9.3 Series

Series

In a sequence, the sum of the directed terms is called a series.

For example, If \(1,4,7,10,13,16, \ldots\) is a sequence, then its sum i.e., \(1+4+7+10+13+16+\ldots\) is a series.
In general, if \(T_1, T_2, T_3, \ldots, T_n, \ldots\) denote a sequence, then the symbolic expression \(T_1+T_2+T_3+\ldots+T_n+\ldots\) is called a series associated with the given sequence.
Each member of the series is called its term.
In a series \(T_1+T_2+T_3+\ldots+T_r+\ldots\), the sum of first \(n\) terms is denoted by \(S_n\). Thus,
\(
S_n=T_1+T_2+T_3+\ldots+T_n=\sum_{r=1}^n T_r=\sum T_n
\)
If \(S_n\) denotes the sum of \(n\) terms of a sequence.
Then,
\(
\begin{aligned}
S_n-S_{n-1}=\left(T_1+T_2\right. & \left.+T_3+\ldots+T_n\right) -\left(T_1+T_2+\ldots+T_{n-1}\right)=T_n
\end{aligned}
\)
Thus,
\(
T_n=S_n-S_{n-1}
\)

Illustration

Let \(a_1, a_2, a_3, \ldots, a_n\), be a given sequence. Then, the expression
\(
a_1+a_2+a_3+\ldots+a_n+\ldots
\)
is called the series associated with the given sequence. The series is finite or infinite according as the given sequence is finite or infinite. Series are often represented in compact form, called sigma notation, using the Greek letter \(\Sigma\) (sigma) as means of indicating the summation involved. Thus, the series \(a_1+a_2+a_3+\ldots+a_n\) is abbreviated
as \(\sum_{k=1}^n a_k\).

Remark

When the series is used, it refers to the indicated sum not to the sum itself. For example, \(1+3+5+7\) is a finite series with four terms. When we use the phrase “sum of a series,” we will mean the number that results from adding the terms, the sum of the series is 16. 

Progression

If the terms of a sequence can be described by an explicit formula, then the sequence is called a progression.

                                 Or
A sequence is said to be progression, if its terms increases (respectively decreases) numerically.
For example, The following sequences are progression :
(i) \(1,3,5,7, \ldots\)
(ii) \(\frac{1}{2}, \frac{1}{6}, \frac{1}{18}, \frac{1}{54}, \ldots\)
(iii) \(1,-\frac{1}{3}, \frac{1}{9},-\frac{1}{27}, \ldots\)
(iv) \(1,8,27,256, \ldots\)
(v) \(8,-4,2,-1, \frac{1}{2}, \ldots\)
The sequences (iii) and (v) are progressions, because
\(
\begin{aligned}
& |1|>\left|-\frac{1}{3}\right|>\left|\frac{1}{9}\right|>\left|-\frac{1}{27}\right|>\ldots \\
& \text { i.e. } \quad 1>\frac{1}{3}>\frac{1}{9}>\frac{1}{27}>\ldots \\
&
\end{aligned}
\)
and \(\quad|8|>|-4|>|2|>|-1|>\left|\frac{1}{2}\right|>\ldots\)
i.e. \(8>4>2>1>\frac{1}{2}>\ldots\)

Remark

All the definitions and formulae are valid for complex numbers in the theory of progressions but it should be assumed (if not otherwise stated) that the terms of the progressions are real numbers.

Example 1: Let the sequence \(a_n\) be defined as follows:
\(
a_1=1, a_n=a_{n-1}+2 \text { for } n \geq 2 \text {. }
\)
Find first five terms and write corresponding series.

Solution: We have
\(
\begin{aligned}
& a_1=1, a_2=a_1+2=1+2=3, a_3=a_2+2=3+2=5, \\
& a_4=a_3+2=5+2=7, a_5=a_4+2=7+2=9 .
\end{aligned}
\)
Hence, the first five terms of the sequence are \(1,3,5,7\) and 9 . The corresponding series is \(1+3+5+7+9+\ldots\)

Types of Series

There are two types of series

Finite Series

A series having finite number of terms is called a finite series.
For example,
(i) \(3+5+7+9+\ldots+21\)
(ii) \(2+6+18+54+\ldots+4374\)

Infinite Series

A series having an infinite number of terms is called an infinite series.
For example,
(i) \(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\ldots\)
(ii) \(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots\)

Example 2: If the sum of \(n\) terms of a series is \(2 n^2+5 n\) for all values of \(n\), find its 7 th term.

Solution: 

\(
\begin{aligned}
& \text { Given, } \quad S_n=2 n^2+5 n \\
& \Rightarrow \quad S_{n-1}=2(n-1)^2+5(n-1)=2 n^2+n-3
\end{aligned}
\)
\(
\begin{array}{ll}
\therefore \quad T_n=S_n-S_{n-1}=\left(2 n^2+5 n\right)-\left(2 n^2+n-3\right)=4 n+3 \\
\text { Hence, } T_7=4 \times 7+3=31
\end{array}
\)

Example 3: (i) Write \(\sum_{r=1}^n\left(r^2+2\right)\) in expanded form.
(ii) Write the series \(\frac{1}{3}+\frac{2}{4}+\frac{3}{5}+\frac{4}{6}+\ldots+\frac{n}{n+2}\) in sigma form.

Solution: (i) On putting \(r=1,2,3,4, \ldots, n\) in \(\left(r^2+2\right)\),
we get \(3,6,11,18, \ldots,\left(n^2+2\right)\)
Hence, \(\sum_{r=1}^n\left(r^2+2\right)=3+6+11+18+\ldots+\left(n^2+2\right)\)
(ii) The \(r\) th term of series \(=\frac{r}{r+2}\).
Hence, the given series can be written as
\(
\frac{1}{3}+\frac{2}{4}+\frac{3}{5}+\frac{4}{6}+\ldots+\frac{n}{n+2}=\sum_{r=1}^n\left(\frac{r}{r+2}\right)
\)