Download SECTION 8-1 Sequences and Series

552
8 Sequences and Series
If someone asked you to list all natural numbers that are perfect squares,
you might begin by writing
1,
4,
9,
16,
25,
36
But you would soon realize that it is impossible to actually list all the perfect squares, since there are an infinite number of them. However, this
collection of numbers can be represented in several different ways. One
common method is to write
1, 4, 9, . . . , n2, . . .
nN
where N is the set of natural numbers. A list of numbers such as this is
generally called a sequence. Sequences and related topics form the subject matter of this chapter. One of the related topics involves a method of
proof we have referred to several times earlier in this book+mathematical induction. This method enables us to prove conjectures involving infinite sets of successive integers.
SECTION
8-1
Sequences and Series
• Sequences
• Series
In this section we introduce special notation and formulas for representing and generating sequences and sums of sequences.
• Sequences
Consider the function f given by
f(n) 2n 1
(1)
where the domain of f is the set of natural numbers N. Note that
f (1) 1, f (2) 3, f(3) 5, . . .
The function f is an example of a sequence. A sequence is a function with domain
a set of successive integers. However, a sequence is hardly ever represented in the
form of equation (1). A special notation for sequences has evolved, which we
describe here.
To start, the range value f (n) is usually symbolized more compactly with a symbol such as an. Thus, in place of equation (1) we write
an 2n 1
The domain is understood to be the set of natural numbers N unless stated to the contrary or the context indicates otherwise. The elements in the range are called terms
8-1
Sequences and Series
553
of the sequence: a1 is the first term, a2 the second term, and an the nth term, or the
general term:
a1 2(1) 1 1
First term
a2 2(2) 1 3
Second term
a3 2(3) 1 5
.
.
.
.
.
.
Third term
The ordered list of elements
1, 3, 5, . . . , 2n 1, . . .
in which the terms of a sequence are written in their natural order with respect to the
domain values, is often informally referred to as a sequence. A sequence is also represented in the abbreviated form {an}, where a symbol for the nth term is placed
between braces. For example, we can refer to the sequence
1, 3, 5, . . . , 2n 1, . . .
as the sequence {2n 1}.
If the domain of a function is a finite set of successive integers, then the sequence
is called a finite sequence. If the domain is an infinite set of successive integers, then
the sequence is called an infinite sequence. The sequence {2n 1} above is an example of an infinite sequence.
EXPLORE-DISCUSS 1
The sequence {2n 1} is a function whose domain is the set of natural numbers,
and so it may be graphed in the same way as any function whose domain and range
are sets of real numbers (see Fig. 1).
FIGURE 1 Graph of {2n 1}.
y
20
18
16
14
12
10
8
6
4
2
x
1 2 3 4 5 6 7 8 9 10
(A) Explain why the graph of the sequence {2n 1} is not continuous.
(B) Explain why the points on the graph of {2n 1} lie on a line. Find an equation for that line.
(C) Graph the sequence
2n2 n 1
. How are the graphs of {2n 1} and
n
2n2 n 1
related?
n
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8 Sequences and Series
Some sequences are specified by a recursion formula—that is, a formula that
defines each term in terms of one or more preceding terms. The sequence we have
chosen to illustrate a recursion formula is a very famous sequence in the history of
mathematics called the Fibonacci sequence. It is named after the most celebrated
mathematician of the thirteenth century, Leonardo Fibonacci from Italy (1180?–1250?).
EXAMPLE 1
Fibonacci Sequence
List the first six terms of the sequence specified by
a1 1
a2 1
an an1 an2
Solution
n3
a1
1
a2
1
a3
a2 a1 1 1
2
a4
a3 a2 2 1
3
a5
a4 a3 3 2
5
a6
a5 a4 5 3
8
The formula an an1 an2 is a recursion formula that can be used to generate the terms of a sequence in terms of preceding terms. Of course, starting terms
a1 and a2 must be provided in order to use the formula. Recursion formulas are particularly suitable for use with calculators and computers (see Problems 57 and 58 in
Exercise 8-1).
Matched Problem 1
List the first five terms of the sequence specified by
a1 4
an 12an1
n2
Now we consider the reverse problem. That is, can a sequence be defined just by
listing the first three or four terms of the sequence? And can we then use these initial terms to find a formula for the nth term? In general, without other information,
the answer to the first question is no. Many different sequences may start off with the
same terms. For example, each of the following sequences starts off with the same
three terms:
1, 3, 9, . . . , 3n1, . . .
1, 3, 9, . . . , 1 2(n 1)2, . . .
1, 3, 9, . . . , 8n 12
19, . . .
n
8-1
Sequences and Series
555
However, these are certainly different sequences. You should verify that these
sequences agree for the first three terms and differ in the fourth term by evaluating
the general term for each sequence at n 1, 2, 3, 4. Thus, simply listing the first
three terms, or any other finite number of terms, does not specify a particular
sequence. In fact, it can be shown that given any list of m numbers, there are an infinite number of sequences whose first m terms agree with these given numbers. What
about the second question? That is, given a few terms, can we find the general formula for at least one sequence whose first few terms agree with the given terms? The
answer to this question is a qualified yes. If we can observe a simple pattern in the
given terms, then we may be able to construct a general term that will produce the
pattern. The next example illustrates this approach.
EXAMPLE 2
Finding the General Term of a Sequence
Find the general term of a sequence whose first four terms are:
(A) 5, 6, 7, 8, . . .
Solutions
(B) 2, 4, 8, 16, . . .
(A) Since these terms are consecutive integers, one solution is an n, n 5. If we
want the domain of the sequence to be all natural numbers, then another solution is bn n 4.
(B) Each of these terms can be written as the product of a power of 2 and a power
of 1:
2 (1)021
4 (1)122
8 (1)223
16 (1)324
If we choose the domain to be all natural numbers, then a solution is
an (1)n12n
Matched Problem 2
Find the general term of a sequence whose first four terms are:
(A) 2, 4, 6, 8, . . .
(B) 1, 12, 14, 18, . . .
In general, there is usually more than one way of representing the nth term of a
given sequence. This was seen in the solution of Example 2A. However, unless stated
to the contrary, we assume the domain of the sequence is the set of natural numbers N.
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8 Sequences and Series
EXPLORE-DISCUSS 2
5 1 5 n
is closely related to the
5
2
Fibonacci sequence. Compute the first 20 terms of both sequences and discuss the
relationship. (The values of b1, b2, and b3 are shown in Fig. 2.)
The sequence with general term bn FIGURE 2
• Series
If a1, a2, a3, . . . , an, . . . is a sequence, then the expression a1 a2 a3 . . . an . . . is called a series. If the sequence is finite, the corresponding series is
a finite series. If the sequence is infinite, the corresponding series is an infinite series.
For example,
1, 2, 4, 8, 16
Finite sequence
1 2 4 8 16
Finite series
We restrict our discussion to finite series in this section.
Series are often represented in a compact form called summation notation using
the symbol , which is a stylized version of the Greek letter sigma. Consider the
following examples:
4
a
k
a1 a2 a3 a4
k1
7
b
k
b3 b4 b5 b6 b7
k3
n
c
k
c0 c1 c2 . . . cn
k0
Domain is the set of integers
greater than or equal to 0.
The terms on the right are obtained from the expression on the left by successively
replacing the summing index k with integers, starting with the first number indicated below
and ending with the number that appears above . Thus, for example, if we are given the sequence
1 1 1
1
, , ,..., n
2 4 8
2
the corresponding series is
1 1 1 ...
1
n
2 4 8
2
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Sequences and Series
557
or, more compactly,
n
1
2
k
k1
EXAMPLE 3
Writing the Terms of a Series
k1
k
k1
5
Write without summation notation:
k1 11 21 31 41 51
k
1
2
3
4
5
k1
5
Solution
0
1 2 3 4
2 3 4 5
5
Matched Problem 3
Write without summation notation:
(1)k
2k 1
k0
If the terms of a series are alternately positive and negative, it is called an alternating series. Example 4 deals with the representation of such a series.
EXAMPLE 4
Writing a Series in Summation Notation
Write the following series using summation notation:
1
1 1 1 1 1
2 3 4 5 6
(A) Start the summing index at k 1.
(B) Start the summing index at k 0.
Solutions
(A) (1)k1 provides the alternation of sign, and 1/k provides the other part of each
term. Thus, we can write
6
(1)k1
k
k1
as can be easily checked.
(B) (1)k provides the alternation of sign, and 1/(k 1) provides the other part of
each term. Thus, we write
558
8 Sequences and Series
5
(1)k
k0 k 1
as can be checked.
Matched Problem 4
Write the following series using summation notation:
1
(A) Start with k 1.
EXPLORE-DISCUSS 3
2 4
8
16
3 9 27 81
(B) Start with k 0.
(A) Find the smallest number of terms of the infinite series
1
1 1 ... 1 ...
2 3
n
that, when added together, give a number greater than 3.
(B) Find the smallest number of terms of the infinite series
1 1 ...
1
n...
2 4
2
that, when added together, give a number greater than 0.99. Greater than 0.999.
Can the sum ever exceed 1? Explain.
Answers to Matched Problems
2. (A) an 2n
1. 4, 2, 1, 12 , 14
5
4. (A)
(1)
k1
k1
EXERCISE
3 2
k1
(B) an (1)n1( 12 )n1
(1) 3 4
(B)
k
2
1
3. 1 13 15 17 19 11
k
k0
8-1
A
9. Write the two-hundredth term of the sequence in Problem 3.
10. Write the ninety-ninth term of the sequence in Problem 4.
Write the first four terms for each sequence in
Problems 1–6.
1. an 2n 5
2. an n 1
3n 1
3. an n1
4. an (n 1)
5. an 3n (2)n
6. an In Problems 11–16, write each series in expanded form
without summation notation.
2
5
n
(1)n
n3
11.
12.
k1
14.
k1
(1)
k
k1
2
5
1
k
10
k1
4
15.
k
k1
3
13.
7. Write the one-hundredth term of the sequence in Problem 1.
8. Write the fiftieth term of the sequence in Problem 2.
4
k
1
3
k
6
16.
(1)
k1
k1
k
8-1
3 4 ... n1
2 3
n
B
54. 2 Write the first five terms of each sequence in Problems
17–26.
55. 1 4 9 . . . (1)n1n2
17. an (1)
n
1
1
1 n
3
10
19. an 18. an (1)
n1 2
21. an ( 12)n 1
n1
1
2n
20. an n[1 (1)n]
22. an ( 32)n 1
24. a1 a2 1; an an1 an2, n 3
26. a1 2; an 2an1, n 2
In Problems 27–38, find the general term of a sequence
whose first four terms are given.
27. 3, 5, 7, 9, . . .
28. 2, 1, 4, 7, . . .
29. 4, 9, 16, 25, . . .
30. 24, 35, 48, 63, . . .
31. 4, 8, 16, 32, . . .
32. 1, 4, 27, 256, . . .
...
34.
35. 7, 7, 7, 7, . . .
37. x,
x2 x3 x4
, , ,...
2 4 8
1 3 7 15
2 , 4 , 8 , 16 ,
36. 1,
C
an 25. a1 4; an 14an1, n 2
33. 1,
1 1 1 . . . (1)n 1
2 4 8
2n
The sequence
23. a1 7; an an1 4, n 2
3 5 7
2, 3, 4,
56.
a 2n 1 M
2an 1
n 2, M a positive real number
can be used to find M to any decimal-place accuracy
desired. To start the sequence, choose a1 arbitrarily from the
positive real numbers. Problems 57 and 58 are related to
this sequence.
57. (A) Find the first four terms of the sequence
a1 3
an ...
1
1
15, 25
,125
,
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Sequences and Series
...
38. x, x4, x7, x10, . . .
a 2n 1 2
2an 1
n2
(B) Compare the terms with 2 from a calculator.
(C) Repeat parts A and B letting a1 be any other positive
number, say 1.
58. (A) Find the first four terms of the sequence
Some graphing utilities use special routines to graph
sequences (consult your manual). In Problems 39–42, use
such routines to graph the first 20 terms of each sequence.
39. an 1/n
41. an (0.9)
40. an 2 n
n
42. a1 1, an 23 an 1 12
In Problems 43–48, write each series in expanded form
without summation notation.
4
43.
(2)k1
k
k1
3
45.
1
kx
5
44.
(2k 1)2
k1
k1
5
k1
46.
k1
(1)k1 k
x
k
k1
x
k1
k1
5
47.
(1)
4
48.
(1)kx 2k1
2k 1
k0
In Problems 49–56, write each series using summation
notation with the summing index k starting at k 1.
49. 12 22 32 42
51.
1
1
1
1
1
2 22 23 24 25
53. 1 1
1
1
... 2
22 32
n
50. 2 3 4 5 6
52. 1 1 1 1
2 3 4
a1 2
an a 2n 1 5
2an 1
n2
(B) Find 5 with a calculator, and compare with the results of part A.
(C) Repeat parts A and B letting a1 be any other positive
number, say 3.
59. Let {an} denote the Fibonacci sequence and let {bn} denote
the sequence defined by b1 1, b2 3, bn bn1 bn2
for n 3. Compute 10 terms of the sequence {cn}, where
cn bn /an. Describe the terms of {cn} for large values of n.
60. Define sequences {un} and {vn} by u1 1, v1 0, un un1 vn1 and vn un1 for n 2. Find the first 10 terms
of each sequence, and explain their relationship to the Fibonacci sequence.
In calculus, it can be shown that
ex xk
x
x2
x3
xn
k! 1 1! 2! 3! . . . n!
k0
where the larger n is, the better the approximation. Problems
61 and 62 refer to this series. Note that n!, read “n factorial,” is defined by 0! 1 and n! 1 2 3 . . . n for
n N.
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8 Sequences and Series
61. Approximate e0.2 using the first five terms of the series.
Compare this approximation with your calculator evaluation of e0.2.
0.5
62. Approximate e
using the first five terms of the series.
Compare this approximation with your calculator evaluation of e0.5.
SECTION
8-2
ca
k
n
c
k1
a
n
64. Show that:
(a
k
k
k1
n
bk ) k1
n
a
k
k1
b
k
k1
Mathematical Induction
•
•
•
•
• Introduction
n
63. Show that:
Introduction
Mathematical Induction
Additional Examples of Mathematical Induction
Three Famous Problems
In common usage, the word “induction” means the generalization from particular
cases or facts. The ability to formulate general hypotheses from a limited number of
facts is a distinguishing characteristic of a creative mathematician. The creative
process does not stop here, however. These hypotheses must then be proved or disproved. In mathematics, a special method of proof called mathematical induction
ranks among the most important basic tools in a mathematician’s toolbox. In this section mathematical induction will be used to prove a variety of mathematical statements, some new and some that up to now we have just assumed to be true.
We illustrate the process of formulating hypotheses by an example. Suppose we
are interested in the sum of the first n consecutive odd integers, where n is a positive
integer. We begin by writing the sums for the first few values of n to see if we can
observe a pattern:
11
n 1
134
n 2
1359
n 3
1 3 5 7 16
n 4
1 3 5 7 9 25
n 5
Is there any pattern to the sums 1, 4, 9, 16, and 25? You no doubt observed that each
is a perfect square and, in fact, each is the square of the number of terms in the sum.
Thus, the following conjecture seems reasonable:
Conjecture Pn: For each positive integer n,
1 3 5 . . . (2n 1) n2
That is, the sum of the first n odd integers is n2 for each positive integer n.
So far ordinary induction has been used to generalize the pattern observed in the
first few cases listed above. But at this point conjecture Pn is simply that—a conjecture. How do we prove that Pn is a true statement? Continuing to list specific cases will
never provide a general proof—not in your lifetime or all your descendants’ lifetimes!
Mathematical induction is the tool we will use to establish the validity of conjecture Pn.