Title: Default for 16.1 Sequences and Series
Course: HELM Workbooks 2023-2024

Introduction

An infinite sequence is any succession of numbers. For example the sequence \[ 1,\ 1,\ 2,\ 3,\ 5,\ 8,\ \ldots\] which is known as the Fibonacci sequence, is formed by adding two consecutive terms together to obtain the next term. The numbers in this sequence continually increase without bound and we say this sequence diverges. An example of a convergent sequence is the harmonic sequence \[1,\ \dfrac{1}{2},\ \dfrac{1}{3},\ \dfrac{1}{4},\ \ldots\] Here we see the magnitude of these numbers continually decrease and it is obvious that the sequence converges to the number zero. The related alternating harmonic sequence \[1,\ -\dfrac{1}{2},\ \dfrac{1}{3},\ -\dfrac{1}{4},\ \ldots\] is also convergent to the number zero. Whether or not a sequence is convergent is often easy to deduce by graphing the individual terms. The diagrams below show how the individual terms of the harmonic and alternating harmonic series behave as the number of terms increase.

[Image: Graph of the individual terms of the harmonic sequence, showing how these behave as the number of terms increases.]

[Image: Graph of the individual terms of the alternating harmonic sequence, showing how these behave as the number of terms incr]

Q: Exercise

Which of the following graphs correctly shows the sequence \(\displaystyle 1,-1,1,-1,\ldots\) ?
- [Image]
- [Image]
A: [Image]

Q: Exercise

Is the sequence \(\displaystyle 1,-1,1,-1,\ldots\) convergent or divergent?
- Convergent.
- Divergent.
A: Divergent.

Convergence or divergence of sequences

A general infinite sequence is denoted by \[ a_1,\ a_2,\ \ldots,\ a_n,\ldots\]in which \(a_1\) is the first term, \(a_2\) is the second term and \(a_n\) is the \(n^{th}\) term in the sequence. For example, in the harmonic sequence \[ a_1=1,\ \ a_2=\dfrac{1}{2},\ldots , a_n=\dfrac{1}{n}\]whilst for the alternating harmonic sequence the \(n^{th}\) term is: \[a_n=\dfrac{(-1)^{n+1}}{n}\]in which \((-1)^n=+1\) if \(n\) is an even number and \((-1)^n=-1\) if \(n\) is an odd number.

Key Point

The sequence \(a_1,a_2,\ldots ,a_n, \ldots\) is said to be convergent if the limit of \(a_n\) as \(n\) increases can be found. (Mathematically we say that \(\displaystyle \lim_{n \to \infty} a_n\) exists.)

If the sequence is not convergent it is said to be divergent.

Exercise

Verify that the following sequence is convergent.\[\dfrac{3}{1\times 2}, \dfrac{4}{2\times 3}, \dfrac{5}{3\times 4}, \ldots\]

1. First, find the expression for the \(n^{th}\) term:
\(a_n =\)
2. Now, find the limit of \(a_n\) as \(n\) increases:
\(\displaystyle \lim_{n\to \infty} a_n = \)
Feedback: 1. The \(n^{th}\) term of the sequence is given by \[\displaystyle a_n = [value].\] (Note that by substituting \(n=1\) in the expression above, we obtain \(a_1 = \dfrac{3}{1\times 2}\) for the first term, as required. Similarly, for the second and third terms.) 2. Then, as \(n\) increases, \[[value] = \left[ \dfrac{1+\dfrac{2}{n}}{n+1} \right] \to \dfrac{1}{n+1} \to [value].\]Hence, \(\displaystyle \lim_{n\to \infty} a_n = [value]\). Thus, the series is convergent, since this limit can be found.

Arithmetic and geometric progressions

Consider the sequences: \[ 1,\ 4,\ 7,\ 10,\ \ldots \quad \text{and} \quad 3,\ 1,\ -1,\ -3,\ \ldots\]In both, any particular term is obtained from the previous term by the addition of a constant value (\(3\) and \(-2\) respectively). Each of these sequences are said to be an arithmetic sequence or arithmetic progression and has general form: \[ a,\ a+d,\ a+2d,\ a+3d,\ \ldots ,\ a+(n-1)d,\ \ldots\] in which \(a,d\) are given numbers. In the first example above \(a=1,\ d=3\) whereas, in the second example, \(a=3,\ d=-2\). The difference between any two successive terms of a given arithmetic sequence gives the value of \(d\) which is called the common difference.

Two sequences which are not arithmetic sequences are: \[ 1,\ 2,\ 4,\ 8,\ \ldots\] \[ -1,\ -\dfrac{1}{3},\ -\dfrac{1}{9},\ -\dfrac{1}{27},\ \ldots\]In each case a particular term is obtained from the previous term by multiplying by a constant factor (\(2\) and \(\dfrac{1}{3}\) respectively). Each is an example of a geometric sequence or geometric progression with the general form: \[ a,\ ar,\ ar^2,\ ar^3,\ \ldots\]where ‘\(a\)’ is the first term and \(r\) is called the common ratio, being the ratio of two successive terms. In the first geometric sequence above \(a=1\), \(r=2\) and in the second geometric sequence \(a=-1\), \(r=\dfrac{1}{3}\).

Exercise

Find \(a, \ d\) for the arithmetic sequence \([value],\ [value],\ [value],\dots\)

\(a = \)

\(d = \)
Feedback: We are given the arithmetic sequence \([value],\ [value],\ [value],\dots\) For an arithmetic sequence: The value of \(a\) is always given by the first term of the sequence. Hence, \(a = [value]\). The value of \(d\) is given by the difference between any two consecutive terms in the sequence. Hence, \(d = [value]\).

Exercise

Find \(a,\ r\) for the geometric sequence \(\displaystyle{[value],\ [value],\ [value],\dots}\)

\(a = \)

\(r = \)
Feedback: We are given the geometric sequence \(\displaystyle{[value],\ [value],\ [value],\dots}\) For a geometric sequence: The value of \(a\) is always given by the first term of the sequence. Hence, \(a = [value]\). The value of \(r\) is given by the ratio between any two consecutive terms in the sequence. Hence, \(r = \displaystyle[value]\).

Exercise

Write out the first four terms of the geometric series with \(a=[value]\), \(r=[value].\)

\(a_1 =\)

\(a_2 =\)

\(a_3 =\)

\(a_4 =\)
Feedback: We have the geometric series with \(a=[value]\), \(r=[value].\) The terms are given as follows: \(a_1 = a = [value].\) \(a_2 = ar = [value].\) \(a_3 = ar^2 = [value].\) \(a_4 = ar^3 = [value].\)

The examples of arithmetic and geometric sequences are particularly nice and relatively easy to work with. It is important to note that a general sequence will not fall into either category. One easily defined example is the harmonic sequence

\[1,\frac12,\frac13,\frac14\,\dots,\]

whose \(n\)th term is \(\dfrac{1}{n}\), is neither arithmetic nor geometric.

Finite Series

A series is the sum of the terms of a sequence. For example, the harmonic series is \[ 1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\cdots\]and the alternating harmonic series is \[ 1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\cdots.\]

We will mainly consider finite series in this section, that is, series which are the sum of the first \(k\) terms of a sequence for some natural number \(k\).

The summation notation

If we consider a general sequence \[ a_1,\ a_2,\ \ldots ,\ a_n,\ \ldots\] then the sum of the first \(k\) terms \(a_1+a_2+a_3+\cdots +a_k\) is concisely denoted by \(\displaystyle \sum_{p=1}^k\ a_p\).

That is, \[ a_1+a_2+a_3+\cdots +a_k= \sum_{p=1}^ka_p.\]

When we encounter the expression \(\displaystyle \sum_{p=1}^k a_p\) we let the index ‘\(p\)’ in the term \(a_p\) take, in turn, the values \(1,2,\ldots,k\) and then add all these terms together. So, for example \[ \sum_{p=1}^3a_p= a_1+a_2+a_3\qquad\qquad \sum_{p=2}^7a_p= a_2+a_3+a_4+a_5+a_6+a_7\] Note that \(p\) is a dummy index; any letter could be used as the index. For example \(\displaystyle \sum_{i=1}^6 a_i\), and \(\displaystyle \sum_{m=1}^6 a_m\) each represent the same collection of terms: \(a_1+a_2+a_3+a_4+a_5+a_6\).

In order to be able to use this ‘summation notation’ we need to obtain a suitable expression for the ‘typical term’ in the series. For example, the finite series \[ 1^2+2^2+\cdots +k^2\]may be written as \(\displaystyle \sum_{p=1}^k p^2\) since the typical term is clearly \(p^2\) in which \(p=1,2,3,\ldots, k\) in turn.

In the same way \[ 1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\cdots -\dfrac{1}{16}= \sum_{p=1}^{16}\dfrac{(-1)^{p+1}}{p}\]since an expression for the typical term in this alternating harmonic series is \(a_p= \dfrac{(-1)^{p+1}}{p}\).

Note that we are using a convention that the terms of the series start with index \(k=1\). When working with power series we will more often see starting index \(k=0\). It is very important to pay attention to this value when summing a series.

Exercise

Complete the summation form of the series \[ \sum_{p=1}^n = \dfrac{1}{1\times [value]}+\dfrac{1}{2\times [value]}+\dfrac{1}{3 \times [value]}+\cdots+\dfrac{1}{[value]\times[value]}\]below. Start by finding an expression for the typical term, “the \(p^{th}\) term” \(a_p\) and the determine the upper index \(n\).

\(a_p=\)

\(n=\)
Feedback: The summation index \(p\) is set to start from \(p=1\). Hence, the typical term must be given by \(\displaystyle [value]\). Then, the summation is written as \[\sum_{p=1}^[value] [value].\] (Note: the question set \(p=1\) for the first term. This is a choice. If you start the summation at \(p=2\), then the upper bound is [value] and the typical term must be given by \(a_p = \dfrac{1}{(p-1)([value])}\).)

Exercise

Write out all terms of the series \[\sum_{p=[value]}^[value] [value].\]

Give \(p\) the values \([value], [value], [value], [value], [value]\) in the typical term \(\displaystyle [value]\).

\(\displaystyle \sum_{p=[value]}^[value] [value] = \) \(+\) \(+\) \(+\) \(+\)

Write the terms in order.
Feedback: The terms of the series are given by \[\begin{aligned} \sum_{p=[value]}^[value] [value] &= \dfrac{(-1)^{[value]}}{([value]+[value])^2} + \dfrac{(-1)^{[value]}}{([value]+[value])^2} + \dfrac{(-1)^{[value]}}{([value]+[value])^2} + \dfrac{(-1)^{[value]}}{([value]+[value])^2} + \dfrac{(-1)^{[value]}}{([value]+[value])^2}, \\ &= \left([value]\right) + \left([value]\right) + \left([value]\right) + \left([value]\right) + \left([value]\right), \\ &= \dfrac{1}{([value])^2} - \dfrac{1}{([value])^2} + \dfrac{1}{([value])^2} - \dfrac{1}{([value])^2} + \dfrac{1}{([value])^2}. \end{aligned}\]

Summing series

The arithmetic series

Consider the finite arithmetic series with 14 terms \[ 1+3+5+\cdots +23+25+27\]

A simple way of working out the value of the sum is to create a second series which is the first written in reverse order. Thus we have two series, each with the same value \(A\): \[ A=1+3+5+\cdots +23+25+27\]and \[ A=27+25+23+\cdots +5+3+1\]Now, adding the terms of these series in pairs \[ 2A=28+28+28+\cdots +28+28+28=28\times 14=392\qquad \text{so}\quad A=196.\]We can use this approach to find the sum of \(n\) terms of a general arithmetic series.

If \[ A=[a]+[a+d]+[a+2d]+\cdots +[a+(n-2)d]+[a+(n-1)d]\]then again simply writing the terms in reverse order: \[ A=[a+(n-1)d]+[a+(n-2)d]+\cdots+[a+2d]+[a+d]+[a]\]Adding these two identical equations together we have \[ 2A=[2a+(n-1)d]+[2a+(n-1)d]+\cdots +[2a+(n-1)d]\]That is, every one of the \(n\) terms on the right-hand side has the same value: \([2a+(n-1)d].\) Hence \[ 2A=n[2a+(n-1)d]\qquad \text{so}\qquad A=\dfrac{1}{2}n[2a+(n-1)d].\]

Key Point

The arithmetic series \[ [a] +[a+d] + [a+2d] + \ldots + [a+(n-1)d]\]having \(n\) terms has sum \(A\) where: \[ A = \dfrac{1}{2} n [2a + (n-1)d]. \]

As an example \[ 1+3+5+\cdots+27\ \ \ \ \text{has}\ \ \ a=1,\ d=2,\ n=14\]So \(A=1+3+\cdots+27=\dfrac{14}{2}[2+(13)2]=196\).

The geometric series

We can also sum a general geometric series.

Let \[ G=a+ar+ar^2+\cdots+ar^{n-1}\]be a geometric series having exactly \(n\) terms. To obtain the value of \(G\) in a more convenient form we first multiply through by the common ratio \(r\): \[ rG=ar+ar^2+ar^3+\cdots +ar^n\]Now, writing the two series together: \[ G=a+ar+ar^2+\cdots+ar^{n-1}\]\[ rG=ar+ar^2+ar^3+\cdots ar^{n-1}+ar^n\]

Subtracting the second expression from the first we see that all terms on the right-hand side cancel out, except for the first term of the first expression and the last term of the second expression so that \[ G-rG= (1-r)G=a-ar^n\]Hence (assuming \(r\ne 1\)) \[ G=\dfrac{a(1-r^n)}{1-r}\](Of course, if \(r=1\) the geometric series simplifies to a simple arithmetic series with \(d=0\) and has sum \(G=na\).)

Key Point

The geometric series \[ a +ar + ar^2 + \ldots + ar^{n-1}\]having \(n\) terms has sum \(G\) where: \[ G = \dfrac{a(1-r^n)}{1-r}, \quad \text{if} \quad r \ne 1 \qquad \text{and} \qquad G = na, \quad \text{if} \quad r = 1.\]

Exercise

Find the sum of each of the following series:

1. \([value]\)

In this arithmetic series state the values of \(a,d,n\):
\(a = \)
\(d = \)
\(n = \)
Now find the sum:
\([value] =\)
2. \(\displaystyle [value]\)

In this geometric series state the values of \(a,r,n\):
\(a = \)
\(r = \)
\(n = \)
Now find the sum:
\(\displaystyle [value]=\)
Feedback: 1. In an arithmetic series \(a\) is first term, \(d \) is the difference between terms, \(n\) is the number of terms and the sum \(A\) of the series is calculated as \[A = \frac{1}{2} n \left[2a+(n-1)d\right] .\]Then, we have \(a = [value]\), \(d=[value]\), \(n = [value]\) and \[A = [value] = [value].\] 2. In an geometric series \(a\) is first term, \(r \) is the multiplication factor between terms, \(n\) is the number of terms and the sum \(G\) of the series is calculated as \[G = \frac{a(1-r^n)}{1-r}\] if \(r\neq 1\). Then, we have \(a = [value]\), \(r=[value]\), \(n = [value]\) and \[G = [value] = [value].\]

Exercise

Which of the following sequences are convergent?

1. \(\displaystyle [value],[value],[value],[value],\ldots\)
2. \(\displaystyle [value],[value],[value],[value],\ldots\)

Select one:
Feedback: 1. This sequence is \[ \displaystyle [value],[value],[value],[value],\ldots =\displaystyle [value],[value],[value],[value],[value],\ldots\] which does not converge. 2. This sequence is \[\displaystyle [value],[value],[value],[value],\ldots = \displaystyle [value],[value],[value],[value],[value],\ldots\] which converges to zero. Remember, a sequence is convergent if the limit of \(a_n\) as \(n\) increases can be found.

Exercise

Write the following finite series in summation form:

1. \(\displaystyle [value] =\)

\(\displaystyle \sum_{p ~=~ 1}\) |

2. \(\displaystyle - \frac{1}{[value] \times (1 + (100)^2)} +\frac{1}{[value] \times (1 - (200)^2)} -\frac{1}{[value] \times (1 + (300)^2)} + \ldots +\frac{1}{[value] \times (1 - (800)^2)}=\)

\(\displaystyle \sum_{p ~=~ 1}\) |
Feedback: To find the summation forms of each series, we need to obtain suitable expressions for the 'typical term'. For us this is the \(p^{th}\)-term, since both series are indexed by \(p\). And \(p\) will take values \(1,\, 2,\, 3,\, \ldots,\, n\), for some integer \(n\) that we must also find. 1. To find an expression for the typical term, we can look at the first three terms of the series, where \(p=1,\,2\) and \(3\), respectively. Let's start with the numerator. For \(p=1\) we have \([value]\), for \(p=2\) we have \([value]\), and for \(p=3\) we have \([value]\). The expression \([value]\) is then appropriate for the numerator of the typical term. Now, for the denominator we have \([value]\) for \(p=1\), \([value]\) for \(p=2\) and \([value]\) for \(p=3\). Then \([value]\) is appropriate for the denominator. Finally, to find \(n\) we look at the final term and compare it to the typical term we have found \(\displaystyle [value]\). We can see that by letting \(p=[value]\) we obtain \(\displaystyle [value]\), so \(n=[value]\). Hence, the summation form of this series is \[\displaystyle \sum_{p~=~1}^{[value]} [value]\] 2. Similarly, looking at the numerators for the first three terms we see that it is either \([value]\) or \([value]\) for odd and even values of \(p\), respectively, which can be expressed as \([value]\). And looking at the denominators we find that they can be expressed as \([value]\). By letting \(p=[value]\), in our newly found typical term \(\displaystyle [value]\) we obtain the final term \(\displaystyle [value]\). So \(n=[value]\). Hence, the summation form of this series is \[\displaystyle \sum_{p~=~1}^{[value]} [value]\]

Exercise

Write out the first three terms and the last term of the following series, note you do not need to simplify the expressions.

1. \(\displaystyle \sum_{p~=~[value]}^{[value]} [value]=\) \(+\) \(+\) \(+\ldots +\)
2. \(\displaystyle \sum_{p~=~[value]}^{[value]} [value]=\) \(+\) \(+\) \(+\ldots +\)
Feedback: For each term, simply substitute in the value of \(p\). The terms then are: 1. \(\displaystyle [value], [value], [value], \ldots, [value]\) 2. \(\displaystyle [value], [value], [value], \ldots, [value]\) Note that for part (b), the count starts from \(p~=~[value]\) for the first term. So for the second term we have \(p~=~[value]\), and so on.

Exercise

Sum the series:

1. \( [value] =\)
2. \( [value] =\)
3. \( \displaystyle [value] = \)
Feedback: 1. This is an arithmetic series with \(a~=~[value]\), \(d~=~[value]\), \(n~=~[value]\). So \(A~=~[value]\) 2. This is an arithmetic series with \(a~=~[value]\), \(d~=~[value]\), \(n~=~[value]\). So \(A~=~[value]\) 3. This is a geometric series with \(a~=~[value]\), \(r~=~[value]\), \(n~=~[value]\). So \(G~=~[value]\)