1.3 Series | Sequences and series

1.3 Series (EMCDV)

It is often important and valuable to determine the sum of the terms of an arithmetic or geometric sequence. The sum of any sequence of numbers is called a series.

Finite series

We use the symbol \({S}_{n}\) for the sum of the first \(n\) terms of a sequence \(\left\{{T}_{1}; {T}_{2}; {T}_{3}; \ldots;{T}_{n}\right\}\):

\[{S}_{n}={T}_{1}+{T}_{2}+{T}_{3}+\cdots + {T}_{n}\]

If we sum only a finite number of terms, we get a finite series.

For example, consider the following sequence of numbers

\[1; 4; 9; 16; 25; 36; 49; \ldots\]

We can calculate the sum of the first four terms:

\[{S}_{4}=1+4+9+16=30\]

This is an example of a finite series since we are only summing four terms.

Infinite series

If we sum infinitely many terms of a sequence, we get an infinite series:

\[{S}_{\infty }={T}_{1}+{T}_{2}+{T}_{3}+ \cdots\]

Sigma notation (EMCDW)

Sigma notation is a very useful and compact notation for writing the sum of a given number of terms of a sequence.

A sum may be written out using the summation symbol \(\sum\) (Sigma), which is the capital letter “S” in the Greek alphabet. It indicates that you must sum the expression to the right of the summation symbol:

For example,

\[\sum _{n=1}^{5}{2n} = 2 + 4 + 6 + 8 + 10 = 30\]

In general,

\[\sum _{i=m}^{n}{T}_{i}={T}_{m}+{T}_{m+1}+\cdots +{T}_{n-1}+{T}_{n}\]

where

\(i\) is the index of the sum;
\(m\) is the lower bound (or start index), shown below the summation symbol;
\(n\) is the upper bound (or end index), shown above the summation symbol;
\({T}_{i}\) is a term of a sequence;
the number of terms in the series \(= \text{end index} - \text{start index} + \text{1}\).

The index \(i\) increases from \(m\) to \(n\) by steps of \(\text{1}\).

Note that this is also sometimes written as:

\[\sum _{i=m}^{n}{a}_{i}={a}_{m}+{a}_{m+1}+\cdots +{a}_{n-1}+{a}_{n}\]

When we write out all the terms in a sum, it is referred to as the expanded form.

If we are summing from \(i=1\) (which implies summing from the first term in a sequence), then we can use either \({S}_{n}\) or \(\sum\) notation:

\[{S}_{n}=\sum _{i=1}^{n}{a}_{i}={a}_{1}+{a}_{2}+\cdots +{a}_{n} \quad (n \text{ terms})\]

Worked example 4: Sigma notation

Expand the sequence and find the value of the series:

\[\sum _{n=1}^{6}{2}^{n}\]

Expand the formula and write down the first six terms of the sequence

\begin{align*} \sum _{n=1}^{6}{2}^{n} &= 2^{1} + 2^{2} + 2^{3} + 2^{4} + 2^{5} + 2^{6} \quad (\text{6} \text{ terms}) \\ &= 2 + 4 + 8 + 16 + 32 + 64 \end{align*}

This is a geometric sequence \(2; 4; 8; 16; 32; 64\) with a constant ratio of \(\text{2}\) between consecutive terms.

Determine the sum of the first six terms of the sequence

\begin{align*} S _{6} &= 2 + 4 + 8 + 16 + 32 + 64 \\ &= 126 \end{align*}

Worked example 5: Sigma notation

Find the value of the series:

\[\sum _{n=3}^{7}{2an}\]

Expand the sequence and write down the five terms

\begin{align*} \sum _{n=3}^{7}{2an} &= 2a(3) + 2a(4) + 2a(5) + 2a(6) + 2a(7) \quad (5 \text{ terms}) \\ &= 6a + 8a + 10a +12a + 14a \end{align*}

Determine the sum of the five terms of the sequence

\begin{align*} S _{5} &= 6a + 8a + 10a +12a + 14a \\ &= 50a \end{align*}

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Worked example 6: Sigma notation

Write the following series in sigma notation:

\[31 + 24 + 17 + 10 + 3\]

Consider the series and determine if it is an arithmetic or geometric series

First test for an arithmetic series: is there a common difference?

We let:

\[\begin{array}{rll} T_{1} &= 31; &T_{4} = 10; \\ T_{2} &= 24; &T_{5} = 3; \\ T_{3} &= 17; & \end{array}\]

We calculate:

\begin{align*} d &= T_{2} - T_{1} \\ &= 24 - 31 \\ &= -7 \\ d &= T_{3} - T_{2} \\ &= 17 - 24 \\ &= -7 \end{align*}

There is a common difference of \(-7\), therefore this is an arithmetic series.

Determine the general formula of the series

\begin{align*} T_{n} &= a + (n-1)d \\ &= 31 + (n-1)(-7) \\ &= 31 -7n + 7 \\ &= -7n + 38 \end{align*}

Be careful: brackets must be used when substituting \(d = -7\) into the general term. Otherwise the equation would be \(T_{n} = 31 + (n-1) - 7\), which would be incorrect.

Determine the sum of the series and write in sigma notation

\begin{align*} 31 + 24 + 17 + 10 + 3 &= 85 \\ \therefore \sum _{n=1}^{5}{(-7n + 38)} &= 85 \end{align*}

Rules for sigma notation

Given two sequences, \({a}_{i}\) and \({b}_{i}\):
\[\sum _{i=1}^{n}\left({a}_{i}+{b}_{i}\right) = \sum _{i=1}^{n}{a}_{i}+\sum _{i=1}^{n}{b}_{i}\]
For any constant \(c\) that is not dependent on the index \(i\):
\begin{align*} \sum _{i=1}^{n} (c \cdot {a}_{i}) & = c\cdot{a}_{1}+c\cdot{a}_{2}+c\cdot{a}_{3}+\cdots +c\cdot{a}_{n} \\ & = c \left({a}_{1}+{a}_{2}+{a}_{3}+\cdots +{a}_{n}\right) \\ & = c\sum _{i=1}^{n}{a}_{i} \end{align*}
Be accurate with the use of brackets:

Example 1:
\begin{align*} \sum _{n=1}^{3}{(2n + 1)}& = 3 + 5 + 7 \\ & = 15 \end{align*}
Example 2:
\begin{align*} \sum _{n=1}^{3}{(2n) + 1}& = (2 + 4 + 6) + 1 \\ & = 13 \end{align*}
Note: the series in the second example has the general term \(T_{n} = 2n\) and the \(\text{+1}\) is added to the sum of the three terms. It is very important in sigma notation to use brackets correctly.
\[\sum_{i = m}^{n}{a_{i}}\]
The values of \(i\):
- start at \(m\) (\(m\) is not always \(\text{1}\));
- increase in steps of \(\text{1}\);
- and end at \(n\).

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Sigma notation

Textbook Exercise 1.7

\[\sum _{k=1}^{4}{2}\]

\(2 + 2 + 2 + 2 = 8\)

\[\sum _{i=-1}^{3}i\]

\begin{align*} \sum _{i=-1}^{3}i & = -1 + 0 + 1 + 2 + 3 \\ & = 5 \end{align*}

\[\sum _{n=2}^{5}(3n - 2)\]

\begin{align*} \sum _{n=2}^{5}(3n - 2) & = [3(2) - 2] + [3(3) - 2] + [3(4) - 2] + [3(5) - 2] \\ & = 4 + 7 + 10 + 13 \\ &= 34 \end{align*}

\[\sum _{k=1}^{6}{0^{k}}\]

\(0^{1} + 0^{2} + 0^{3} + 0^{4} + 0^{5} + 0^{6} = 0\)

\[\sum_{n=-3}^{0}{8}\]

\(8 + 8 + 8 + 8 = 32\)

\[\sum _{k=1}^{5}(ak)\]

\begin{align*} \sum _{k=1}^{5}(ak) & = a + 2a + 3a + 4a + 5a \\ & = 15a \end{align*}

\[\sum _{k=1}^{3} \left( a \cdot {2}^{k-1} \right) =28\]

\begin{align*} \sum _{k=1}^{3} \left( a \cdot {2}^{k-1} \right) &=28\\ \therefore a+ 2a + 4a & = 28 \\ 7a &= 28 \\ \therefore a &= 4 \end{align*}

\[\sum _{j=1}^{4} \left( 2^{-j} \right) = a\]

\begin{align*} \sum _{j=1}^{4} \left( 2^{-j} \right) &= a \\ \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} & = a \\ \therefore \frac{15}{16} &= a \end{align*}

Write the following in sigma notation:

\[\frac{1}{9} + \frac{1}{3} + 1 + 3\]

Geometric series with \(a = \frac{1}{9}\), constant ratio \(r = 3\) and general formula \(T_{n} = ar^{n-1}\).

\begin{align*} T_{n} &= ar^{n-1} \\ & = \frac{1}{9}(3)^{n-1} \\ &= 3^{-2} \cdot 3^{n-1} \\ &= 3^{n-3} \\ \therefore & \sum _{n=1}^{4} \left( 3^{n-3} \right) \end{align*}

Write the sum of the first \(\text{25}\) terms of the series below in sigma notation:

\[11 + 4 - 3 - 10 \ldots\]

Arithmetic series with \(a = 11\), common difference \(d = -7\) and general formula \(T_{n} = a + (n-1)d\).

\begin{align*} T_{n} &= a + (n-1)d \\ & = 11 + (n-1)(-7)\\ &= 11 -7n + 7 \\ &= 18 - 7n \\ \therefore & \sum _{n = 1}^{25} \left( 18 - 7n \right) \end{align*}

Write the sum of the first \(\text{1 000}\) natural, odd numbers in sigma notation.

\[1 + 3 + 5 + 7 + \ldots\]

Arithmetic series with \(a = 1\), common difference \(d = 2\) and general formula \(T_{n} = a + (n-1)d\).

\begin{align*} T_{n} &= a + (n-1)d \\ & = 1 + (n-1)(2)\\ &= 1 + 2n - 2 \\ &= 2n - 1 \\ \therefore & \sum _{n=1}^{1000} \left( 2n - 1 \right) \\ \text{ or } & \sum _{n=0}^{999} \left( 2n + 1 \right) \end{align*}

1.2 Geometric sequences

Table of Contents

1.4 Finite arithmetic series