<< стр. 12(всего 14)СОДЕРЖАНИЕ >>
Then F (b) в€’ F (a) represents the percentage of normally distributed data that
lies between a and b. This percentage is given by
b
f (x)dx.
a

Furthermore,
Вµ+bПѓ b
1 2
eв€’x /2 dx.
в€љ
f (x)dx =
2ПЂ
Вµ+aПѓ a

Proof. The proof of this theorem is omitted.

Exercises 7.1 None available.

7.2 Discontinuities at End Points
Deп¬Ѓnition 7.2.1 (i) Suppose that f is continuous on [a, b) and

lim f (x) = +в€ћ or в€’ в€ћ.
xв†’bв€’

Then, we deп¬Ѓne
b x
f (x)dx = lim f (x)dx.
в€’xв†’b
a a
If the limit exists, we say that the improper integral converges; otherwise we
say that it diverges.
(ii) Suppose that f is continuous on (a, b] and

lim f (x) = +в€ћ or в€’ в€ћ.
xв†’a+

Then we deп¬Ѓne,
b b
f (x)dx = lim+ f (x)dx.
xв†’a
a x
If the limit exists, we say that the improper integral converges; otherwise we
say that it diverges.

Exercises 7.2

1. Suppose that f is continuous on (в€’в€ћ, в€ћ) and g (x) = f (x). Then deп¬Ѓne
each of the following improper integrals:
300CHAPTER 7. IMPROPER INTEGRALS AND INDETERMINATE FORMS

+в€ћ
(a) f (x)dx
a
b
(b) f (x)dx
в€’в€ћ
+в€ћ
(c) f (x)dx
в€’в€ћ

2. Suppose that f is continuous on the open interval (a, b) and g (x) = f (x)
on (a, b). Deп¬Ѓne each of the following improper integrals if f is not
continuous at a or b:
x
f (x)dx, a в‰¤ x < b
(a)
a
b
f (x)dx, a < x в‰¤ b
(b)
x
b
(c) f (x)dx
a

+в€ћ
eв€’x dx = 1
3. Prove that
0

1
ПЂ
1
в€љ dx =
4. Prove that
2
1 в€’ x2
0

+в€ћ
1
5. Prove that dx = ПЂ
1 + x2
в€’в€ћ

в€ћ
1 1
6. Prove that dx = , if and only if p > 1.
xp pв€’1
1

в€ћ
+в€ћ
в€’x2 2
eв€’x dx. Use the comparison between
7. Show that e dx = 2
в€’в€ћ 0
+в€ћ
2
2
eв€’x dx exists.
eв€’x and eв€’x . Show that
в€’в€ћ

1
dx
8. Prove that converges if and only if p < 1.
xp
0
7.2. DISCONTINUITIES AT END POINTS 301

+в€ћ
eв€’x sin(2x)dx.
9. Evaluate
0
+в€ћ
eв€’4x cos(3x)dx.
10. Evaluate
0
+в€ћ
x2 eв€’x dx.
11. Evaluate
0
+в€ћ
xeв€’x dx.
12. Evaluate
0
в€ћ
13. Prove that sin(2x)dx diverges.
0
в€ћ
14. Prove that cos(3x)dx diverges.
0

15. Compute the volume of the solid generated when the area between the
2
graph of y = eв€’x and the x-axis is rotated about the y-axis.

16. Compute the volume of the solid generated when the area between the
graph of y = eв€’x , 0 в‰¤ x < в€ћ and the x-axis is rotated

1
17. Let A represent the area bounded by the graph y = , 1 в‰¤ x < в€ћ
x
and the x-axis. Let V denote the volume generated when the area A is

(a) show that A is +в€ћ
(b) show that V = ПЂ
(c) show that the surface area of V is +в€ћ.
(d) Is it possible to п¬Ѓll the volume V with paint and not be able to paint
its surface? Explain.

18. Let A represent the area bounded by the graph of y = eв€’2x , 0 в‰¤ x < в€ћ,
and y = 0.
302CHAPTER 7. IMPROPER INTEGRALS AND INDETERMINATE FORMS

(a) Compute the area of A.
(b) Compute the volume generated when A is rotated about the x-axis.

(c) Compute the volume generated when A is rotated about the y-axis.

+в€ћ +в€ћ
sin x
2
в€љ dx.
19. Assume that sin(x )dx = (ПЂ/8). Compute
x
0 0

+в€ћ в€љ
2
eв€’x = ПЂ.
20. It is known that
в€’в€ћ

+в€ћ
2
eв€’x dx.
(a) Compute
0
+в€ћ
eв€’x
в€љ dx.
(b) Compute
x
0
+в€ћ
2
eв€’4x dx.
(c) Compute
0

Deп¬Ѓnition 7.2.2 Suppose that f (t) is continuous on [0, в€ћ) and there exist
some constants a > 0, M > 0 and T > 0 such that |f (t)| < M eat for all
t в‰Ґ T . Then we deп¬Ѓne the Laplace transform of f (t), denoted L{f (t)}, by
в€ћ
eв€’st f (t)dt
L{f (t)} =
0

for all s в‰Ґ s0 . In problems 21вЂ“34, compute L{f (t)} for the given f (t).

1 if t в‰Ґ 0
22. f (t) = t
21. f (t) =
0 if t < 0

23. f (t) = t2 24. f (t) = t3

25. f (t) = tn , n = 1, 2, 3, В· В· В· 26. f (t) = ebt

27. f (t) = tebt 28. f (t) = tn ebt , n = 1, 2, 3, В· В· В·
7.2. DISCONTINUITIES AT END POINTS 303

eat в€’ ebt aeat в€’ bebt
29. f (t) = 30. f (t) =
aв€’b aв€’b
1
31. f (t) = sin(bt) 32. f (t) = cos(bt)
b
1
33. f (t) = sinh(bt) 34. f (t) = cosh(bt)
b

Deп¬Ѓnition 7.2.3 For x > 0, we deп¬Ѓne the Gamma function О“(x) by
+в€ћ
txв€’1 eв€’t dt.
О“(x) =
0

+в€ћ
1в€љ
2
eв€’x =
In problems 35вЂ“40 assume that О“(x) exists for x > 0 and ПЂ.
2
0
в€љ
35. Show that О“(1/2) = ПЂ 36. Show that О“(1) = 1
в€љ
3 ПЂ
37. Prove that О“(x + 1) = xО“(x) 38. Show that О“ =
2 2

3в€љ
5
39. Show that О“ = ПЂ 40. Show that О“(n + 1) = n!
2 4

In problems 41вЂ“60, evaluate the given improper integrals.
+в€ћ +в€ћ
dx
в€’x2
41. 2xe dx 42.
x3/2
0 1

+в€ћ +в€ћ
dx 4x
43. 44. dx
x5/2 1 + x2
4 1

+в€ћ +в€ћ
x 4
45. dx 46. dx
(1 + x2 )3/2 x2 в€’ 4
1 16

+в€ћ +в€ћ
1 1
47. dx 48. dx, p > 1
x(ln x)2 x(ln x)p
2 2
304CHAPTER 7. IMPROPER INTEGRALS AND INDETERMINATE FORMS

1 2
в€’x2
ex dx
49. 3xe dx 50.
в€’в€ћ в€’в€ћ

в€ћ в€ћ
2 dx
51. dx 52.
ex + eв€’x x2 + 9
в€’в€ћ
0

2 4
dx x
в€љ в€љ
53. 54. dx
4 в€’ x2 16 в€’ x2
0 0

+в€ћ
5
dx
x
в€љ
dx 56.
55.
(25 в€’ x2 )2/3 x x2 в€’ 4
2
0
в€љ
в€ћ
+в€ћ в€’x
e dx
в€љ dx в€љ
57. 58.
x x(x + 25)
0 0

в€ћ +в€ћ
eв€’x 3
x2 eв€’x dx
59. dx 60.
1 в€’ (eв€’x )2
0 0

7.3
Theorem 7.3.1 (Cauchy Mean Value Theorem) Suppose that two functions
f and g are continuous on the closed interval [a, b], diп¬Ђerentiable on the open
interval (a, b) and g (x) = 0 on (a, b). Then there exists at least one number
c such that a < c < b and
f (b) в€’ f (a)
f (c)
= .
g(b) в€’ g(a)
g (c)

Proof. See the proof of Theorem 4.1.6.

Theorem 7.3.2 Suppose that f and g are continuous and diп¬Ђerentiable on
an open interval (a, b) and a < c < b. If f (c) = g(c) = 0, g (x) = 0 on (a, b)
and
f (x)
=L
lim
xв†’c g (x)

then
f (x)
lim = L.
g(x)
xв†’c
7.3. 305

Proof. See the proof of Theorem 4.1.7.

Theorem 7.3.3 (LвЂ™HЛ†pitalвЂ™s Rule) Let lim represent one of the limits
o
lim, lim , lim , lim , or lim .
в€’
xв†’c xв†’+в€ћ xв†’в€’в€ћ
+
xв†’c xв†’c

Suppose that f and g are continuous and diп¬Ђerentiable on an open interval
(a, b) except at an interior point c, a < c < b. Suppose further that g (x) = 0
on (a, b), lim f (x) = lim g(x) = 0 or lim f (x) = lim g(x) = +в€ћ or в€’в€ћ. If
f (x)
= L, +в€ћ or в€’ в€ћ
lim
g (x)
then
f (x) f (x)
lim = lim .
g(x) g (x)
Proof. The proof of this theorem is omitted.

Deп¬Ѓnition 7.3.1 (Extended Arithmetic) For the sake of convenience in deal-
ing with indeterminate forms, we deп¬Ѓne the following arithmetic operations
with real numbers, +в€ћ and в€’в€ћ. Let c be a real number and c > 0. Then
we deп¬Ѓne
+ в€ћ + в€ћ = +в€ћ, в€’в€ћ в€’ в€ћ = в€’в€ћ, c(+в€ћ) = +в€ћ, c(в€’в€ћ) = в€’в€ћ
в€’c
c c
(в€’c)(+в€ћ) = в€’в€ћ, (в€’c)(в€’в€ћ) = +в€ћ, = 0, = 0, = 0,
в€’в€ћ
+в€ћ +в€ћ
в€’c
= 0, (+в€ћ)c = +в€ћ, (+в€ћ)в€’c = 0, (+в€ћ)(+в€ћ) = +в€ћ, (+в€ћ)(в€’в€ћ) = в€’в€ћ,
в€’в€ћ
(в€’в€ћ)(в€’в€ћ) = +в€ћ.

Deп¬Ѓnition 7.3.2 The following operations are indeterminate:
0 +в€ћ +в€ћ в€’в€ћ в€’в€ћ
, в€ћ в€’ в€ћ, 0 В· в€ћ, 00 , 1в€ћ , в€ћ0 .
, , ,
0 +в€ћ в€’в€ћ в€’в€ћ +в€ћ

Remark 23 The LвЂ™HЛ†pitalвЂ™s Rule can be applied directly to the 0 and В±в€ћ 0
o В±в€ћ
В±в€ћ
0
forms. The forms в€ћ в€’ в€ћ and 0 В· в€ћ can be changed to the 0 or В±в€ћ by
using arithmetic operations. For the 00 and 1в€ћ forms we use the following
procedure:
ln(f (x))
lim (1/g(x))
g(x) g(x) ln(f (x))
.
lim(f (x)) = lim e =e
It is best to study a lot of examples and work problems.
306CHAPTER 7. IMPROPER INTEGRALS AND INDETERMINATE FORMS

Exercises 7.3

1. Prove the Theorem of the Mean: Suppose that a function f is continuous
on a closed and bounded interval [a, b] and f exists on the open interval
(a, b). Then there exists at least one number c such that a < c < b and

f (b) в€’ f (a)
(2) f (b) = f (a) + f (c)(b в€’ a).
= f (c)
(1)
bв€’a
2. Prove the Generalized Theorem of the Mean: Suppose that f and g are
continuous on a closed and bounded interval [a, b] and f and g exist
on the open interval (a, b) and g (x) = 0 for any x in (a, b). Then there
exists some c such that a < c < b and

f (b) в€’ f (a) f (c)
= .
g(b) в€’ g(a) g (c)

3. Prove the following theorem known as lвЂ™HЛ†pitalвЂ™s Rule: Suppose that f
o
and g are diп¬Ђerentiable functions, except possibly at a, such that

f (x)
= L.
lim f (x) = 0, lim g(x) = 0, and lim
g(x)
xв†’a xв†’a xв†’a

Then
f (x) f (x)
lim = lim = L.
g(x) xв†’a g (x)
xв†’a

4. Prove the following theorem known as an alternate form of lвЂ™HЛ†pitalвЂ™s
o
Rule: Suppose that f and g are diп¬Ђerentiable functions, except possibly
at a, such that

f (x)
lim f (x) = в€ћ, lim g(x) = в€ћ, = L.
and lim
g (x)
xв†’a xв†’a xв†’a

Then
f (x) f (x)
lim = lim = L.
g(x) xв†’a g (x)
xв†’a
7.3. 307

5. Prove that if f and g exist and

f (x)
lim f (x) = 0, lim g(x) = 0, and lim = L,
g (x)
xв†’+в€ћ xв†’+в€ћ xв†’+в€ћ

then
f (x)
lim = L.
g(x)
xв†’+в€ћ

6. Prove that if f and g exist and

f (x)
lim f (x) = 0, lim g(0) = 0, and lim = L,
xв†’в€’в€ћ g (x)
xв†’в€’в€ћ xв†’+в€ћ

then
f (x)
lim = L.
g(x)
xв†’в€’в€ћ

7. Prove that if f and g exist and

f (x)
lim f (x) = в€ћ, lim g(x) = в€ћ, and lim = L,
g (x)
xв†’+в€ћ xв†’+в€ћ xв†’+в€ћ

then
f (x)
lim = L.
g(x)
xв†’+в€ћ

8. Prove that if f and g exist and

f (x)
lim f (x) = в€ћ, lim g(x) = в€ћ, and lim = L,
g (x)
xв†’в€’в€ћ xв†’в€’в€ћ xв†’в€’в€ћ

then
f (x)
lim = L.
g(x)
xв†’+в€ћ

9. Suppose that f and f exist in an open interval (a, b) containing c. Then
prove that
f (c + h) в€’ 2f (c) + f (c в€’ h)
lim = f (c).
h2
hв†’0
308CHAPTER 7. IMPROPER INTEGRALS AND INDETERMINATE FORMS

10. Suppose that f is continuous in an open interval (a, b) containing c.
Then prove that
f (c + h) в€’ f (c в€’ h)
lim = f (c).
2h
hв†’0

11. Suppose that f (x) and g(x) are two polynomials such that
f (x) = a0 xn + a1 xnв€’1 + В· В· В· + anв€’1 x + an , a0 = 0,
g(x) = b0 xm + b1 xmв€’1 + В· В· В· + bmв€’1 x + bm , b0 = 0.
Then prove that
пЈ±
пЈґ0 if m > n
f (x) пЈІ
= +в€ћ or в€’ в€ћ if m < n
lim
g(x) пЈґ
xв†’+в€ћ
a0 /b0 if m = n
пЈі

12. Suppose that f and g are diп¬Ђerentiable functions, except possibly at c,
and
lim f (x) = 0, lim g(x) = 0 and lim g(x) ln(f (x)) = L.
xв†’c xв†’c xв†’c

Then prove that
lim (f (x))g(x) = eL .
xв†’c

13. Suppose that f and g are diп¬Ђerentiable functions, except possibly at c,
and
lim f (x) = +в€ћ, lim g(x) = 0 and lim g(x) ln(f (x)) = L.
xв†’c xв†’c xв†’c

Then prove that
lim (f (x))g(x) = eL .
xв†’c

14. Suppose that f and g are diп¬Ђerentiable functions, except possibly at c,
and
lim f (x) = 1, lim g(x) = +в€ћ and lim g(x) ln(f (x)) = L.
xв†’c xв†’c xв†’c

Then prove that
lim (f (x))g(x) = eL .
xв†’c
7.3. 309

15. Suppose that f and g are diп¬Ђerentiable functions, except possibly at c,
and
f (x)
lim f (x) = 0, lim g(x) = +в€ћ and lim = L.
(1/g(x))
xв†’c xв†’c xв†’c

Then prove that
lim f (x)g(x) = L.
xв†’c

1
16. Prove that lim (1 + x) x = e.
xв†’0

1
1
17. Prove that lim (1 в€’ x) x = .
e
xв†’0

xn
18. Prove that lim = 0 for each natural number n.
ex
xв†’+в€ћ

sin x в€’ x
19. Prove that lim = 0.
x sin x
+xв†’0

ПЂ
в€’ x tan x = 1.
20. Prove that lim
2
ПЂ
xв†’ 2

In problems 21вЂ“50 evaluate each of the limits.

sin(x2 ) 1 в€’ cos x2
21. lim 22. lim
x2 x2
xв†’0 xв†’0

sin(ax) tan(mx)
23. lim 24. lim
sin(bx) tan(nx)
xв†’0 xв†’0

e3x в€’ 1
26. lim (1 + 2x)3/x
25. lim
x
xв†’0 xв†’0

ex+h в€’ ex
ln(x + h) в€’ ln(x)
27. lim 28. lim
h h
hв†’0 hв†’0

ln(100 + x)
29. lim (1 + mx)n/x 30. lim
x
xв†’0 xв†’в€ћ
310CHAPTER 7. IMPROPER INTEGRALS AND INDETERMINATE FORMS

31. lim (1 + sin mx)n/x lim (sin x)x
32.
xв†’0 +
xв†’0

x4 в€’ 2x3 + 10
sin x
33. lim (x) 34. lim
xв†’в€ћ 3x4 + 2x3 в€’ 7x + 1
xв†’0+

2ПЂ
35. lim tan(2x) ln(x) 36. lim x sin
x
xв†’+в€ћ
+
xв†’0

x
3 + 2x
x 2/x
37. lim (x + e ) 38. lim
4 + 2x
xв†’0 xв†’в€ћ

39. lim (1 + sin mx)n/x lim (x)sin(3x)
40.
xв†’0 +
xв†’0

1 cos 4x
lim (e3x в€’ 1)2/ ln x в€’
41. 42. lim
x2 x2
xв†’0
+
xв†’0

cot(ax) ln x
43. lim 44. lim
cot(bx) x
xв†’+в€ћ
+
xв†’0

x 1 2
в€’
45. lim 46. lim
ln x x ln x
+ +
xв†’0 xв†’0

2x + 3 sin x
lim x(b1/x в€’ 1), b > 0, b = 1
47. lim 48.
4x + 2 sin x
xв†’+в€ћ xв†’+в€ћ

bx+h в€’ bx logb (x + h) в€’ logb x
49. lim , b > 0, b = 1 50. lim , b > 0, b = 1
h h
hв†’0 hв†’0

(ex в€’ 1) sin x x+1
51. lim 52. lim x ln
xв†’0 cos x в€’ cos2 x xв€’1
xв†’+в€ћ

2x в€’ 3x6 + x7
sin 5x
53. lim 54. lim
(1 в€’ x)3
xв†’0+ 1 в€’ cos 4x xв†’1

ex + 1 tan x в€’ sin x
x
55. lim e ln 56. lim
ex x3
xв†’+в€ћ xв†’0
7.3. 311

x3 sin 2x 5x в€’ 3x
57. lim 58. lim
xв†’0 (1 в€’ cos x)2 x2
xв†’0

arctan x в€’ x
1 1+x
59. lim ln 60. lim
x3
1в€’x
x
xв†’0 xв†’0

ln(1 + xe2x )
sin(ПЂ cos x)
61. lim 62. lim
x2
x sin x
xв†’0 xв†’+в€ћ

(ln x)n x + e2x
1
в€љ ln
, n = 1, 2, В· В· В·
63. lim 64. lim
x x
x
xв†’+в€ћ xв†’+в€ћ

ln x ln(tan 3x)
65. lim 66. lim
(1 + x3 )1/2 ln(tan 4x)
xв†’+в€ћ +
xв†’0

1/x2
sin x
lim (1 в€’ 3в€’x )в€’2x
67. 68. lim
x
xв†’0
+
xв†’0

x2
3
lim (eв€’x + eв€’2x )1/x
69. 70. lim cos
x
xв†’+в€ћ xв†’+в€ћ

x2
x
1 1
71. lim ln 72. lim 1+
x 2x
xв†’+в€ћ
xв†’0+

3x+ln x
1 1 1
в€’
73. lim 1+ 74. lim
2x x sin 2x
xв†’+в€ћ xв†’0

в€љ 1 1
x2 + b 2 в€’ x в€’2
75. lim x 76. lim
x sin x x
xв†’+в€ћ xв†’0

5 1 1
1
в€’2 в€’ ln
78. lim
77. lim
xв€’2 x +xв€’6 x x
xв†’2 +
xв†’0

1 1 1
cot x в€’ в€’
79. lim 80. lim
x2 tan2 x
x
xв†’0 xв†’0
312CHAPTER 7. IMPROPER INTEGRALS AND INDETERMINATE FORMS

eв€’x x в€’ sin x
1
в€’x
81. lim 82. lim
e в€’1
x x
xв†’0 xв†’в€ћ

1
x2 sin x 1
83. lim 84. lim x sin
sin x x
xв†’0 xв†’в€ћ

e в€’ (1 + x)1/x ln(ln x)
85. lim 86. lim
ln(x в€’ ln x)
x
xв†’0 xв†’+в€ћ

x
1 1 1 ln t
в€’
87. lim 88. lim dt
x2 x ln x x 1+t
xв†’+в€ћ
+
xв†’0 1

x
1
x
sin2 x dx
lim (ln(1 + e ) в€’ x)
89. 90. lim
x2
xв†’+в€ћ xв†’+в€ћ 0

91. Suppose that f is deп¬Ѓned and diп¬Ђerentiable in an open interval (a, b).
Suppose that a < c < b and f (c) exists. Prove that

f (x) в€’ f (c) в€’ (x в€’ c)f (c)
f (c) = lim .
((x в€’ c)2 /2!)
xв†’c

92. Suppose that f is deп¬Ѓned and f , f , В· В· В· , f (nв€’1) exist in an open interval
(a, b). Also, suppose that a < c < b and f (n) (c) exists

(a) Prove that
(xв€’c)nв€’1 nв€’1
f (x) в€’ f (c) в€’ (x в€’ c)f (c) в€’ В· В· В· в€’ f (c)
(nв€’1)!
f (n) (c) = lim .
(xв€’c)n
xв†’c
n!

(b) Show that there is a function En (x) deп¬Ѓned on (a, b), except possibly
at c, such that

(x в€’ c)nв€’1 (nв€’1)
f (x) = f (c) + (x в€’ c)f (c) + В· В· В· + f (x)
(n в€’ 1)!
(x в€’ c)n (n) (x в€’ c)n En (x)
+ f (c) + En (x)
n! n!
7.3. 313

and lim En (x) = 0. Find E2 (x) if c = 0 and
nв†’c

1
x4 sin , x=0
x
f (x) =
0 , x=0

(c) If f (c) = В· В· В· = f (nв€’1) (c) = 0, n is even, and f has a relative mini-
mum at x = c, then show that f (n) (c) в‰Ґ 0. What can be said if f has
a relative maximum at c? What are the suп¬ѓcient conditions for a rel-
ative maximum or minimum at c when f (c) = В· В· В· = f (nв€’1) (c) = 0?
What can be said if n is odd and f (c) = В· В· В· = f (nв€’1) (c) = 0 but
f (n) (c) = 0.

93. Suppose that f and g are deп¬Ѓned, have derivatives of order 1, 2, В· В· В· , nв€’1
in an open interval (a, b), a < c < b, f (n) (c) and g (n) (c) exist and g (n) (c) =
0. Prove that if f and g, as well as their п¬Ѓrst n в€’ 1 derivatives are 0,
then
f (n) (c)
f (x)
= (n) .
lim
xв†’c g(x) g (c)

Evaluate the following limits:

ПЂ
1
x2 sin x cos cos x
2
94. lim 95. lim
sin2 x
x
xв†’0 xв†’0

1
96. lim x( 1в€’x ) lim x(ln(x))n , n = 1, 2, 3, В· В· В·
97.
xв†’1 +
xв†’0

x3/2 ln x
xx в€’ x
99. lim
98. lim
(1 + x4 )1/2
xв†’1+ 1 в€’ x + ln x xв†’+в€ћ

1 + ex
n
, n = 1, 2, В· В· В·
100. lim x ln
ex
xв†’+в€ћ

x в€’t2
x e dx
0
101. lim
1 в€’ eв€’x2
xв†’0
314CHAPTER 7. IMPROPER INTEGRALS AND INDETERMINATE FORMS

7.4 Improper Integrals
1. Suppose that f is continuous on (в€’в€ћ, в€ћ) and g (x) = f (x). Then deп¬Ѓne
each of the following improper integrals:
Chapter 8

Inп¬Ѓnite Series

8.1 Sequences
Deп¬Ѓnition 8.1.1 An inп¬Ѓnite sequence (or sequence) is a function, say f ,
whose domain is the set of all integers greater than or equal to some integer
m. If n is an integer greater than or equal to m and f (n) = an , then we
express the sequence by writing its range in any of the following ways:

1. f (m), f (m + 1), f (m + 2), . . .

2. am , am+1 , am+2 , . . .

3. {f (n) : n в‰Ґ m}

4. {f (n)}в€ћ
n=m

5. {an }в€ћ
n=m

Deп¬Ѓnition 8.1.2 A sequence {an }в€ћ is said to converge to a real number
n=m
L (or has limit L) if for each > 0 there exists some positive integer M such
that |an в€’ L| < whenever n в‰Ґ M . We write,

lim an = L or an в†’ L as n в†’ в€ћ.
nв†’в€ћ

If the sequence does not converge to a п¬Ѓnite number L, we say that it diverges.

315
316 CHAPTER 8. INFINITE SERIES

Theorem 8.1.1 Suppose that c is a positive real number, {an }в€ћ and {bn }в€ћ
n=m n=m
are convergent sequences. Then

(i) lim (can ) = c lim an
nв†’в€ћ nв†’в€ћ

(ii) lim (an + bn ) = lim an + lim bn
nв†’в€ћ nв†’в€ћ nв†’в€ћ

(iii) lim (an в€’ bn ) = lim an в€’ lim bn
nв†’в€ћ nв†’в€ћ nв†’в€ћ

(iv) lim (an bn ) = lim an lim bn
nв†’в€ћ nв†’в€ћ nв†’в€ћ

an limnв†’в€ћ an
(v) lim = , if lim bn = 0.
bn limnв†’в€ћ bn
nв†’в€ћ nв†’в€ћ

c
(vi) lim (an )c = lim an
nв†’в€ћ nв†’в€ћ

(vii) lim (ean ) = elimnв†’в€ћ an
nв†’в€ћ

(viii) Suppose that an в‰¤ bn в‰¤ cn for all n в‰Ґ m and

lim an = lim cn = L.
nв†’в€ћ nв†’в€ћ

Then
lim bn = L.
nв†’в€ћ

Proof. Suppose that {an }в€ћ converges to a and {bn }в€ћ converges to b.
n=m n=m
Let 1 > 0 be given. Then there exist natural numbers N and M such that

|an в€’ a| < if n в‰Ґ N, (1)
1
|bn в€’ b| < if n в‰Ґ M. (2)
1

and n в‰Ґ N + M . Then
Part (i) Let > 0 be given and c = 0. Let =
1
2|c|
by the inequalities (1) and (2), we get

|can в€’ ca| = |c| |an в€’ a|
< |c| 1
<.
8.1. SEQUENCES 317

This completes the proof of Part (i).
. Let m в‰Ґ N + M . Then by the
Part (ii) Let > 0 be given and =
1
2
inequalities (1) and (2), we get
|(an + bn ) в€’ (a + b)| = |(an в€’ a) + (bn в€’ b)|
в‰¤ |an в€’ a| + |bn в€’ b|
< 1+ 1
=.
This completes the proof of Part (ii).
Part (iii)
lim (an в€’ bn ) = lim (an + (в€’1)bn )
nв†’в€ћ nв†’в€ћ
= lim an + lim [(в€’1)bn ] (by Part (ii))
nв†’в€ћ nв†’в€ћ
= lim an + (в€’1) lim bn (by Part (i))
nв†’в€ћ nв†’в€ћ
= a + (в€’1)b
= a в€’ b.

. If n в‰Ґ N + M ,
Part (iv) Let > 0 be given and = min 1,
1
1 + |a| + |b|
then by the inequalities (1) and (2) we have
|an bn в€’ ab| = |[(an в€’ a) + a][(bn в€’ b) + b] в€’ ab|
= |(an в€’ a)(bn в€’ b) + (an в€’ a)b + a(bn в€’ b|
в‰¤ |an в€’ a| |bn в€’ b| + |b| |an в€’ a| + |a| |bn в€’ b|
< 2 + |b| 1 + |a| 1
1
= 1 ( 1 + |b| + |a|)
в‰¤ 1 (1 + |b| + |a|)
в‰¤.

Part (v) First we assume that b > 0 and prove that
1 1
lim =.
bn b
nв†’в€ћ
318 CHAPTER 8. INFINITE SERIES

1
b and using inequality (2) for n в‰Ґ M , we get
By taking =
1
2
1 1 1
|bn в€’ b| < b, в€’ b < bn в€’ b < b,
2 2 2
1 3 2 1 2
b < bn < b, 0 < < <.
2 2 3b bn b
Then, for n в‰Ґ M , we get
b в€’ bn
1 1
в€’ =
b в€’ nb
bn b
11
= |bn в€’ b| В· В·
b bn
2
< |bn в€’ b| В· 2 . (3)
b
b b2
Let > 0 be given. Choose = min , . There exists some natural
2
22
number N such that if n в‰Ґ N , then

|bn в€’ b| < 2. (4)

If n в‰Ґ N + M , then the inequalities (3) and (4) imply that
2
1 1
< |bn в€’ b| 2
в€’
bn b b
2
<22
b
в‰¤.

It follows that
1
1
=
lim
bn b
nв†’в€ћ

an 1
= lim (an ) В· lim
lim
bn bn
nв†’в€ћ nв†’в€ћ nв†’в€ћ

1
=aВ·
b
a
=.
b
8.1. SEQUENCES 319

If b < 0, then

an 1
= lim (в€’an ) В· lim
lim
в€’bn
bn
nв†’в€ћ nв†’в€ћ nв†’в€ћ

1
= (в€’a)
в€’b
a
=.
b

This completes the proof of Part (v).
Part (vi) Since f (x) = xc is a continuous function,
c
c
= ac .
lim (an ) = lim an
nв†’в€ћ nв†’в€ћ

Part (vii) Since f (x) = ex is a continuous function,

lim ean = elimnв†’в€ћ an
= ea .
nв†’в€ћ

Part (viii) Suppose that an в‰¤ bn в‰¤ cn for all n в‰Ґ m and

lim an = L = lim cn = L.
nв†’в€ћ nв†’в€ћ

Let > 0 be given. Then there exists natural numbers N and M such that

в€’
< an в€’ L < for n в‰Ґ N,
|an в€’ L| < ,
2 2 2
в€’
|cn в€’ L| < , < cn в€’ L < for n в‰Ґ M.
2 2 2
If n в‰Ґ N + M , then n > N and n > M and, hence,

в€’ < an в€’ L в‰¤ bn в€’ L в‰¤ cn в€’ L < .
2 2
It follows that
lim bn = L.
nв†’в€ћ

This completes the proof of this theorem.
320 CHAPTER 8. INFINITE SERIES

8.2 Monotone Sequences
Deп¬Ѓnition 8.2.1 Let {tn }в€ћ be a given sequence. Then {tn }в€ћ is said
n=m n=m
to be

(a) increasing if tn < tn+1 for all n в‰Ґ m;

(b) decreasing if tn+1 < tn for all n в‰Ґ m;

(c) nondecreasing if tn в‰¤ tn+1 for all n в‰Ґ m;

(d) nonincreasing if tn+1 в‰¤ tn for all n в‰Ґ m;

(e) bounded if a в‰¤ tn в‰¤ b for some constants a and b and all n в‰Ґ m;

(f) monotone if {tn }в€ћ is increasing, decreasing, nondecreasing or nonin-
n=m
creasing.

(g) a Cauchy sequence if for each > 0 there exists some M such that
|an1 в€’ an2 | < whenever n1 в‰Ґ M and n2 в‰Ґ M .

Theorem 8.2.1 (a) A monotone sequence converges to some real number if
and only if it is a bounded sequence.
(b) A sequence is convergent if and only if it is a Cauchy sequence.
Proof.
Part (a) Suppose that an в‰¤ an+1 в‰¤ B for all n в‰Ґ M and some B. Let L be
the least upper bound of the sequence {an }в€ћ . Let > 0 be given. Then
n=m
there exists some natural number N such that

L в€’ < aN в‰¤ L.

Then for each n в‰Ґ N , we have

L в€’ < aN в‰¤ an в‰¤ L.

By deп¬Ѓnition {an }в€ћ converges to L.
n=m
Similarly, suppose that B в‰¤ an+1 в‰¤ an for all n в‰Ґ M . Let L be the
greatest lower bound of {an }в€ћ . Then {an }в€ћ converges to L. It follows
n=m n=m
that a bounded monotone sequence converges. Conversely, suppose that a
8.2. MONOTONE SEQUENCES 321

monotone sequence {an }в€ћ converges to L. Let = 1. Then there exists
n=m
some natural number N such that if n в‰Ґ N , then

|an в€’ L| <
в€’ < an в€’ L <
L в€’ < an < L + .

The set {an : m в‰¤ n в‰¤ N } is bounded and the set {an : n в‰Ґ N } is bounded.
It follows that {an }в€ћ is bounded. This completes the proof of Part (a) of
n=m
the theorem.

Part (b) First, let us suppose that {an }в€ћ converges to L. Let > 0 be
n=m
given. Then > 0 and hence there exists some natural number N such that
2
for all natural numbers p в‰Ґ N and q в‰Ґ N , we have

|ap в€’ L| < and |aq в€’ L| <
2 2
|ap в€’ aq | = |(ap в€’ L) + (L + aq )|
в‰¤ |ap в€’ L| + |a1 в€’ L|
< +
2 2
=.

It follows that {an }в€ћ is a Cauchy sequence.
n=m
Next, we suppose that {an }в€ћ is a Cauchy sequence. Let S = {an : m в‰¤
n=m
n < в€ћ}. Suppose > 0. Then there exists some natural number N such
that for all p в‰Ґ 1

|aN +p в€’ aN | < , aN в€’ < aN +p < aN + (1)
2 2 2

It follows that S is a bounded set. If S is an inп¬Ѓnite set, then S has some
limit point q and some subsequence {ank }в€ћ of {an }в€ћ that converges to
n=m
k=1
q. Since > 0, there exists some natural number M such that for all k в‰Ґ M ,
we have
|ank в€’ q| < (2)
2
322 CHAPTER 8. INFINITE SERIES

Also, for all k в‰Ґ N + M , we get nk в‰Ґ k в‰Ґ N + M and
|ak в€’ q| = |ak в€’ ank + ank в€’ q|
в‰¤ |ank в€’ ak | + |ank в€’ q|
< +2 (by (1) and (2))
2
=.
It follows that the sequence {an }в€ћ converges to q. If S is a п¬Ѓnite set, then
n=m
some ak is repeated inп¬Ѓnite number of times and hence some subsequences of
{an }в€ћ converges to ak . By the preceding argument {an }в€ћ also converges
n=m n=m
to ak . This completes the proof of this theorem.

Theorem 8.2.2 Let {f (n)}в€ћ be a sequence where f is a diп¬Ђerentiable
n=m
function deп¬Ѓned for all real numbers x в‰Ґ m. Then the sequence {f (n)}в€ћ
n=m
 << стр. 12(всего 14)СОДЕРЖАНИЕ >>