Lista de sequências

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UNIVERSIDADE FEDERAL DE GOIA´S
INSTITUTO DE MATEMA´TICA E ESTATI´STICA
DISCIPLINA: Ca´lculo 2A Profa. Marina Tuyako Mizukoshi
1. Dado uma fo´rmula para o n−e´simo termo an de uma sequeˆncia {an}. Obtenha os valores de a1, a2, a3
e a4.
2. Dado o primeiro ou os dois primeiros termos de uma sequeˆncia com uma fo´rmula de recurssa˜o para
os termos restantes. Escreva explicitamente os dez primeiros termos de cada sequeˆncia.
10.1 Sequences 559
It is important to realize that Theorem 6 does not say that convergent sequences are
monotonic. The sequence converges and is bounded, but it is not monotonic
since it alternates between positive and negative values as it tends toward zero. What the
theorem does say is that a nondecreasing sequence converges when it is bounded from
above, but it diverges to infinity otherwise.
{s-1dn+1>n}
Exercises 10.1
Finding Terms of a Sequence
Each of Exercises 1–6 gives a formula for the nth term of a se-
quence Find the values of and 
1. 2.
3. 4.
5. 6.
Each of Exercises 7–12 gives the first term or two of a sequence along
with a recursion formula for the remaining terms. Write out the first
ten terms of the sequence.
7.
8.
9.
10.
11.
12.
Finding a Sequence’s Formula
In Exercises 13–26, find a formula for the nth term of the sequence.
13. The sequence 1’s with alternating signs
14. The sequence 1’s with alternating signs
15. The sequence 
16. The sequence 
17.
18.
19. The sequence 
20. The sequence 
21. The sequence 
22. The sequence 
23. 5
1
, 
8
2
, 11
6
, 14
24
, 
17
120
, Á
2, 6, 10, 14, 18, Á
1, 5, 9, 13, 17, Á
-3, -2, -1, 0, 1, Á
0, 3, 8, 15, 24, Á
- 3
2
, - 1
6
, 1
12
, 
3
20
, 
5
30
, Á
1
9
, 2
12
, 2
2
15
, 2
3
18
, 2
4
21
, Á
1, - 1
4
, 1
9
, - 1
16
, 1
25
, Á
1, -4, 9, -16, 25, Á
-1, 1, -1, 1, -1, Á
1, -1, 1, -1, 1, Á
a1 = 2, a2 = -1, an+2 = an+1>ana1 = a2 = 1, an+2 = an+1 + an
a1 = -2, an+1 = nan>sn + 1da1 = 2, an+1 = s-1dn+1an>2
a1 = 1, an+1 = an>sn + 1da1 = 1, an+1 = an + s1>2nd
an =
2n - 1
2n
an =
2n
2n+1
an = 2 + s-1dnan =
s-1dn+1
2n - 1
an =
1
n!
an =
1 - n
n2
a4 .a1, a2, a3 ,5an6 . an 24.
25. The sequence 
26. The sequence 
Convergence and Divergence
Which of the sequences in Exercises 27–90 converge, and which
diverge? Find the limit of each convergent sequence.
27. 28.
29. 30.
31. 32.
33. 34.
35. 36.
37. 38.
39. 40.
41. 42.
43. 44.
45. 46.
47. 48.
49. 50.
51. 52.
53. 54.
55. 56.
57. 58.
59. 60. an = ln n - ln sn + 1dan =
ln n
n1>n
an = sn + 4d1>sn+4dan = a3n b1>n an = 2n n2an = 2n 10n
an = a1 - 1n bnan = a1 + 7n bn
an = s0.03d1>nan = 81>n
an =
ln n
ln 2n
an =
ln sn + 1d2n an =
3n
n3
an =
n
2n
an =
sin2 n
2n
an =
sin n
n
an = np cos snpdan = sin ap2 + 1n b
an =
1
s0.9dnan = A 2nn + 1 an = a- 12 b
n
an =
s-1dn+1
2n - 1
an = a2 - 12n b a3 + 12n ban = an + 12n b a1 - 1n b
an = s-1dn a1 - 1n ban = 1 + s-1dn
an =
1 - n3
70 - 4n2
an =
n2 - 2n + 1
n - 1
an =
n + 3
n2 + 5n + 6
an =
1 - 5n4
n4 + 8n3
an =
2n + 1
1 - 32nan = 1 - 2n1 + 2n an =
n + s-1dn
nan = 2 + s0.1d
n
5an6
0, 1, 1, 2, 2, 3, 3, 4, Á
1, 0, 1, 0, 1, Á
1
25
, 
8
125
, 
27
625
, 
64
3125
, 
125
15,625
, Á
Squares of the positive
integers, with
alternating signs
Reciprocals of squares
of the positive integers,
with alternating signs
Powers of 2 divided by
multiples of 3
Integers differing by 2
divided by products of
consecutive integers
Squares of the positive
integers diminished by 1
Integers, beginning with -3
Every other odd positive
integer
Every other even positive
integer
Integers differing by 3
divided by factorials
Alternating 1’s and 0’s
Cubes of positive integers
divided by powers of 5
Each positive integer
repeated
7001_ThomasET_ch10p550-627.qxd 10/30/09 8:21 AM Page 559
3. Nos exerc´ıcios de 13-26, obtenha uma fo´rmula para o n−e´simo termo da sequeˆncia.
10.1 Sequences 559
It is important to realize that Theorem 6 does not say that convergent sequences are
monotonic. The sequence converges and is bounded, but it is not monotonic
since it alternates between positive and negative values as it tends toward zero. What the
theorem does say is that a nondecreasing sequence converges when it is bounded from
above, but it diverges to infinity otherwise.
{s-1dn+1>n}
Exercises 10.1
Finding Terms of a Sequence
Each of Exercises 1–6 gives a formula for the nth term of a se-
quence Find the values of and 
1. 2.
3. 4.
5. 6.
Each of Exercises 7–12 gives the first term or two of a sequence along
with a recursion formula for the remaining terms. Write out the first
ten terms of the sequence.
7.
8.
9.
10.
11.
12.
Finding a Sequence’s Formula
In Exercises 13–26, find a formula for the nth term of the sequence.
13. The sequence 1’s with alternating signs
14. The sequence 1’s with alternating signs
15. The sequence 
16. The sequence 
17.
18.
19. The sequence 
20. The sequence 
21. The sequence 
22. The sequence 
23. 5
1
, 
8
2
, 11
6
, 14
24
, 
17
120
, Á
2, 6, 10, 14, 18, Á
1, 5, 9, 13, 17, Á
-3, -2, -1, 0, 1, Á
0, 3, 8, 15, 24, Á
- 3
2
, - 1
6
, 1
12
, 
3
20
, 
5
30
, Á
1
9
, 2
12
, 2
2
15
, 2
3
18
, 2
4
21
, Á
1, - 1
4
, 1
9
, - 1
16
, 1
25
, Á
1, -4, 9, -16, 25, Á
-1, 1, -1, 1, -1, Á
1, -1, 1, -1, 1, Á
a1 = 2, a2 = -1, an+2 = an+1>ana1 = a2 = 1, an+2 = an+1 + an
a1 = -2, an+1 = nan>sn + 1da1 = 2, an+1 = s-1dn+1an>2
a1 = 1, an+1 = an>sn + 1da1 = 1, an+1 = an + s1>2nd
an =
2n - 1
2n
an =
2n
2n+1
an = 2 + s-1dnan =
s-1dn+1
2n - 1
an =
1
n!
an =
1 - n
n2
a4 .a1, a2, a3 ,5an6 . an 24.
25. The sequence 
26. The sequence 
Convergence and Divergence
Which of the sequences in Exercises 27–90 converge, and which
diverge? Find the limit of each convergent sequence.
27. 28.
29. 30.
31. 32.
33. 34.
35. 36.
37. 38.
39. 40.
41. 42.
43. 44.
45. 46.
47. 48.
49. 50.
51. 52.
53. 54.
55. 56.
57. 58.
59. 60. an = ln n - ln sn + 1dan =
ln n
n1>n
an = sn + 4d1>sn+4dan = a3n b1>n an = 2n n2an = 2n 10n
an = a1 - 1n bnan = a1 + 7n bn
an = s0.03d1>nan = 81>n
an =
ln n
ln 2n
an =
ln sn + 1d2n an =
3n
n3
an =
n
2n
an =
sin2 n
2n
an =
sin n
n
an = np cos snpdan = sin ap2 + 1n b
an =
1
s0.9dnan = A 2nn + 1 an = a- 12 b
n
an =
s-1dn+1
2n - 1
an = a2 - 12n b a3 + 12n ban = an + 12n b a1 - 1n b
an = s-1dn a1 - 1n ban = 1 + s-1dn
an =
1 - n3
70 - 4n2
an =
n2 - 2n + 1
n - 1
an =
n + 3
n2 + 5n + 6
an =
1 - 5n4
n4 + 8n3
an =
2n + 1
1 - 32nan = 1 - 2n1 + 2n an =
n + s-1dn
nan = 2 + s0.1d
n
5an6
0, 1, 1, 2, 2, 3, 3, 4, Á
1, 0, 1, 0, 1, Á
1
25
, 
8
125
, 
27
625
, 
64
3125
, 
125
15,625
, Á
Squares of the positive
integers, with
alternating signs
Reciprocals of squares
of the positive integers,
with alternating signs
Powers of 2 divided by
multiples of 3
Integers differing by 2
divided by products of
consecutive integers
Squares of the positive
integers diminished by 1
Integers, beginning with -3
Every other odd positive
integer
Every other even positive
integer
Integers differing by 3
divided by factorials
Alternating 1’s and 0’s
Cubes of positive integers
divided by powers of 5
Each positive integer
repeated
7001_ThomasET_ch10p550-627.qxd 10/30/09 8:21 AM Page 559
10.1 Sequences 559
It is important to realize that Theorem 6 does not say that convergent sequences
are
monotonic. The sequence converges and is bounded, but it is not monotonic
since it alternates between positive and negative values as it tends toward zero. What the
theorem does say is that a nondecreasing sequence converges when it is bounded from
above, but it diverges to infinity otherwise.
{s-1dn+1>n}
Exercises 10.1
Finding Terms of a Sequence
Each of Exercises 1–6 gives a formula for the nth term of a se-
quence Find the values of and 
1. 2.
3. 4.
5. 6.
Each of Exercises 7–12 gives the first term or two of a sequence along
with a recursion formula for the remaining terms. Write out the first
ten terms of the sequence.
7.
8.
9.
10.
11.
12.
Finding a Sequence’s Formula
In Exercises 13–26, find a formula for the nth term of the sequence.
13. The sequence 1’s with alternating signs
14. The sequence 1’s with alternating signs
15. The sequence 
16. The sequence 
17.
18.
19. The sequence 
20. The sequence 
21. The sequence 
22. The sequence 
23. 5
1
, 
8
2
, 11
6
, 14
24
, 
17
120
, Á
2, 6, 10, 14, 18, Á
1, 5, 9, 13, 17, Á
-3, -2, -1, 0, 1, Á
0, 3, 8, 15, 24, Á
- 3
2
, - 1
6
, 1
12
, 
3
20
, 
5
30
, Á
1
9
, 2
12
, 2
2
15
, 2
3
18
, 2
4
21
, Á
1, - 1
4
, 1
9
, - 1
16
, 1
25
, Á
1, -4, 9, -16, 25, Á
-1, 1, -1, 1, -1, Á
1, -1, 1, -1, 1, Á
a1 = 2, a2 = -1, an+2 = an+1>ana1 = a2 = 1, an+2 = an+1 + an
a1 = -2, an+1 = nan>sn + 1da1 = 2, an+1 = s-1dn+1an>2
a1 = 1, an+1 = an>sn + 1da1 = 1, an+1 = an + s1>2nd
an =
2n - 1
2n
an =
2n
2n+1
an = 2 + s-1dnan =
s-1dn+1
2n - 1
an =
1
n!
an =
1 - n
n2
a4 .a1, a2, a3 ,5an6 . an 24.
25. The sequence 
26. The sequence 
Convergence and Divergence
Which of the sequences in Exercises 27–90 converge, and which
diverge? Find the limit of each convergent sequence.
27. 28.
29. 30.
31. 32.
33. 34.
35. 36.
37. 38.
39. 40.
41. 42.
43. 44.
45. 46.
47. 48.
49. 50.
51. 52.
53. 54.
55. 56.
57. 58.
59. 60. an = ln n - ln sn + 1dan =
ln n
n1>n
an = sn + 4d1>sn+4dan = a3n b1>n an = 2n n2an = 2n 10n
an = a1 - 1n bnan = a1 + 7n bn
an = s0.03d1>nan = 81>n
an =
ln n
ln 2n
an =
ln sn + 1d2n an =
3n
n3
an =
n
2n
an =
sin2 n
2n
an =
sin n
n
an = np cos snpdan = sin ap2 + 1n b
an =
1
s0.9dnan = A 2nn + 1 an = a- 12 b
n
an =
s-1dn+1
2n - 1
an = a2 - 12n b a3 + 12n ban = an + 12n b a1 - 1n b
an = s-1dn a1 - 1n ban = 1 + s-1dn
an =
1 - n3
70 - 4n2
an =
n2 - 2n + 1
n - 1
an =
n + 3
n2 + 5n + 6
an =
1 - 5n4
n4 + 8n3
an =
2n + 1
1 - 32nan = 1 - 2n1 + 2n an =
n + s-1dn
nan = 2 + s0.1d
n
5an6
0, 1, 1, 2, 2, 3, 3, 4, Á
1, 0, 1, 0, 1, Á
1
25
, 
8
125
, 
27
625
, 
64
3125
, 
125
15,625
, Á
Squares of the positive
integers, with
alternating signs
Reciprocals of squares
of the positive integers,
with alternating signs
Powers of 2 divided by
multiples of 3
Integers differing by 2
divided by products of
consecutive integers
Squares of the positive
integers diminished by 1
Integers, beginning with -3
Every other odd positive
integer
Every other even positive
integer
Integers differing by 3
divided by factorials
Alternating 1’s and 0’s
Cubes of positive integers
divided by powers of 5
Each positive integer
repeated
7001_ThomasET_ch10p550-627.qxd 10/30/09 8:21 AM Page 559
4. Quais das sequeˆncias {an} nos exerc´ıcios 27-90 convergem, e quais divergem? Obtenha o limite de
cada uma das sequeˆncias.
10.1 Sequences 559
It is important to realize that Theorem 6 does not say that convergent sequences are
monotonic. The sequence converges and is bounded, but it is not monotonic
since it alternates between positive and negative values as it tends toward zero. What the
theorem does say is that a nondecreasing sequence converges when it is bounded from
above, but it diverges to infinity otherwise.
{s-1dn+1>n}
Exercises 10.1
Finding Terms of a Sequence
Each of Exercises 1–6 gives a formula for the nth term of a se-
quence Find the values of and 
1. 2.
3. 4.
5. 6.
Each of Exercises 7–12 gives the first term or two of a sequence along
with a recursion formula for the remaining terms. Write out the first
ten terms of the sequence.
7.
8.
9.
10.
11.
12.
Finding a Sequence’s Formula
In Exercises 13–26, find a formula for the nth term of the sequence.
13. The sequence 1’s with alternating signs
14. The sequence 1’s with alternating signs
15. The sequence 
16. The sequence 
17.
18.
19. The sequence 
20. The sequence 
21. The sequence 
22. The sequence 
23. 5
1
, 
8
2
, 11
6
, 14
24
, 
17
120
, Á
2, 6, 10, 14, 18, Á
1, 5, 9, 13, 17, Á
-3, -2, -1, 0, 1, Á
0, 3, 8, 15, 24, Á
- 3
2
, - 1
6
, 1
12
, 
3
20
, 
5
30
, Á
1
9
, 2
12
, 2
2
15
, 2
3
18
, 2
4
21
, Á
1, - 1
4
, 1
9
, - 1
16
, 1
25
, Á
1, -4, 9, -16, 25, Á
-1, 1, -1, 1, -1, Á
1, -1, 1, -1, 1, Á
a1 = 2, a2 = -1, an+2 = an+1>ana1 = a2 = 1, an+2 = an+1 + an
a1 = -2, an+1 = nan>sn + 1da1 = 2, an+1 = s-1dn+1an>2
a1 = 1, an+1 = an>sn + 1da1 = 1, an+1 = an + s1>2nd
an =
2n - 1
2n
an =
2n
2n+1
an = 2 + s-1dnan =
s-1dn+1
2n - 1
an =
1
n!
an =
1 - n
n2
a4 .a1, a2, a3 ,5an6 . an 24.
25. The sequence 
26. The sequence 
Convergence and Divergence
Which of the sequences in Exercises 27–90 converge, and which
diverge? Find the limit of each convergent sequence.
27. 28.
29. 30.
31. 32.
33. 34.
35. 36.
37. 38.
39. 40.
41. 42.
43. 44.
45. 46.
47. 48.
49. 50.
51. 52.
53. 54.
55. 56.
57. 58.
59. 60. an = ln n - ln sn + 1dan =
ln n
n1>n
an = sn + 4d1>sn+4dan = a3n b1>n an = 2n n2an = 2n 10n
an = a1 - 1n bnan = a1 + 7n bn
an = s0.03d1>nan = 81>n
an =
ln n
ln 2n
an =
ln sn + 1d2n an =
3n
n3
an =
n
2n
an =
sin2 n
2n
an =
sin n
n
an = np cos snpdan = sin ap2 + 1n b
an =
1
s0.9dnan = A 2nn + 1 an = a- 12 b
n
an =
s-1dn+1
2n - 1
an = a2 - 12n b a3 + 12n ban = an + 12n b a1 - 1n b
an = s-1dn a1 - 1n ban = 1 + s-1dn
an =
1 - n3
70 - 4n2
an =
n2 - 2n + 1
n - 1
an =
n + 3
n2 + 5n + 6
an =
1 - 5n4
n4 + 8n3
an =
2n + 1
1 - 32nan = 1 - 2n1 + 2n an =
n + s-1dn
nan = 2 + s0.1d
n
5an6
0, 1, 1, 2, 2, 3, 3, 4, Á
1, 0, 1, 0, 1, Á
1
25
, 
8
125
, 
27
625
, 
64
3125
, 
125
15,625
, Á
Squares of the positive
integers, with
alternating signs
Reciprocals of squares
of the positive integers,
with alternating signs
Powers of 2 divided by
multiples of 3
Integers differing by 2
divided by products of
consecutive integers
Squares of the positive
integers diminished by 1
Integers, beginning with -3
Every other odd positive
integer
Every other even positive
integer
Integers differing by 3
divided by factorials
Alternating 1’s and 0’s
Cubes of positive integers
divided by powers of 5
Each positive integer
repeated
7001_ThomasET_ch10p550-627.qxd 10/30/09 8:21 AM Page 559
61. 62.
63. (Hint: Compare with 1 n.)
64. 65.
66. 67.
68. 69.
70. 71.
72. 73.
74. 75.
76. 77.
78. 79.
80. 81.
82. 83.
84. 85.
86. 87.
88.
89. 90.
Recursively Defined Sequences
In Exercises 91–98, assume that each sequence converges and find its
limit.
91.
92.
93.
94.
95.
96.
97.
98. 51 + 41 + 31 + 21, Á21, 31 + 21, 41 + 31 + 21,
2, 2 + 1
2
, 2 + 1
2 + 1
2
, 2 + 1
2 + 1
2 + 1
2
, Á
a1 = 3, an+1 = 12 - 2ana1 = 5, an+1 = 25ana1 = 0, an+1 = 28 + 2an
a1 = -4, an+1 = 28 + 2ana1 = -1, an+1 = an + 6an +
2
a1 = 2, an+1 =
72
1 + an
an = L
n
1
 
1
xp
 dx, p 7 1an =
1
nL
n
1
 
1
x dx
an =
12n2 - 1 - 2n2 + n an = n - 2n
2 - nan =
sln nd52n an = sln nd
200
nan = 2n n2 + n an = a13 b
n
+ 122nan = 12n tan-1 n an = tan-1 nan = s3n + 5nd1>n
an = 2n sin 12nan = n a1 - cos 1n b an = n22n - 1 sin 1nan = sinh sln nd
an = tanh nan =
s10>11dn
s9/10dn + s11/12dn
an =
3n # 6n
2-n # n!an = a1 - 1n2 bn
an = a xn2n + 1 b1>n, x 7 0an = a nn + 1 bn
an = a3n + 13n - 1 bnan = ln a1 + 1n bn
an = a1n b1>sln ndan = n!2n # 3n
an =
n!
106n
an =
s-4dn
n!
>an = n!nn
an = 2n 32n+1an = 2n 4nn560 Chapter 10: Infinite Sequences and Series Theory and Examples
99. The first term of a sequence is Each succeeding term is
the sum of all those that come before it:
Write out enough early terms of the sequence to deduce a gen-
eral formula for that holds for 
100. A sequence of rational numbers is described as follows:
.
Here the numerators form one sequence, the denominators form
a second sequence, and their ratios form a third sequence. Let 
and be, respectively, the numerator and the denominator of
the nth fraction 
a. Verify that and, more
generally, that if or then
respectively.
b. The fractions approach a limit as n increases.
What is that limit? (Hint: Use part (a) to show that
and that is not less than n.)
101. Newton’s method The following sequences come from the re-
cursion formula for Newton’s method,
Do the sequences converge? If so, to what value? In each case,
begin by identifying the function ƒ that generates the sequence.
a.
b.
c.
102. a. Suppose that ƒ(x) is differentiable for all x in [0, 1] and that
Define sequence by the rule 
Show that Use the result in part (a) to
find the limits of the following sequences 
b. c.
d.
103. Pythagorean triples A triple of positive integers a, b, and c is
called a Pythagorean triple if Let a be an odd
positive integer and let
be, respectively, the integer floor and ceiling for a2>2.b = j
a2
2
k and c = l a2
2
m
a2 + b2 = c2 .
an = n ln a1 + 2n b
an = nse1>n - 1dan = n tan-1 1n
5an6 .lim n:q an = ƒ¿s0d.
nƒs1>nd .an =5an6ƒs0d = 0.
x0 = 1, xn+1 = xn - 1
x0 = 1, xn+1 = xn -
tan xn - 1
sec2 xn
x0 = 1, xn+1 = xn -
xn2 - 2
2xn
=
xn
2
+ 1xn
xn+1 = xn -
ƒsxnd
ƒ¿sxnd
.
ynrn2 - 2 = ; s1>ynd2rn = xn>yn
sa + 2bd2 - 2sa + bd2 = +1 or -1,
+1,a2 - 2b2 = -1
x12 - 2y12 = -1, x22 - 2y22 = +1
rn = xn>yn .yn xn
1
1
, 
3
2
, 
7
5, 
17
12
, Á , a
b
, 
a + 2b
a + b , Á
n Ú 2.xn
xn+1 = x1 + x2 + Á + xn .
x1 = 1.
7001_ThomasET_ch10p550-627.qxd 10/30/09 8:21 AM Page 560
5. Nos exercc¸ios 91-98, suponha que cada uma das sequeˆncias e´ convergente e obtenha o limite.
61. 62.
63. (Hint: Compare with 1 n.)
64. 65.
66. 67.
68. 69.
70. 71.
72. 73.
74. 75.
76. 77.
78. 79.
80. 81.
82. 83.
84. 85.
86. 87.
88.
89. 90.
Recursively Defined Sequences
In Exercises 91–98, assume that each sequence converges and find its
limit.
91.
92.
93.
94.
95.
96.
97.
98. 51 + 41 + 31 + 21, Á21, 31 + 21, 41 + 31 + 21,
2, 2 + 1
2
, 2 + 1
2 + 1
2
, 2 + 1
2 + 1
2 + 1
2
, Á
a1 = 3, an+1 = 12 - 2ana1 = 5, an+1 = 25ana1 = 0, an+1 = 28 + 2an
a1 = -4, an+1 = 28 + 2ana1 = -1, an+1 = an + 6an + 2
a1 = 2, an+1 =
72
1 + an
an = L
n
1
 
1
xp
 dx, p 7 1an =
1
nL
n
1
 
1
x dx
an =
12n2 - 1 - 2n2 + n an = n - 2n
2 - nan =
sln nd52n an = sln nd
200
nan = 2n n2 + n an = a13 b
n
+ 122nan = 12n tan-1 n an = tan-1 nan = s3n + 5nd1>n
an = 2n sin 12nan = n a1 - cos 1n b an = n22n - 1 sin 1nan = sinh sln nd
an = tanh nan =
s10>11dn
s9/10dn + s11/12dn
an =
3n # 6n
2-n # n!an = a1 - 1n2 bn
an = a xn2n + 1 b1>n, x 7 0an = a nn + 1 bn
an = a3n + 13n - 1 bnan = ln a1 + 1n bn
an = a1n b1>sln ndan = n!2n # 3n
an =
n!
106n
an =
s-4dn
n!
>an = n!nn
an = 2n 32n+1an = 2n 4nn560 Chapter 10: Infinite Sequences and Series Theory and Examples
99. The first term of a sequence is Each succeeding term is
the sum of all those that come before it:
Write out enough early terms of the sequence to deduce a gen-
eral formula for that holds for 
100. A sequence of rational numbers is described as follows:
.
Here the numerators form one sequence, the denominators form
a second sequence, and their ratios form a third sequence. Let 
and be, respectively, the numerator and the denominator of
the nth fraction 
a. Verify that and, more
generally, that if or then
respectively.
b. The fractions approach a limit as n increases.
What is that limit? (Hint: Use part (a) to show that
and that is not less than n.)
101. Newton’s method The following sequences come from the re-
cursion formula for Newton’s method,
Do the sequences converge? If so, to what value? In each case,
begin by identifying the function ƒ that generates the sequence.
a.
b.
c.
102. a. Suppose that ƒ(x) is differentiable for all x in [0, 1] and that
Define sequence by the rule 
Show that Use the result in part (a) to
find the limits of the following sequences 
b. c.
d.
103. Pythagorean triples A triple of positive integers a, b, and c is
called a Pythagorean triple if Let a be an odd
positive integer and let
be, respectively, the integer floor and ceiling for a2>2.b = j
a2
2
k and c = l a2
2
m
a2 + b2 = c2 .
an = n ln a1 + 2n b
an = nse1>n - 1dan = n tan-1 1n
5an6 .lim n:q an = ƒ¿s0d.
nƒs1>nd .an =5an6ƒs0d = 0.
x0 = 1, xn+1 = xn - 1
x0 = 1, xn+1 = xn -
tan xn - 1
sec2 xn
x0 = 1, xn+1 = xn -
xn2 - 2
2xn
=
xn
2
+ 1xn
xn+1 = xn -
ƒsxnd
ƒ¿sxnd
.
ynrn2 - 2 = ; s1>ynd2rn = xn>yn
sa + 2bd2 - 2sa + bd2 = +1 or -1,
+1,a2 - 2b2 = -1
x12 - 2y12 = -1, x22 - 2y22 = +1
rn = xn>yn .yn xn
1
1
, 
3
2
, 
7
5, 
17
12
, Á , a
b
, 
a + 2b
a + b , Á
n Ú 2.xn
xn+1 = x1 + x2 + Á + xn .
x1 = 1.
7001_ThomasET_ch10p550-627.qxd 10/30/09 8:21 AM Page 560
6. Verifique se a sequeˆncia e´ mono´tona e se e´ limitada
10.1 Sequences 561
a. Show that (Hint: Let and express
b and c in terms of n.)
b. By direct calculation, or by appealing to the accompanying
figure, find
104. The nth root of n!
a. Show that and hence, using Stirling’s
approximation (Chapter 8, Additional Exercise 32a), that
b. Test the approximation in part (a) for as
far as your calculator will allow.
105. a. Assuming that if c is any positive con-
stant, show that
if c is any positive constant.
b. Prove that if c is any positive constant.
(Hint: If and how large should N be to
ensure that if )
106. The zipper theorem Prove the “zipper theorem” for se-
quences: If and both converge to L, then the sequence
converges to L.
107. Prove that 
108. Prove that 
109. Prove Theorem 2. 110. Prove Theorem 3.
In Exercises 111–114, determine if the sequence is monotonic and if it
is bounded.
111. 112.
113. 114.
Which of the sequences in Exercises 115–124 converge, and which di-
verge? Give reasons for your answers.
115. 116. an = n -
1
nan = 1 -
1
n
an = 2 -
2
n -
1
2n
an =
2n3n
n!
an =
s2n + 3d!
sn + 1d!an =
3n + 1
n + 1
limn:q x1>n = 1, sx 7 0d .limn:q2n n = 1.
a1, b1, a2 , b2 , Á , an , bn , Á
5bn65an6
n 7 N?ƒ 1>nc - 0 ƒ 6 Pc = 0.04 ,P = 0.001limn:q s1>n
cd = 0
lim
n:q 
ln n
nc
= 0
limn:q s1>ncd = 0
n = 40, 50, 60, Á ,
2n n! L ne for large values of n .limn:q s2npd1>s2nd = 1
lim
a:q 
j a2
2
k
l a2
2
m .
a = 2n + 1a2 + b2 = c2 .
a
⎡⎢⎢ a22
⎢⎢⎣
⎢⎢⎣
⎡⎢⎢
a2
2
!
117. 118.
119.
120. The first term of a sequence is The next terms are
or cos (2), whichever is larger; and or cos (3),
whichever is larger (farther to the right). In general,
121.
122.
123.
124.
125. The sequence { ( )} has a least upper bound of 1
Show that if M is a number less than 1, then the terms of
eventually exceed M. That is, if there is an
integer N such that whenever Since
for every n, this proves that 1 is a least upper
bound for 
126. Uniqueness of least upper bounds Show that if and 
are least upper bounds for the sequence then 
That is, a sequence cannot have two different least upper bounds.
127. Is it true that a sequence of positive numbers must con-
verge if it is bounded from above? Give reasons for your answer.
128. Prove that if is a convergent sequence, then to every posi-
tive number there corresponds an integer N such that for all m
and n,
129. Uniqueness of limits Prove that limits of sequences are
unique. That is, show that if and are numbers such that
and then 
130. Limits and subsequences If the terms of one sequence appear
in another sequence in their given order, we call the first se-
quence a subsequence of the second. Prove that if two sub-
sequences of a sequence have different limits 
then diverges.
131. For a sequence the terms of even index are denoted by 
and the terms of odd index by Prove that if and
then 
132. Prove that a sequence converges to 0 if and only if the se-
quence of absolute values converges to 0.
133. Sequences generated by Newton’s method Newton’s method,
applied to a differentiable function ƒ(x), begins with a starting
value and constructs from it a sequence of numbers that
under favorable circumstances converges to a zero of ƒ. The
recursion formula for the sequence is
a. Show that the recursion formula for 
can be written as 
b. Starting with and calculate successive terms
of the sequence until the display begins to repeat. What num-
ber is being approximated? Explain.
a = 3,x0 = 1
xn+1 = sxn + a>xnd>2.ƒsxd = x2 - a, a 7 0,
xn+1 = xn -
ƒsxnd
ƒ¿sxnd
.
5xn6x0
5ƒ an ƒ65an6
an: L .a2k+1: L ,
a2k: La2k+1 .
a2k5an65an6
L1 Z L2 ,5an6
L1 = L2 .an: L2 ,an: L1
L2L1
m 7 N and n 7 N Q ƒ am - an ƒ 6 P .
P
5an6
5an6
M1 = M2 .5an6 , M2M1
5n>sn + 1d6 .n>sn + 1d 6 1 n 7 N .n>sn + 1d 7 M
M 6 15n>sn + 1d6 n + 1n/
a1 = 1, an+1 = 2an - 3
an =
4n+1 + 3n
4n
an =
n + 1
nan =
1 + 22n2nxn+1 = max 5xn , cos sn + 1d6 .
x3 = x2x2 = x1
x1 = cos s1d .
an = ss-1dn + 1d an + 1n b
an =
2n - 1
3n
an =
2n - 1
2n
T
T
7001_ThomasET_ch10p550-627.qxd 10/30/09 8:21 AM Page 561
7. Me´todo de Newton: A seguinte sequeˆncia segue da fo´rmula do Me´todo de Newton
xn+1 = xn − f(xn)
f ′(xn)
. As sequeˆncias convergem? Se sim, para qual valor?Em cada caso, identifique a
2
func¸a˜o f que gera a sequeˆncia.
(a)x + 0 = 1, xn+1 = xn − x
2
n − 2
2xn
=
xn
2
+
1
xn
; (b)x0 = 1, xn+1 = xn − tan(xn)− 1
sec2(xn)
;
(c)x0 = 1, xn+1 = xn − 1.
8. Respostas
Chapter 10: Answers to Odd-Numbered Exercises A-39
(c) when or
when or
By picking and , we insure that an equilib-
rium point exists inside the first quadrant.
Practice Exercises, pp. 547–548
1.
3. 5.
7. 9.
11. 13. 15.
17. 19.
21.
23.
25.
27.
(a)
(b) Note that we choose a small interval of x-values because the
y-values decrease very rapidly and our calculator cannot
handle the calculations for (This occurs because
the analytic solution is which has
an asymptote at Obviously, the Euler
approximations are misleading for ) 
29. exact value is .
31. exact value is
-e3>2 L -4.4817.ysexactd = -e sx
2-1d>2; y(2) L -3.4192;
1
2
ysexactd = 1
2
 x2 - 3
2
; y(2) L 0.4 ;
[–1, 0.2] by [–10, 2]
x … -0.7 .
x = - ln 2 L -0.69 .
y = -2 + lns2 - e-xd ,
x … -1.
[–0.2, 4.5] by [–2.5, 0.5]
y(3) L 0.8981
y = e-x s3x3 - 3x2d
y = 1
3
 s1 - 4e-x3dy = 2x
3 + 3x2 + 6
6sx + 1d2
xy + y3 = Cy = e
-x + C
1 + exy =
x2 - 2x + C
2x2
y = x
2
4
 ex>2 + Cex>2y = C x - 1x
s y + 1de-y = - ln ƒ x ƒ + Ctan y = -x sin x - cos x + C
y = - ln aC - 25 sx - 2d5>2 - 43 sx - 2d3>2b
m>n 7 k1a>b 7 k2 y = k2 -
k2n
m x.y = 0
dy
dt
= 0
y = a
b
- a
bk1
x,x = 0dx
dt
= 0 33. (a) is stable and is unstable.
(b)
(c)
Additional and Advanced Exercises, pp. 548–549
1. (a)
(b) Steady-state solution: 
5.
7.
9.
CHAPTER 10
Section 10.1, pp. 559–562
1.
3.
5.
7.
9.
11. 1, 1, 2, 3, 5, 8, 13, 21, 34, 55 13.
15. 17.
19. 21.
23. 25.
27. Converges, 2 29. Converges, 31. Converges, 
33. Diverges 35. Diverges 37. Converges, 1 2
39. Converges, 0 41. Converges, 43. Converges, 1
45. Converges, 0 47. Converges, 0 49. Converges, 0
51. Converges, 1 53. Converges, 55. Converges, 1
57. Converges, 1 59. Diverges 61. Converges, 4
63. Converges, 0 65. Diverges 67. Converges, 
69. Converges, 71. Converges, 
73. Converges, 0 75. Converges, 1 77. Converges, 1 2
79. Converges, 1 81. Converges, 83. Converges, 0
85. Converges, 0 87. Converges, 1 2 89. Converges, 0
91. 8 93. 4 95. 5 97. 99. xn = 2n-21 + 22>p>2 >
x sx 7 0de2>3 e-1
e7
22 > -5-1
an =
1 + s-1dn+1
2
, n Ú 1an =
3n + 2
n!
, n Ú 1
an = 4n - 3, n Ú 1an = n2 - 1, n Ú 1
an =
2n-1
3(n + 2), n Ú 1an = s-1d
n+1snd2, n Ú 1
an = s-1dn+1, n Ú 1
2, 1, - 1
2
, - 1
4
, 1
8
, 1
16
, - 1
32
, - 1
64
, 1
128
, 1
256
1, 
3
2
, 
7
4
, 
15
8
, 
31
16
, 
63
32
, 
127
64
, 
255
128
, 
511
256
, 
1023
512
a1 = 1>2, a2 = 1>2, a3 = 1>2, a4 = 1>2a1 = 1, a2 = -1>3, a3 = 1>5, a4 = -1>7
a1 = 0, a2 = -1>4, a3 = -2>9, a4 = -3>16
ln ƒ x ƒ - ln ƒ sec (y>x - 1) + tan (y>x - 1) ƒ = Cln ƒ x ƒ + e-y>x = C
x2 sx2 + 2y2d = C
yq = c
y = c + s y0 - cde-k sA>V dt
y
2 
1
x
1
0
0.5 1.5 2.52
–1
–2
dy
dx > 0
dy
dx < 0
dydy
dx < 0dx > 0
dx2
> 0d
2y
dx2
> 0 d
2y d2y
dx2
< 0d
2y
dx2
< 0
y
y = 1
y = 0
y = –1
d2y
dx2
= 2y 
dy
dx
= 2ysy2 - 1d
y = 1y = -1
x y
0 0
0.1 0.1000
0.2 0.2095
0.3 0.3285
0.4 0.4568
0.5 0.5946
0.6 0.7418
0.7 0.8986
0.8 1.0649
0.9 1.2411
1.0 1.4273
x y
1.1 1.6241
1.2 1.8319
1.3 2.0513
1.4 2.2832
1.5 2.5285
1.6 2.7884
1.7 3.0643
1.8 3.3579
1.9 3.6709
2.0 4.0057
7001_ThomasET_OddAnsCh8-app.qxd 11/3/09 3:01 PM Page 39
Chapter 10: Answers to Odd-Numbered Exercises A-39
(c) when or
when or
By picking and , we insure that an equilib-
rium point exists inside the first quadrant.
Practice Exercises, pp. 547–548
1.
3. 5.
7. 9.
11. 13. 15.
17. 19.
21.
23.
25.
27.
(a)
(b) Note that we choose a small interval of x-values because the
y-values decrease very rapidly and our calculator cannot
handle the calculations for (This occurs because
the analytic solution is which has
an asymptote at Obviously, the Euler
approximations are misleading for ) 
29. exact value is .
31. exact value is
-e3>2 L -4.4817.ysexactd = -e sx
2-1d>2; y(2) L -3.4192;
1
2
ysexactd = 1
2
 x2 - 3
2
; y(2) L 0.4 ;
[–1, 0.2] by [–10, 2]
x … -0.7 .
x = - ln 2 L -0.69 .
y = -2 + lns2 - e-xd ,
x … -1.
[–0.2, 4.5] by [–2.5, 0.5]
y(3) L 0.8981
y = e-x s3x3 - 3x2d
y = 1
3
 s1 - 4e-x3dy = 2x
3 + 3x2 + 6
6sx + 1d2
xy + y3 = Cy = e
-x + C
1 + exy =
x2 - 2x + C
2x2
y = x
2
4
 ex>2 + Cex>2y = C x - 1x
s y + 1de-y = - ln ƒ x ƒ + Ctan y = -x sin x - cos x + C
y = - ln aC - 25 sx - 2d5>2 - 43 sx - 2d3>2b
m>n 7 k1a>b 7 k2 y = k2 -
k2n
m x.y = 0
dy
dt
= 0
y = a
b
- a
bk1
x,x = 0dx
dt
= 0 33. (a) is stable and is unstable.
(b)
(c)
Additional and Advanced Exercises, pp. 548–549
1. (a)
(b) Steady-state solution: 
5.
7.
9.
CHAPTER 10
Section 10.1, pp. 559–562
1.
3.
5.
7.
9.
11. 1, 1, 2, 3, 5, 8, 13, 21, 34, 55 13.
15. 17.
19. 21.
23. 25.
27. Converges, 2 29. Converges, 31. Converges, 
33.
Diverges 35. Diverges 37. Converges, 1 2
39. Converges, 0 41. Converges, 43. Converges, 1
45. Converges, 0 47. Converges, 0 49. Converges, 0
51. Converges, 1 53. Converges, 55. Converges, 1
57. Converges, 1 59. Diverges 61. Converges, 4
63. Converges, 0 65. Diverges 67. Converges, 
69. Converges, 71. Converges, 
73. Converges, 0 75. Converges, 1 77. Converges, 1 2
79. Converges, 1 81. Converges, 83. Converges, 0
85. Converges, 0 87. Converges, 1 2 89. Converges, 0
91. 8 93. 4 95. 5 97. 99. xn = 2n-21 + 22>p>2 >
x sx 7 0de2>3 e-1
e7
22 > -5-1
an =
1 + s-1dn+1
2
, n Ú 1an =
3n + 2
n!
, n Ú 1
an = 4n - 3, n Ú 1an = n2 - 1, n Ú 1
an =
2n-1
3(n + 2), n Ú 1an = s-1d
n+1snd2, n Ú 1
an = s-1dn+1, n Ú 1
2, 1, - 1
2
, - 1
4
, 1
8
, 1
16
, - 1
32
, - 1
64
, 1
128
, 1
256
1, 
3
2
, 
7
4
, 
15
8
, 
31
16
, 
63
32
, 
127
64
, 
255
128
, 
511
256
, 
1023
512
a1 = 1>2, a2 = 1>2, a3 = 1>2, a4 = 1>2a1 = 1, a2 = -1>3, a3 = 1>5, a4 = -1>7
a1 = 0, a2 = -1>4, a3 = -2>9, a4 = -3>16
ln ƒ x ƒ - ln ƒ sec (y>x - 1) + tan (y>x - 1) ƒ = Cln ƒ x ƒ + e-y>x = C
x2 sx2 + 2y2d = C
yq = c
y = c + s y0 - cde-k sA>V dt
y
2 
1
x
1
0
0.5 1.5 2.52
–1
–2
dy
dx > 0
dy
dx < 0
dydy
dx < 0dx > 0
dx2
> 0d
2y
dx2
> 0 d
2y d2y
dx2
< 0d
2y
dx2
< 0
y
y = 1
y = 0
y = –1
d2y
dx2
= 2y 
dy
dx
= 2ysy2 - 1d
y = 1y = -1
x y
0 0
0.1 0.1000
0.2 0.2095
0.3 0.3285
0.4 0.4568
0.5 0.5946
0.6 0.7418
0.7 0.8986
0.8 1.0649
0.9 1.2411
1.0 1.4273
x y
1.1 1.6241
1.2 1.8319
1.3 2.0513
1.4 2.2832
1.5 2.5285
1.6 2.7884
1.7 3.0643
1.8 3.3579
1.9 3.6709
2.0 4.0057
7001_ThomasET_OddAnsCh8-app.qxd 11/3/09 3:01 PM Page 39
111-114. Resposta: Na˜o decrescente e limitada
3