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

554


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


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.

556


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

8-1


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 ,

559


...

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.

560


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.