is an integer, this means $n≥5n\geq5$ . Therefore $u_{n+1}>u_n$ for $n≥5n\geq5$ . The sequence starts increasing from $u_5$ to $u_6$ .

Common Mistake

Ignoring integer values of n

After solving an inequality, convert the result to the appropriate integer values. Here, $n>92n>\frac{9}{2}$ means $n≥5n\geq5$ , not $n≥92n\geq\frac{9}{2}$ .

Using a recurrence

For a recursively defined sequence, whether it increases can depend on the current term rather than directly on $n$ .

Suppose

u_{n+1}=f(u_n).

Then

u_{n+1}-u_n=f(u_n)-u_n.

The sign of this expression tells you whether the next term is above or below the current one.

Example

Determining behaviour from a recurrence

A sequence satisfies $u_{n+1}=3u_n-4$ .

Compare consecutive terms by subtracting $u_n$ :
$u_{n+1}-u_n=2u_n-4=2(u_n-2).$
If $u_n>2$ , then $2(u_n-2)>0$ , so $u_{n+1}>u_n$ . The sequence increases at that step.
If $u_n<2$ , then $2(u_n-2)<0$ , so $u_{n+1}<u_n$ . If $u_n=2$ , then $u_{n+1}=2$ , so the next term is unchanged.

Periodic sequences

Definition

Periodic sequence

A sequence is periodic if its terms repeat after a fixed positive integer number of positions. In symbols, there is a positive integer $p$ such that

u_{n+p}=u_n

for every relevant $n$ . The number $p$ is called a period.

The least period, sometimes called the fundamental period, is the smallest positive value of $p$ for which the repetition occurs.

For example, the sequence

…1,\ 3,\ 2,\ 1,\ 3,\ 2,\ \ldots

has least period 3. It also repeats after 6 or 9 positions, but these are not its least period.

Example

Finding the least period

Consider the sequence defined by $un=cos⁡(πn2)u_n=\cos\left(\frac{\pi n}{2}\right)$ .

Calculate enough terms to reveal a possible repeating block:
$u1=0,u2=−1,u3=0,u4=1,u5=0.u_1=0,\quad u_2=-1,\quad u_3=0,\quad u_4=1,\quad u_5=0.$
This suggests the repeating block $0, - 1, 0, 1$ .
Use the periodicity of cosine:
$un+4=cos⁡(π(n+4)2)=cos⁡(πn2+2π)=cos⁡(πn2)=un.\begin{aligned} u_{n+4} &=\cos\left(\frac{\pi(n+4)}{2}\right)\\ &=\cos\left(\frac{\pi n}{2}+2\pi\right)\\ &=\cos\left(\frac{\pi n}{2}\right)=u_n. \end{aligned}$
Therefore 4 is a period.
Smaller positive integers do not reproduce every term: for instance, $u_1=0$ but $u_2=-1$ , and $u_2=-1$ but $u_4=1$ . Hence the least period is 4.

Common Mistake

Giving any period instead of the least period

If a sequence has least period 3, then 6 is also a period. However, when asked to identify “the period”, questions usually expect the smallest positive period.

Periodic does not mean constant

A constant sequence satisfies $u_{n+1}=u_n$ , so it is periodic with least period 1. However, most periodic sequences move repeatedly through a block of different values.

Key Idea

Classifying the behaviour

Increasing and decreasing describe how consecutive terms compare.
Periodic describes repetition: $u_{n+p}=u_n$ .
A sequence may be neither increasing, decreasing nor periodic.
A non-constant periodic sequence cannot be strictly increasing or strictly decreasing.

Exam technique

In the exam

Write down $u_{n+1}$ carefully by replacing every occurrence of $n$ with $n + 1$ .
Simplify $u_{n+1}-u_n$ and determine its sign over the stated integer domain.
For a periodic sequence, identify a candidate repeating block and then verify $u_{n+p}=u_n$ ; check that no smaller positive period works.
State the conclusion explicitly, including any restriction such as “increasing for $n≥5n\geq5$ ”.

Self review

Check yourself

How would you prove that $u_n=3n^2-n$ is increasing for positive integer values of $n$ ?
For which values of $n$ is the sequence $u_n=n^2-14n$ increasing?
What is the least period of the sequence $2,−1,4,2,−1,4,…2,-1,4,2,-1,4,\ldots$ ?

What you'll learn

Sequence notation

Increasing sequences

Decreasing sequences

Strict and non-strict behaviour

When the behaviour depends on the term number

Using a recurrence

Periodic sequences

Periodic does not mean constant

Review 1.4.6 Increasing, decreasing and periodic sequences (A-level only) by teaching Genie

1.4.6 Increasing, decreasing and periodic sequences (A-level only)

What you'll learn

Sequence notation

Increasing sequences

Decreasing sequences

Strict and non-strict behaviour

When the behaviour depends on the term number

Using a recurrence

Periodic sequences

Periodic does not mean constant

Review 1.4.6 Increasing, decreasing and periodic sequences (A-level only) by teaching Genie

1.4.6 Increasing, decreasing and periodic sequences (A-level only)

What you'll learn

Sequence notation

Increasing sequences

Decreasing sequences

Strict and non-strict behaviour

When the behaviour depends on the term number

Using a recurrence

Periodic sequences

Periodic does not mean constant

Review 1.4.6 Increasing, decreasing and periodic sequences (A-level only) by teaching Genie

1.4.6 Increasing, decreasing and periodic sequences (A-level only) Revision Guide

1.4.6 Increasing, decreasing and periodic sequences (A-level only)

What you'll learn

Sequence notation

Increasing sequences

Decreasing sequences

Strict and non-strict behaviour

When the behaviour depends on the term number

Using a recurrence

Periodic sequences

Periodic does not mean constant

Review 1.4.6 Increasing, decreasing and periodic sequences (A-level only) by teaching Genie

1.4.6 Increasing, decreasing and periodic sequences (A-level only) Revision Guide