Terminology of Sequences

Learning Outcomes

Find the formula for the general term of a sequence

To work with this new topic, we need some new terms and definitions. First, an infinite sequence is an ordered list of numbers of the form

a_{1}, a_{2}, a_{3}, \dots, a_{n}, \dots .

${a}_{1},{a}_{2},{a}_{3}\text{,}\ldots,{a}_{n}\text{,}\ldots\text{.}$

Each of the numbers in the sequence is called a term. The symbol $n$ is called the index variable for the sequence. We use the notation

{a_{n}}_{n = 1}^{\infty}, or simply {a_{n}}

${\left\{{a}_{n}\right\}}_{n=1}^{\infty },\text{or simply}\left\{{a}_{n}\right\}$ ,

to denote this sequence. A similar notation is used for sets, but a sequence is an ordered list, whereas a set is not ordered. Because a particular number ${a}_{n}$ exists for each positive integer $n$ , we can also define a sequence as a function whose domain is the set of positive integers.

Let’s consider the infinite, ordered list

2, 4, 8, 16, 32, \dots

$2,4,8,16,32\text{,}\ldots$ .

This is a sequence in which the first, second, and third terms are given by ${a}_{1}=2$ , ${a}_{2}=4$ , and ${a}_{3}=8$ . You can probably see that the terms in this sequence have the following pattern:

a_{1} = 2^{1}, a_{2} = 2^{2}, a_{3} = 2^{3}, a_{4} = 2^{4}, and a_{5} = 2^{5}

${a}_{1}={2}^{1},{a}_{2}={2}^{2},{a}_{3}={2}^{3},{a}_{4}={2}^{4},\text{and }{a}_{5}={2}^{5}$ .

Assuming this pattern continues, we can write the $n\text{th}$ term in the sequence by the explicit formula ${a}_{n}={2}^{n}$ . Using this notation, we can write this sequence as

{2^{n}}_{n = 1}^{\infty} or {2^{n}}

${\left\{{2}^{n}\right\}}_{n=1}^{\infty }\text{or}\left\{{2}^{n}\right\}$ .

Alternatively, we can describe this sequence in a different way. Since each term is twice the previous term, this sequence can be defined recursively by expressing the $n\text{th}$ term ${a}_{n}$ in terms of the previous term ${a}_{n - 1}$ . In particular, we can define this sequence as the sequence $\left\{{a}_{n}\right\}$ where ${a}_{1}=2$ and for all $n\ge 2$ , each term ${a}_{n}$ is defined by the ${a}_{n}=2{a}_{n - 1}$ .

Definition

An infinite sequence $\left\{{a}_{n}\right\}$ is an ordered list of numbers of the form

a_{1}, a_{2}, \dots, a_{n}, \dots .

${a}_{1},{a}_{2}\text{,}\ldots,{a}_{n}\text{,}\ldots\text{.}$

The subscript $n$ is called the index variable of the sequence. Each number ${a}_{n}$ is a of the sequence. Sometimes sequences are defined by explicit formulas, in which case ${a}_{n}=f\left(n\right)$ for some function $f\left(n\right)$ defined over the positive integers. In other cases, sequences are defined by using a recurrence relation. In a recurrence relation, one term (or more) of the sequence is given explicitly, and subsequent terms are defined in terms of earlier terms in the sequence.

Note that the index does not have to start at $n=1$ but could start with other integers. For example, a sequence given by the explicit formula ${a}_{n}=f\left(n\right)$ could start at $n=0$ , in which case the sequence would be

a_{0}, a_{1}, a_{2}, \dots .

${a}_{0},{a}_{1},{a}_{2}\text{,}\ldots\text{.}$

Similarly, for a sequence defined by a recurrence relation, the term ${a}_{0}$ may be given explicitly, and the terms ${a}_{n}$ for $n\ge 1$ may be defined in terms of ${a}_{n - 1}$ . Since a sequence $\left\{{a}_{n}\right\}$ has exactly one value for each positive integer $n$ , it can be described as a function whose domain is the set of positive integers. As a result, it makes sense to discuss the graph of a sequence. The graph of a sequence $\left\{{a}_{n}\right\}$ consists of all points $\left(n,{a}_{n}\right)$ for all positive integers $n$ . Figure 1 shows the graph of $\left\{{2}^{n}\right\}$ .

A graph in quadrant one containing the following points: (1, 2), (2, 4), (3, 8), (4, 16).

Two types of sequences occur often and are given special names: arithmetic sequences and geometric sequences. In an arithmetic sequence, the difference between every pair of consecutive terms is the same. For example, consider the sequence

3, 7, 11, 15, 19, \dots .

$3,7,11,15,19\text{,}\ldots\text{.}$

You can see that the difference between every consecutive pair of terms is $4$ . Assuming that this pattern continues, this sequence is an arithmetic sequence. It can be described by using the recurrence relation

{\begin{matrix} a_{1} = 3 \\ a_{n} = a_{n - 1} + 4 for n \geq 2 \end{matrix}

$\bigg\{ \begin{array}{c} {a}_{1}=3\hfill \\ {a}_{n}={a}_{n-1}+4 \text{ for } n\ge 2 \end{array}$

Note that

\begin{array}{l} a_{2} = 3 + 4 \\ a_{3} = 3 + 4 + 4 = 3 + 2 \cdot 4 \\ a_{4} = 3 + 4 + 4 + 4 = 3 + 3 \cdot 4. \end{array}

$\begin{array}{l}{a}_{2}=3+4\\ {a}_{3}=3+4+4=3+2\cdot 4\\ {a}_{4}=3+4+4+4=3+3\cdot 4.\end{array}$

Thus the sequence can also be described using the explicit formula

\begin{array}{cc} a_{n} & = 3 + 4 (n - 1) \\ = 4 n - 1. \end{array}

$\begin{array}{cc}\hfill {a}_{n}& =3+4\left(n - 1\right)\hfill \\ & =4n - 1.\hfill \end{array}$

In general, an arithmetic sequence is any sequence of the form ${a}_{n}=cn+b$ .

In a geometric sequence, the ratio of every pair of consecutive terms is the same. For example, consider the sequence

2, - \frac{2}{3}, \frac{2}{9}, - \frac{2}{27}, \frac{2}{81}, \dots .

$2,-\frac{2}{3},\frac{2}{9},-\frac{2}{27},\frac{2}{81}\text{,}\ldots\text{.}$

We see that the ratio of any term to the preceding term is $-\frac{1}{3}$ . Assuming this pattern continues, this sequence is a geometric sequence. It can be defined recursively as

\begin{matrix} a_{1} = 2 \\ a_{n} = - \frac{1}{3} \cdot a_{n - 1} for n \geq 2. \end{matrix}

$\begin{array}{}\\ {a}_{1}=2\hfill \\ {a}_{n}=-\frac{1}{3}\cdot {a}_{n - 1}\text{ for }n\ge 2.\hfill \end{array}$

Alternatively, since

\begin{matrix} a_{2} = - \frac{1}{3} \cdot 2 \\ a_{3} = (- \frac{1}{3}) (- \frac{1}{3}) (2) = {(- \frac{1}{3})}^{2} \cdot 2 \\ a_{4} = (- \frac{1}{3}) (- \frac{1}{3}) (- \frac{1}{3}) (2) = {(- \frac{1}{3})}^{3} \cdot 2, \end{matrix}

$\begin{array}{}\\ \\ {a}_{2}=-\frac{1}{3}\cdot 2\hfill \\ {a}_{3}=\left(-\frac{1}{3}\right)\left(-\frac{1}{3}\right)\left(2\right)={\left(-\frac{1}{3}\right)}^{2}\cdot 2\hfill \\ {a}_{4}=\left(-\frac{1}{3}\right)\left(-\frac{1}{3}\right)\left(-\frac{1}{3}\right)\left(2\right)={\left(-\frac{1}{3}\right)}^{3}\cdot 2,\hfill \end{array}$

we see that the sequence can be described by using the explicit formula

a_{n} = 2 {(- \frac{1}{3})}^{n - 1}

${a}_{n}=2{\left(-\frac{1}{3}\right)}^{n - 1}$ .

The sequence $\left\{{2}^{n}\right\}$ that we discussed earlier is a geometric sequence, where the ratio of any term to the previous term is $2$ . In general, a geometric sequence is any sequence of the form ${a}_{n}=c{r}^{n}$ .

Example: Finding Explicit Formulas

For each of the following sequences, find an explicit formula for the $n\text{th}$ term of the sequence.

$-\frac{1}{2},\frac{2}{3},-\frac{3}{4},\frac{4}{5},-\frac{5}{6}\text{,}\ldots$
$\frac{3}{4},\frac{9}{7},\frac{27}{10},\frac{81}{13},\frac{243}{16}\text{,}\ldots$

Show Solution

First, note that the sequence is alternating from negative to positive. The odd terms in the sequence are negative, and the even terms are positive. Therefore, the $n\text{th}$ term includes a factor of ${\left(-1\right)}^{n}$ . Next, consider the sequence of numerators $\left\{1,2,3\text{,}\ldots\right\}$ and the sequence of denominators $\left\{2,3,4\text{,}\ldots\right\}$ . We can see that both of these sequences are arithmetic sequences. The $n\text{th}$ term in the sequence of numerators is $n$ , and the $n\text{th}$ term in the sequence of denominators is $n+1$ . Therefore, the sequence can be described by the explicit formula

${a}_{n}=\frac{{\left(-1\right)}^{n}n}{n+1}$ .
The sequence of numerators $3,9,27,81,243\text{,}\ldots$ is a geometric sequence. The numerator of the $n\text{th}$ term is ${3}^{n}$ The sequence of denominators $4,7,10,13,16\text{,}\ldots$ is an arithmetic sequence. The denominator of the $n\text{th}$ term is $4+3\left(n - 1\right)=3n+1$ . Therefore, we can describe the sequence by the explicit formula ${a}_{n}=\frac{{3}^{n}}{3n+1}$ .

try it

Find an explicit formula for the $n\text{th}$ term of the sequence $\left\{\frac{1}{5},-\frac{1}{7},\frac{1}{9},-\frac{1}{11}\text{,}\ldots\right\}$ .

Hint

Show Solution

${a}_{n}=\frac{{\left(-1\right)}^{n+1}}{3+2n}$

Watch the following video to see the worked solution to the above Try IT.

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Example: Defined by Recurrence Relations

For each of the following recursively defined sequences, find an explicit formula for the sequence.

${a}_{1}=2$ , ${a}_{n}=-3{a}_{n - 1}$ for $n\ge 2$
${a}_{1}=\frac{1}{2}$ , ${a}_{n}={a}_{n - 1}+{\left(\frac{1}{2}\right)}^{n}$ for $n\ge 2$

Show Solution

Writing out the first few terms, we have

$\begin{array}{c}{a}_{1}=2\hfill \\ {a}_{2}=-3{a}_{1}=-3\left(2\right)\hfill \\ {a}_{3}=-3{a}_{2}={\left(-3\right)}^{2}2\hfill \\ {a}_{4}=-3{a}_{3}={\left(-3\right)}^{3}2.\hfill \end{array}$

In general,

${a}_{n}=2{\left(-3\right)}^{n - 1}$ .
Write out the first few terms:

$\begin{array}{}\\ \\ {a}_{1}=\frac{1}{2}\hfill \\ {a}_{2}={a}_{1}+{\left(\frac{1}{2}\right)}^{2}=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}\hfill \\ {a}_{3}={a}_{2}+{\left(\frac{1}{2}\right)}^{3}=\frac{3}{4}+\frac{1}{8}=\frac{7}{8}\hfill \\ {a}_{4}={a}_{3}+{\left(\frac{1}{2}\right)}^{4}=\frac{7}{8}+\frac{1}{16}=\frac{15}{16}.\hfill \end{array}$

From this pattern, we derive the explicit formula

${a}_{n}=\frac{{2}^{n}-1}{{2}^{n}}=1-\frac{1}{{2}^{n}}$ .

try it

Find an explicit formula for the sequence defined recursively such that ${a}_{1}=-4$ and ${a}_{n}={a}_{n - 1}+6$ .

Hint

Show Solution

${a}_{n}=6n - 10$

Watch the following video to see the worked solution to the above Try IT.

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Module 5: Sequences and Series