9.2 Sequences

Sequence

A sequence can be regarded as a function whose domain is the set of natural numbers or some subset of it. Sometimes, we use the functional notation \(a(n)\) for \(a_n\).

Let us consider the following examples:
Assume that there is a generation gap of 30 years, we are asked to find the number of ancestors, i.e., parents, grandparents, great grandparents, etc. that a person might have over 300 years.

Here, the total number of generations \(=\frac{300}{30}=10\)
The number of person’s ancestors for the first, second, third, …, tenth generations are \(2,4,8,16,32, \ldots, 1024\). These numbers form what we call a sequence.

Consider the successive quotients that we obtain in the division of 10 by 3 at different steps of division. In this process we get \(3,3.3,3.33,3.333, \ldots\) and so on. These quotients also form a sequence. The various numbers occurring in a sequence are called its terms. We denote the terms of a sequence by \(a_{1}, a_{2}, a_{3}, \ldots, a_{n}, \ldots\), etc., the subscripts denote the position of the term. The \(n^{\text {th }}\) term is the number at the \(n^{\text {th }}\) position of the sequence and is denoted by \(a_n\). The \(n^{\text {th }}\) term is also called the general term of the sequence.

Thus, the terms of the sequence of person’s ancestors mentioned above are:
\(
a_1=2, a_2=4, a_3=8, \ldots, a_{10}=1024 .
\)
Similarly, in the example of successive quotients
\(
a_1=3, a_2=3.3, a_3=3.33, \ldots, a_6=3.33333 \text {, etc. }
\)

\(Or\)
A sequence is a function of natural numbers \((N)\) with codomain is the set of real numbers (R) [complex numbers (C)]. If range is subset of real numbers (complex numbers), it is called a real sequence (complex sequence).
\(Or\)

A mapping \(f: N \rightarrow C\), then \(f(n)=t_n, n \in N\) is called a sequence to be denoted it by \(\{f(1), f(2), f(3), \ldots\}=\left\{t_1, t_2, t_3, \ldots\right\}=\left\{t_n\right\}\).
The \(n\)th term of a sequence is denoted by \(T_n, t_n, a_n, a(n), u_n\), etc.

Remark

The sequence \(a_1, a_2, a_3, \ldots\) is generally written as \(\left\{a_n\right\}\).
For example ,
(i) \(1,3,5,7, \ldots\) is a sequence, because each term (except first) is obtained by adding 2 to the previous term and \(T_n=2 n-1, n \in N\).
Or
If \(T_1=1, T_{n+1}=T_{n+2}, n \geq 1\)
(ii) \(1,2,3,5,8,13, \ldots\) is a sequence, because each term (except first two) is obtained by taking the sum of preceding two terms.
Or
If \(T_1=1, T_2=2, T_{n+2}=T_n+T_{n+1}, n \geq 1\)
(iii) \(2,3,5,7,11,13,17,19, \ldots\) is a sequence.
Here, we cannot express \(T_n, n \in N\) by an algebraic formula.

Recursive Formula

A formula to determine the other terms of the sequence in terms of its preceding terms is known as recursive formula.
For example,
If \(\quad T_1=1\) and \(T_{n+1}=6 T_n, n \in N\).
Then,
\(
\begin{aligned}
& T_2=6 T_1=6 \cdot 1=6 \\
& T_3=6 T_2=6 \cdot 6=36 \\
& T_4=6 T_3=6 \cdot 36=216 \ldots
\end{aligned}
\)
Then, sequence is \(1,6,36,216, \ldots\)

Types of Sequences

There are two types of sequences

Finite Sequence

A sequence is said to be finite sequence, if it has finite number of terms. A finite sequence is described by \(a_1, a_2, a_3, \ldots, a_n\) or \(T_1, T_2, T_3, \ldots, T_n\), where \(n \in N\).
For example
(i) \(3,5,7,9, \ldots, 37\)
(ii) \(2,6,18,54, \ldots, 4374\)

Infinite Sequence

A sequence is said to be an infinite sequence, if it has infinite number of terms. An infinite sequence is described by \(a_1, a_2, a_3, \ldots\) or \(T_1, T_2, T_3, \ldots\)
For example,
(i) \(1, \frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \ldots\)
(ii) \(1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \ldots\)

Illustration

Often, it is possible to express the rule, which yields the various terms of a sequence in terms of algebraic formula. Consider for instance, the sequence of even natural numbers \(2,4,6, \ldots\)
Here
\(
\begin{array}{ll}
a_1=2=2 \times 1 & a_2=4=2 \times 2 \\
a_3=6=2 \times 3 & a_4=8=2 \times 4
\end{array}
\)
\(\ldots, \quad \quad \ldots \quad \quad \ldots \quad \quad \ldots \quad \quad \ldots \quad \quad\ldots\)
\(\ldots, \quad \quad \ldots \quad \quad \ldots \quad \quad \ldots \quad \quad \ldots \quad \quad\ldots\)
\(a_{23}=46=2 \times 23, a_{24}=48=2 \times 24\), and so on.
In fact, we see that the \(n^{\text {th }}\) term of this sequence can be written as \(a_n=2 n\), where \(n\) is a natural number. Similarly, in the sequence of odd natural numbers \(1,3,5, \ldots\), the \(n^{\text {th }}\) term is given by the formula, \(a_n=2 n-1\), where \(n\) is a natural number.

In some cases, an arrangement of numbers such as \(1,1,2,3,5,8, \ldots\) has no visible pattern, but the sequence is generated by the recurrence relation given by
\(
\begin{aligned}
& a_1=a_2=1 \\
& a_3=a_1+a_2 \\
& a_n=a_{n-2}+a_{n-1}, n>2
\end{aligned}
\)
This sequence is called Fibonacci sequence.
In the sequence of primes \(2,3,5,7, \ldots\), we find that there is no formula for the \(n^{\text {th }}\) prime. Such sequence can only be described by verbal description.

In every sequence, we should not expect that its terms will necessarily be given by a specific formula. However, we expect a theoretical scheme or a rule for generating the terms \(a_1, a_2, a_3, \ldots, a_n, \ldots\) in succession.

Example 1: If \(f: N \rightarrow R\), where \(f(n)=a_n=\frac{n}{(2 n+1)^2}\), write the sequence in ordered pair form.

Solution: Here, \(a_n=\frac{n}{(2 n+1)^2}\)
On putting \(n=1,2,3,4, \ldots\) successively, we get
\(
\begin{aligned}
& a_1=\frac{1}{(2 \cdot 1+1)^2}=\frac{1}{9}, \quad a_2=\frac{2}{(2 \cdot 2+1)^2}=\frac{2}{25} \\
& a_3=\frac{3}{(2 \cdot 3+1)^2}=\frac{3}{49}, \quad a_4=\frac{4}{(2 \cdot 4+1)^2}=\frac{4}{81} \\
& \vdots \quad \vdots \\
&
\end{aligned}
\)

Hence, we obtain the sequence \(\frac{1}{9}, \frac{2}{25}, \frac{3}{49}, \frac{4}{81}, \ldots\)
Now, the sequence in ordered pair form is
\(
\left\{\left(1, \frac{1}{9}\right),\left(2, \frac{2}{25}\right),\left(3, \frac{3}{49}\right),\left(4, \frac{4}{81}\right), \ldots\right\}
\)

Example 2: The Fibonacci sequence is defined by \(a_1=1=a_2, a_n=a_{n-1}+a_{n-2}, n>2\). Find \(\frac{a_{n+1}}{a_n}\) for \(n=1,2,3,4,5\).

Solution:

\(
\begin{aligned}
& a_1=1=a_2 \\
& a_3=a_2+a_1=1+1=2, \\
& a_4=a_3+a_2=2+1=3 \\
& a_5=a_4+a_3=3+2=5
\end{aligned}
\)
and \(\quad a_6=a_5+a_4=5+3=8\)
\(\therefore \quad \frac{a_2}{a_1}=1, \frac{a_3}{a_2}=\frac{2}{1}=2, \frac{a_4}{a_3}=\frac{3}{2}, \frac{a_5}{a_4}=\frac{5}{3}\) and \(\frac{a_6}{a_5}=\frac{8}{5}\)

Example 3: Write the first three terms in each of the following sequences defined by the following:
(i) \(a_n=2 n+5\),
(ii) \(a_n=\frac{n-3}{4}\).

Solution: (i) Here \(a_n=2 n+5\)
Substituting \(n=1,2,3\), we get
\(
a_1=2(1)+5=7, a_2=9, a_3=11
\)
Therefore, the required terms are 7,9 and 11 .
(ii) Here \(a_n=\frac{n-3}{4}\). Thus, \(a_1=\frac{1-3}{4}=-\frac{1}{2}, a_2=-\frac{1}{4}, a_3=0\)
Hence, the first three terms are \(-\frac{1}{2},-\frac{1}{4}\) and 0.

Example 4: What is the \(20^{\text {th }}\) term of the sequence defined by
\(
a_n=(n-1)(2-n)(3+n) \text { ? }
\)

Solution: Putting \(n=20\), we obtain
\(
\begin{aligned}
a_{20} & =(20-1)(2-20)(3+20) \\
& =19 \times(-18) \times(23)=-7866 .
\end{aligned}
\)