Sequences (Senior 5 Mathematics): free lessons | Ganzaa

Sequences

Lessons

Senior 5 · Mathematics

Sequences

4 lessons

Lesson 1 of 4

Sequences: rules, monotonicity and bounds

25 min

A list with a rule

Write down $\frac{1}{4}, \frac{3}{5}, \frac{5}{6}, 1, \frac{9}{8}, \dots$ . The numerators go up by 2 and the denominators by 1, so the 10th number is $\frac{19}{13}$ and the 100th is $\frac{199}{103}$ . A list like this, with a rule that gives the number in every position, is a sequence. Sequences describe anything that happens step by step: a balance at the end of each month, the height of a crop each week, the next approximation that a calculator produces for $\sqrt{2}$ . This unit studies the most useful ones and asks where a sequence is heading in the long run.

Definition

Sequence (of real numbers): a function $u$ whose domain is the set of positive integers $\mathbb{N}^* = \{1, 2, 3, \dots\}$ (sometimes $\mathbb{N}$ , starting at 0). The value $u (n)$ is written $u_{n}$ and called the $n$ th term; the whole sequence is written $(u_{n})$ . A finite sequence stops after a last term.

General term (explicit rule): a formula for $u_{n}$ in terms of $n$ , such as $u_n = \frac{2n - 1}{n + 3}$ .

Recurrence relation (recursive rule): the first term together with a formula for each term in terms of the one before, such as $u_{1} = 2$ , $u_{n+1} = 3u_n - 2$ .

Example: the even numbers $2, 4, 6, \dots$ have general term $u_{n} = 2 n$ , and also the recurrence $u_{1} = 2$ , $u_{n+1} = u_n + 2$ .

Worked example 1: from a recurrence to a formula

$u_{1} = 2$ and $u_{n+1} = 3u_n - 2$ . Find $u_{2}$ to $u_{5}$ , guess a general term and check it.

Solution $u_{2} = 3 (2) - 2 = 4$ , $u_{3} = 10$ , $u_{4} = 28$ , $u_{5} = 82$ . Each term is one more than $1, 3, 9, 27, 81$ , the powers of 3, so guess $u_n = 3^{n-1} + 1$ .

Check: $u_{1} = 3^{0} + 1 = 2$ ✓, and if $u_n = 3^{n-1} + 1$ then $3 u_{n} - 2 = 3^{n} + 3 - 2 = 3^{n} + 1$ , which is the formula for $u_{n+1}$ . ✓ (This is a proof by induction, which you will meet in S6.)

Increasing, decreasing, bounded

To decide whether the terms go up or down, compare each term with the next for every $n$ . Two tests are used:

Difference test: study the sign of $u_{n+1} - u_n$ . It suits sums and fractions.
Ratio test (all terms positive): compare $\frac{u_{n+1}}{u_n}$ with 1. It suits powers and products such as $2^{n}$ or $n!$ .

Figure 1.

u_n = \frac{2n - 1}{n + 3}

: each term is higher than the one before (increasing), and every term lies between

\frac{1}{4}

and 2 (bounded). A sequence is drawn as separate dots, not a curve:

n

takes only whole values.

Definition

Increasing sequence: $u_{n+1} \ge u_n$ for every $n$ (strictly increasing if $u_{n+1} \gt u_n$ ). Decreasing sequence: $u_{n+1} \le u_n$ for every $n$ . A monotone sequence is one that is increasing or decreasing.

Bounded above: there is a number $M$ with $u_n \le M$ for every $n$ . Bounded below: there is $m$ with $u_n \ge m$ for every $n$ . Bounded: both.

Example: $u_n = \frac{1}{n}$ is strictly decreasing and bounded ( $0 \lt u_n \le 1$ ); $u_{n} = (- 1)^{n}$ is bounded but not monotone.

Worked example 2: a difference test

Show that $u_n = \frac{2n - 1}{n + 3}$ is strictly increasing and bounded.

Solution

u_{n+1} - u_n = \frac{2n + 1}{n + 4} - \frac{2n - 1}{n + 3} = \frac{(2n + 1)(n + 3) - (2n - 1)(n + 4)}{(n + 3)(n + 4)} = \frac{(2n^2 + 7n + 3) - (2n^2 + 7n - 4)}{(n + 3)(n + 4)} = \frac{7}{(n + 3)(n + 4)}.

This is positive for every $n \ge 1$ , so $(u_{n})$ is strictly increasing, and its smallest term is $u_1 = \frac{1}{4}$ . Dividing, $u_n = 2 - \frac{7}{n + 3} \lt 2$ . So $\frac{1}{4} \le u_n \lt 2$ for every $n$ : the sequence is bounded.

Worked example 3: a ratio test

Is $v_n = \frac{n^2}{2^n}$ monotone?

Solution All terms are positive, and

\frac{v_{n+1}}{v_n} = \frac{(n + 1)^2}{2^{n+1}} \cdot \frac{2^n}{n^2} = \frac{(n + 1)^2}{2n^2}.

This is less than 1 when $(n + 1)^2 \lt 2n^2$ , that is $n + 1 \lt \sqrt{2}\,n$ , or $n \gt \frac{1}{\sqrt{2} - 1} \approx 2.41$ . So $v_{n+1} \lt v_n$ for $n \ge 3$ , but $v_{n+1} \gt v_n$ for $n = 1, 2$ : $v_1 = \frac{1}{2}$ , $v_{2} = 1$ , $v_3 = \frac{9}{8}$ , $v_{4} = 1$ . The sequence is not monotone; it is decreasing from $n = 3$ on (Figure 2).

Figure 2.

v_n = \frac{n^2}{2^n}

rises for three terms and then falls, so it is not monotone; it is decreasing only from

n = 3

on.

Warning

A few terms prove nothing. $v_1 \lt v_2 \lt v_3$ in Worked example 3, yet $(v_{n})$ is not increasing. "Increasing" is a statement about every $n$ , so it needs an algebraic argument such as the sign of $u_{n+1} - u_n$ . Also, the ratio test only works when all the terms are positive: $-1, -2, -4, \dots$ has ratio 2 but is decreasing.

Tip

Choose the test from the formula. Fractions of polynomials in $n$ : use the difference, and write $u_{n}$ as "number $+ \frac{\text{constant}}{\text{something}}$ " to find a bound. Powers and factorials: use the ratio, because $\frac{2^{n+1}}{2^n} = 2$ and $\frac{(n + 1)!}{n!} = n + 1$ simplify at once.

Khan Academy: Explicit and recursive definitions of sequences. Watch how one sequence can be given both by a formula for the nth term and by a starting term with a rule for the next one.

✦

Practice

The last two are harder.

Write the first four terms of $u_n = (-1)^n\,\frac{n + 1}{n}$ .
Find a general term for (a) $3, 7, 11, 15, \dots$ (b) $\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \dots$ (c) $1, -\frac{1}{2}, \frac{1}{3}, -\frac{1}{4}, \dots$ .
$u_{1} = 5$ and $u_{n+1} = u_n - 2n$ . Find $u_{2}$ , $u_{3}$ , $u_{4}$ and $u_{5}$ .
Show that $u_n = \frac{3n}{n + 2}$ is strictly increasing and that $u_n \lt 3$ for every $n$ .
Show that $w_n = \frac{5^n}{n!}$ satisfies $w_{n+1} \lt w_n$ for every $n \ge 5$ .
Is $z_{n} = (- 1)^{n}$ monotone? Is it bounded?
$u_{1} = 1$ and $u_{n+1} = 2u_n + 1$ . Find the first four terms and show that $u_{n} = 2^{n} - 1$ fits the recurrence.
For $u_n = \frac{n}{n^2 + 4}$ , find the sign of $u_{n+1} - u_n$ , and hence the largest term of the sequence.

Show answersHide answers

$-2, \frac{3}{2}, -\frac{4}{3}, \frac{5}{4}$ .
(a) $4 n - 1$ (b) $\frac{1}{2^n}$ (c) $\frac{(-1)^{n+1}}{n}$ .
$3, - 1, - 7, - 15$ .
$u_{n+1} - u_n = \frac{3(n + 1)(n + 2) - 3n(n + 3)}{(n + 2)(n + 3)} = \frac{6}{(n + 2)(n + 3)} \gt 0$ ; and $u_n = 3 - \frac{6}{n + 2} \lt 3$ .
$\frac{w_{n+1}}{w_n} = \frac{5}{n + 1} \lt 1$ when $n + 1 \gt 5$ , that is $n \ge 5$ . (At $n = 4$ the ratio is exactly 1: $w_5 = w_4 = \frac{625}{24}$ .)
Not monotone (it goes $-1, 1, -1, \dots$ ); bounded, since $-1 \le z_n \le 1$ .
$1, 3, 7, 15$ . If $u_{n} = 2^{n} - 1$ , then $2u_n + 1 = 2^{n+1} - 2 + 1 = 2^{n+1} - 1$ , and $u_{1} = 2 - 1 = 1$ .
$u_{n+1} - u_n = \frac{(n + 1)(n^2 + 4) - n\left((n + 1)^2 + 4\right)}{(\dots)(\dots)} = \frac{-n^2 - n + 4}{\big((n + 1)^2 + 4\big)(n^2 + 4)}$ : positive for $n = 1$ , negative for $n \ge 2$ . So $u_1 \lt u_2$ and then the terms fall: the largest term is $u_2 = \frac{1}{4}$ .

Sources and credits

Written by Ganzaa: the explanations, the graphs, the worked examples, the practice items and the quiz questions. The unit follows the topic list of REB's S5 Core Mathematics (2026–2027); no REB text is reproduced.
Video: Explicit and recursive definitions of sequences, Khan Academy, embedded from YouTube.

1A list with a rule

2Increasing, decreasing, bounded

A list with a rule

Increasing, decreasing, bounded