calculus test bank

7.2. DISCONTINUITIES AT END POINTS 299

Then F (b) − 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,

µ+bσ b √1 e−x2/2dx.
a 2π
f (x)dx =

µ+aσ

Proof. The proof of this theorem is omitted.

Exercises 7.1 None available.

7.2 Discontinuities at End Points

Definition 7.2.1 (i) Suppose that f is continuous on [a, b) and

lim f (x) = +∞ or − ∞.

x→b−

Then, we define

b x

f (x)dx = lim f (x)dx.

a x→b− 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) = +∞ or − ∞.

x→a+

Then we define,

b b

f (x)dx = lim f (x)dx.

a x→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 (−∞, ∞) and g (x) = f (x). Then define
each of the following improper integrals:

300CHAPTER 7. IMPROPER INTEGRALS AND INDETERMINATE FORMS

+∞

(a) f (x)dx

a
b

(b) f (x)dx

−∞
+∞

(c) f (x)dx

−∞

2. Suppose that f is continuous on the open interval (a, b) and g (x) = f (x)
on (a, b). Define each of the following improper integrals if f is not
continuous at a or b:

x

(a) f (x)dx, a ≤ x < b

a

b

(b) f (x)dx, a < x ≤ b

x

b

(c) f (x)dx

a

+∞

3. Prove that e−xdx = 1

0

4. Prove that 1 √ 1 π
dx =
0 1 − x2 2

5. Prove that +∞ 1 dx = π

−∞ 1 + x2

∞1 1
6. Prove that xp dx = p − 1 , if and only if p > 1.

1

+∞ ∞
e−x2dx = 2 e−x2dx. Use the comparison between
7. Show that

−∞ 0
+∞
e−x and e−x2. Show that e−x2dx exists.

−∞

1 dx
8. Prove that 0 xp converges if and only if p < 1.

7.2. DISCONTINUITIES AT END POINTS 301

+∞

9. Evaluate e−x sin(2x)dx.

0

+∞

10. Evaluate e−4x cos(3x)dx.

0

+∞

11. Evaluate x2e−xdx.

0

+∞

12. Evaluate xe−xdx.

0

∞

13. Prove that sin(2x)dx diverges.

0

∞

14. Prove that cos(3x)dx diverges.

0

15. Compute the volume of the solid generated when the area between the
graph of y = e−x2 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 = e−x, 0 ≤ x < ∞ and the x-axis is rotated

(a) about the x-axis
(b) about the y-axis.

17. Let A represent the area bounded by the graph y = 1 1 ≤ x < ∞
,
x
and the x-axis. Let V denote the volume generated when the area A is

rotated about the x-axis.

(a) show that A is +∞
(b) show that V = π
(c) show that the surface area of V is +∞.
(d) Is it possible to fill 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 = e−2x, 0 ≤ x < ∞,
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.

19. Assume that +∞ (π/8). Compute +∞ s√in x dx.

sin(x2)dx =
0 0x

+∞ e−x2 √
π.
20. It is known that =

−∞

(a) Compute +∞
(b) Compute
(c) Compute e−x2 dx.

0

+∞ e√−x dx.
0x

+∞

e−4x2 dx.

0

Definition 7.2.2 Suppose that f (t) is continuous on [0, ∞) and there exist
some constants a > 0, M > 0 and T > 0 such that |f (t)| < M eat for all
t ≥ T . Then we define the Laplace transform of f (t), denoted L{f (t)}, by

∞

L{f (t)} = e−stf (t)dt

0

for all s ≥ s0. In problems 21–34, compute L{f (t)} for the given f (t).

21. f (t) = 1 if t ≥ 0 22. f (t) = t
0 if t < 0

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

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

27. f (t) = tebt 28. f (t) = tnebt, n = 1, 2, 3, · · ·

7.2. DISCONTINUITIES AT END POINTS 303

eat − ebt aeat − bebt
29. f (t) = 30. f (t) =

a−b a−b
1 32. f (t) = cos(bt)
31. f (t) = sin(bt) 34. f (t) = cosh(bt)
b
1
33. f (t) = sinh(bt)
b

Definition 7.2.3 For x > 0, we define the Gamma function Γ(x) by

+∞

Γ(x) = tx−1e−tdt.

0

+∞ e−x2 1 √
π.
In problems 35–40 assume that Γ(x) exists for x > 0 and =
02

√ 36. Show that Γ(1) = 1
35. Show that Γ(1/2) = π

37. Prove that Γ(x + 1) = xΓ(x) √
3π
38. Show that Γ =
22

5 3√ 40. Show that Γ(n + 1) = n!
39. Show that Γ = π

24

In problems 41–60, evaluate the given improper integrals.

+∞ +∞ dx
42.
41. 2xe−x2dx
1 x3/2
0

+∞ dx 44. +∞ 4x
43. 1 + x2 dx
x5/2 1
4

+∞ x +∞ 4
45. (1 + x2)3/2 dx 46. x2 − 4 dx

1 16

+∞ 1 +∞ 1
47. dx 48. dx, p > 1

2 x(ln x)2 2 x(ln x)p

304CHAPTER 7. IMPROPER INTEGRALS AND INDETERMINATE FORMS

1 2

49. 3xe−x2dx 50. ex dx

−∞ −∞

∞2 ∞ dx
51. 0 ex + e−x dx 52.

−∞ x2 + 9

53. 2 √ dx 54. 4 x dx
0 4 − x2
√
0 16 − x2

5x 56. +∞ √ dx
55. dx 2 x x2 − 4

0 (25 − x2)2/3 58. ∞ dx
57. +∞ e√−√x dx
√
0x 0 x(x + 25)

∞ e−x +∞
dx
59. 60. x2e−x3dx
1 − (e−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], differentiable 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 (c) = f (b) − f (a)
g (c) g(b) − .

g(a)

Proof. See the proof of Theorem 4.1.6.

Theorem 7.3.2 Suppose that f and g are continuous and differentiable 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)
lim = L
x→c g (x)

then
f (x)

lim = L.
x→c g(x)

7.3. 305

Proof. See the proof of Theorem 4.1.7.

Theorem 7.3.3 (L’Hˆopital’s Rule) Let lim represent one of the limits

lim, lim , lim , lim , or lim .
x→c x→c+ x→c− x→+∞ x→−∞

Suppose that f and g are continuous and differentiable 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) = +∞ or −∞. If

lim f (x) = L, +∞ or − ∞
g (x)

then
f (x) f (x)

lim = lim .
g(x) g (x)

Proof. The proof of this theorem is omitted.

Definition 7.3.1 (Extended Arithmetic) For the sake of convenience in deal-
ing with indeterminate forms, we define the following arithmetic operations
with real numbers, +∞ and −∞. Let c be a real number and c > 0. Then
we define

+ ∞ + ∞ = +∞, −∞ − ∞ = −∞, c(+∞) = +∞, c(−∞) = −∞

(−c)(+∞) = −∞, (−c)(−∞) = +∞, c −c c
+∞ = 0, +∞ = 0, −∞ = 0,
−c
−∞ = 0, (+∞)c = +∞, (+∞)−c = 0, (+∞)(+∞) = +∞, (+∞)(−∞) = −∞,

(−∞)(−∞) = +∞.

Definition 7.3.2 The following operations are indeterminate:

0 +∞ +∞ −∞ −∞ ∞ − ∞, 0 · ∞, 00, 1∞, ∞0.
, +∞ , −∞ −∞ , +∞ ,

0

Remark 23 The L’Hoˆpital’s Rule can be applied directly to the 0 and ±∞
0 ±∞
0 ±∞
forms. The forms ∞−∞ and 0·∞ can be changed to the 0 or ±∞ by

using arithmetic operations. For the 00 and 1∞ forms we use the following

procedure:

lim(f (x))g(x) = lim eg(x) ln(f(x)) = elim ln(f (x))

.(1/g(x))

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) − f (a) (2) f (b) = f (a) + f (c)(b − a).
(1) b − a = f (c)

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) − f (a) = f (c)
g(b) − g(a) g .

(c)

3. Prove the following theorem known as l’Hˆopital’s Rule: Suppose that f
and g are differentiable functions, except possibly at a, such that

lim f (x) = 0, lim g(x) = 0, and lim f (x) = L.

x→a x→a x→a g(x)

Then

f (x) f (x)
lim = lim = L.
x→a g(x) x→a g (x)

4. Prove the following theorem known as an alternate form of l’Hˆopital’s
Rule: Suppose that f and g are differentiable functions, except possibly
at a, such that

lim f (x) = ∞, lim g(x) = ∞, and lim f (x) = L.

x→a x→a x→a g (x)

Then

f (x) f (x)
lim = lim = L.
x→a g(x) x→a g (x)

7.3. 307

5. Prove that if f and g exist and

lim f (x) = 0, lim g(x) = 0, and lim f (x) = L,

x→+∞ x→+∞ x→+∞ g (x)

then f (x)
lim = L.
x→+∞ g(x)

6. Prove that if f and g exist and

lim f (x) = 0, lim g(0) = 0, and lim f (x) = L,

x→−∞ x→+∞ x→−∞ g (x)

then f (x)
lim = L.
x→−∞ g(x)

7. Prove that if f and g exist and

lim f (x) = ∞, lim g(x) = ∞, and f (x)
lim = L,
x→+∞ x→+∞ x→+∞ g (x)

then f (x)
lim = L.
x→+∞ g(x)

8. Prove that if f and g exist and

lim f (x) = ∞, lim g(x) = ∞, and lim f (x) = L,

x→−∞ x→−∞ x→−∞ g (x)

then f (x)
lim = L.
x→+∞ g(x)

9. Suppose that f and f exist in an open interval (a, b) containing c. Then

prove that

f (c + h) − 2f (c) + f (c − h)
lim = f (c).
h→0 h2

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) − f (c − h)
lim = f (c).
h→0 2h

11. Suppose that f (x) and g(x) are two polynomials such that

f (x) = a0xn + a1xn−1 + · · · + an−1x + an, a0 = 0,
g(x) = b0xm + b1xm−1 + · · · + bm−1x + bm, b0 = 0.

Then prove that

 if m > n
0 if m < n
f (x)  if m = n

lim = +∞ or − ∞
x→+∞ g(x) a0/b0

12. Suppose that f and g are differentiable functions, except possibly at c,
and

lim f (x) = 0, lim g(x) = 0 and lim g(x) ln(f (x)) = L.
x→c x→c
x→c

Then prove that

lim (f (x))g(x) = eL.

x→c

13. Suppose that f and g are differentiable functions, except possibly at c,
and

lim f (x) = +∞, lim g(x) = 0 and lim g(x) ln(f (x)) = L.
x→c x→c
x→c

Then prove that

lim (f (x))g(x) = eL.

x→c

14. Suppose that f and g are differentiable functions, except possibly at c,
and

lim f (x) = 1, lim g(x) = +∞ and lim g(x) ln(f (x)) = L.
x→c x→c
x→c

Then prove that

lim (f (x))g(x) = eL.

x→c

7.3. 309

15. Suppose that f and g are differentiable functions, except possibly at c,
and

lim f (x) = 0, lim g(x) = +∞ and lim f (x) = L.

x→c x→c x→c (1/g(x))

Then prove that

lim f (x)g(x) = L.

x→c

1

16. Prove that lim (1 + x) x = e.
x→0

17. Prove that lim (1 − x) 1 = 1
x .

x→0 e

xn
18. Prove that lim = 0 for each natural number n.

x→+∞ ex

19. Prove that lim sin x − x = 0.

x→0+ x sin x

π
20. Prove that lim − x tan x = 1.
2x→π
2

In problems 21–50 evaluate each of the limits.

21. lim sin(x2) 22. lim 1 − cos x2
x→0 x→0
x2 x2

sin(ax) tan(mx)
23. lim 24. lim

x→0 sin(bx) x→0 tan(nx)

e3x − 1 26. lim (1 + 2x)3/x
25. lim x→0

x→0 x

ln(x + h) − ln(x) 28. lim ex+h − ex
27. lim
h→0 h
h→0 h

29. lim (1 + mx)n/x 30. lim ln(100 + x)
x→0
x→∞ x

310CHAPTER 7. IMPROPER INTEGRALS AND INDETERMINATE FORMS

31. lim (1 + sin mx)n/x 32. lim (sin x)x
x→0 x→0+
x4 − 2x3 + 10
33. lim (x)sin x
x→0+ 34. lim
x→∞ 3x4 + 2x3 − 7x + 1

35. lim tan(2x) ln(x) 36. lim x sin 2π
x→0+
x→+∞ x

37. lim (x + ex)2/x 3 + 2x x
x→0 38. lim

39. lim (1 + sin mx)n/x x→∞ 4 + 2x
x→0
40. lim (x)sin(3x)
x→0+

41. lim (e3x − 1)2/ ln x 42. lim 1 − cos 4x
x→0+ x→0 x2 x2

cot(ax) ln x
43. lim 44. lim

x→0+ cot(bx) x→+∞ x

x 12
45. lim 46. lim −

x→0+ ln x x→0+ x ln x

2x + 3 sin x 48. lim x(b1/x − 1), b > 0, b = 1
47. lim x→+∞

x→+∞ 4x + 2 sin x

49. lim bx+h − bx , b > 0, b = 1 50. lim logb(x + h) − logb x b > 0, b = 1
,
h→0 h h→0 h

(ex − 1) sin x 52. lim x ln x+1
51. lim
x→+∞ x−1
x→0 cos x − cos2 x

sin 5x 2x − 3x6 + x7
53. lim 54. lim

x→0+ 1 − cos 4x x→1 (1 − x)3

55. lim ex ln ex + 1 56. lim tan x − sin x

x→+∞ ex x→0 x3

7.3. 311

x3 sin 2x 5x − 3x
57. lim 58. lim

x→0 (1 − cos x)2 x→0 x2

1 1+x arctan x − x
59. lim ln 60. lim
x→0 x 1−x x→0 x3

sin(π cos x) 62. lim ln(1 + xe2x)
61. lim
x→+∞ x2
x→0 x sin x
63. lim (ln x)n , n = 1, 2, · · · 64. lim √1 ln x + e2x
x→+∞ x x
x→+∞ x

65. lim ln x ln(tan 3x)
66. lim
x→+∞ (1 + x3)1/2
x→0+ ln(tan 4x)

67. lim (1 − 3−x)−2x 68. lim sin x 1/x2
x→0+ x→0 x

3 x2

69. lim (e−x + e−2x)1/x 70. lim cos x
x→+∞ x→+∞

1 x 1 x2

71. lim ln x 72. lim 1 + 2x
x→0+ x→+∞

73. lim 1 + 1 3x+ln x 74. lim 1 − 1
x→0 x sin 2x
x→+∞ 2x

√ 76. lim x 1 x − 1
75. lim x x2 + b2 − x x→0 sin x2

x→+∞

1 − 5 11
77. lim − +x 78. lim − ln
x→2 x 2 x2 − 6 x→0+ x x

1 11
79. lim cot x − 80. lim x2 − tan2 x
x→0
x→0 x

312CHAPTER 7. IMPROPER INTEGRALS AND INDETERMINATE FORMS

81. lim e−x − 1 82. lim x − sin x
x→0 x −
ex 1 x→∞ x

83. lim x2 sin 1 84. lim x sin 1
x

x→0 sin x x→∞ x

e − (1 + x)1/x ln(ln x)
85. lim 86. lim

x→0 x x→+∞ ln(x − ln x)

11 1 x ln t
87. lim x2 − x ln x 88. lim dt
x→0+ x→+∞ x 1 1 + t

89. lim (ln(1 + ex) − x) 1 x
x→+∞ 90. lim
sin2 x dx
x→+∞ x2
0

91. Suppose that f is defined and differentiable in an open interval (a, b).
Suppose that a < c < b and f (c) exists. Prove that

f (c) = lim f (x) − f (c) − (x − c)f (c) .
x→c
((x − c)2/2!)

92. Suppose that f is defined and f , f , · · · , f (n−1) exist in an open interval
(a, b). Also, suppose that a < c < b and f (n)(c) exists

(a) Prove that

f (x) − f (c) − (x − c)f (c) − · · · − (x−c)n−1 f n−1 (c)
(n−1)!
f (n)(c) = lim (x−c)n .
x→c
n!

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

f (x) = f (c) + (x − c)f (c) + · · · + (x − c)n−1 f (n−1)(x)
(n − 1)!

+ (x − c)n f (n)(c) + En(x) (x − c)nEn(x)
n! n!

7.3. 313

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

n→c

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

0 , x=0

(c) If f (c) = · · · = f (n−1)(c) = 0, n is even, and f has a relative mini-
mum at x = c, then show that f (n)(c) ≥ 0. What can be said if f has

a relative maximum at c? What are the sufficient conditions for a rel-
ative maximum or minimum at c when f (c) = · · · = f (n−1)(c) = 0?
What can be said if n is odd and f (c) = · · · = f (n−1)(c) = 0 but
f (n)(c) = 0.

93. Suppose that f and g are defined, have derivatives of order 1, 2, · · · , n−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 first n − 1 derivatives are 0,

then

f (x) f (n)(c)
lim g(x) = g(n)(c) .

x→c

Evaluate the following limits:

94. lim x2 sin 1 95. lim cos π cos x
x 2
x→0
x→0 x sin2 x

96. lim x( 1 ) 97. lim x(ln(x))n, n = 1, 2, 3, · · ·
1−x x→0+
x3/2 ln x
x→1
99. lim
xx − x x→+∞ (1 + x4)1/2
98. lim

x→1+ 1 − x + ln x

100. lim xn ln 1 + ex , n = 1, 2, · · ·
x→+∞ ex

101. lim x x e−t2 dx
0
x→0
1 − e−x2

314CHAPTER 7. IMPROPER INTEGRALS AND INDETERMINATE FORMS

7.4 Improper Integrals

1. Suppose that f is continuous on (−∞, ∞) and g (x) = f (x). Then define
each of the following improper integrals:

Chapter 8
Infinite Series

8.1 Sequences

Definition 8.1.1 An infinite 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 ≥ m}
4. {f (n)}∞n=m
5. {an}n∞=m

Definition 8.1.2 A sequence {an}∞n=m is said to converge to a real number
L (or has limit L) if for each > 0 there exists some positive integer M such

that |an − L| < whenever n ≥ M . We write,

lim an = L or an → L as n → ∞.

n→∞

If the sequence does not converge to a finite 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}∞n=m and {bn}∞n=m
are convergent sequences. Then

(i) nl→im∞(can) = c lim an

n→∞

(ii) nl→im∞(an + bn) = lim an + lim bn

n→∞ n→∞

(iii) nl→im∞(an − bn) = lim an − lim bn

n→∞ n→∞

(iv) lim (anbn) = lim an lim bn

n→∞ n→∞ n→∞

(v) lim an = limn→∞ an , if lim bn = 0.
n→∞ bn limn→∞ bn
n→∞

(an)c c

(vi) lim = lim an

n→∞ n→∞

(vii) lim (ean ) = elimn→∞ an
n→∞

(viii) Suppose that an ≤ bn ≤ cn for all n ≥ m and

lim an = lim cn = L.

n→∞ n→∞

Then

lim bn = L.

n→∞

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

|an − a| < 1 if n ≥ N, (1)
|bn − b| < 1 if n ≥ M. (2)

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

|can − ca| = |c| |an − a|
< |c| 1
<.

8.1. SEQUENCES 317

This completes the proof of Part (i).

Part (ii) Let > 0 be given and 1 = . Let m ≥ N + M . Then by the
2
inequalities (1) and (2), we get

|(an + bn) − (a + b)| = |(an − a) + (bn − b)|
≤ |an − a| + |bn − b|
< 1+ 1
=.

This completes the proof of Part (ii).
Part (iii)

nl→im∞(an − bn) = lim (an + (−1)bn)

n→∞

= lim an + lim [(−1)bn] (by Part (ii))
n→∞
n→∞

= lim an + (−1) lim bn (by Part (i))
n→∞
n→∞

= a + (−1)b

= a − b.

Part (iv) Let > 0 be given and 1 = min 1, 1 + |a| + |b| . If n ≥ N + M ,
then by the inequalities (1) and (2) we have

|anbn − ab| = |[(an − a) + a][(bn − b) + b] − ab|

= |(an − a)(bn − b) + (an − a)b + a(bn − b|

≤ |an − a| |bn − b| + |b| |an − a| + |a| |bn − b|

< 2 + |b| 1 + |a| 1
1

= 1( 1 + |b| + |a|)

≤ 1(1 + |b| + |a|)

≤.

Part (v) First we assume that b > 0 and prove that
11

lim = .
n→∞ bn b

318 CHAPTER 8. INFINITE SERIES

By taking 1 = 1 b and using inequality (2) for n ≥ M, we get
2

|bn − b| < 1 b, −1 b < bn − b < 1 b,
2 2 2

1 3 2 12
b < bn < b, 0< < < .
2 2 3b bn b

Then, for n ≥ M , we get

11 = b − bn
− b − nb

bn b

11
= |bn − b| · b · bn

< |bn − b| · 2 (3)
.

b2

b b2
Let > 0 be given. Choose 2 = min , . There exists some natural
22

number N such that if n ≥ N , then

|bn − b| < 2. (4)
If n ≥ N + M , then the inequalities (3) and (4) imply that

1 −1 < |bn − b| 2
bn b b2

2
< 2 b2

≤.

It follows that

11
lim =
n→∞ bn b

lim an = lim (an) · lim 1
bn bn
n→∞ n→∞ n→∞

=a· 1
b

a
=.

b

8.1. SEQUENCES 319

If b < 0, then

lim an = lim (−an) · lim 1
bn −bn
n→∞ n→∞ n→∞

= (−a) 1
−b

a
=.

b

This completes the proof of Part (v).
Part (vi) Since f (x) = xc is a continuous function,

(an)c c = ac.

lim = lim an

n→∞ n→∞

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

lim ean = elimn→∞ an = ea.

n→∞

Part (viii) Suppose that an ≤ bn ≤ cn for all n ≥ m and

lim an = L = lim cn = L.

n→∞ n→∞

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

|an − L| < 2 , − < an − L < 2 for n ≥ N,
2 for n ≥ M.
−
|cn − L| < 2 , 2 < cn − L < 2

If n ≥ N + M , then n > N and n > M and, hence,

− < an − L ≤ bn − L ≤ cn − L < .
2 2

It follows that

lim bn = L.

n→∞

This completes the proof of this theorem.

320 CHAPTER 8. INFINITE SERIES

8.2 Monotone Sequences

Definition 8.2.1 Let {tn}n∞=m be a given sequence. Then {tn}n∞=m is said
to be

(a) increasing if tn < tn+1 for all n ≥ m;

(b) decreasing if tn+1 < tn for all n ≥ m;

(c) nondecreasing if tn ≤ tn+1 for all n ≥ m;

(d) nonincreasing if tn+1 ≤ tn for all n ≥ m;

(e) bounded if a ≤ tn ≤ b for some constants a and b and all n ≥ m;
(f) monotone if {tn}∞n=m is increasing, decreasing, nondecreasing or nonin-

creasing.

(g) a Cauchy sequence if for each > 0 there exists some M such that
|an1 − an2| < whenever n1 ≥ M and n2 ≥ 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 ≤ an+1 ≤ B for all n ≥ M and some B. Let L be
the least upper bound of the sequence {an}∞n=m. Let > 0 be given. Then
there exists some natural number N such that

L − < aN ≤ L.

Then for each n ≥ N , we have

L − < aN ≤ an ≤ L.

By definition {an}n∞=m converges to L.
Similarly, suppose that B ≤ an+1 ≤ an for all n ≥ M . Let L be the

greatest lower bound of {an}∞n=m. Then {an}∞n=m converges to L. It follows
that a bounded monotone sequence converges. Conversely, suppose that a

8.2. MONOTONE SEQUENCES 321

monotone sequence {an}n∞=m converges to L. Let = 1. Then there exists
some natural number N such that if n ≥ N , then

|an − L| <
− < an − L <
L − < an < L + .

The set {an : m ≤ n ≤ N } is bounded and the set {an : n ≥ N } is bounded.
It follows that {an}n∞=m is bounded. This completes the proof of Part (a) of
the theorem.

Part (b) First, let us suppose that {an}n∞=m converges to L. Let > 0 be
given. Then > 0 and hence there exists some natural number N such that

2
for all natural numbers p ≥ N and q ≥ N , we have

|ap − L| < 2 and |aq − L| < 2
|ap − aq| = |(ap − L) + (L + aq)|

≤ |ap − L| + |a1 − L|

<+
22

=.

It follows that {an}∞n=m is a Cauchy sequence.
Next, we suppose that {an}n∞=m is a Cauchy sequence. Let S = {an : m ≤

n < ∞}. Suppose > 0. Then there exists some natural number N such
that for all p ≥ 1

|aN+p − aN | < , aN − < aN+p < aN + 2 (1)
2 2

It follows that S is a bounded set. If S is an infinite set, then S has some
limit point q and some subsequence {ank}∞k=1 of {an}∞n=m that converges to
q. Since > 0, there exists some natural number M such that for all k ≥ M ,

we have

|ank − q| < 2 (2)

322 CHAPTER 8. INFINITE SERIES

Also, for all k ≥ N + M , we get nk ≥ k ≥ N + M and

|ak − q| = |ak − ank + ank − q|
≤ |ank − ak| + |ank − q|

< +2 (by (1) and (2))
2

=.

It follows that the sequence {an}n∞=m converges to q. If S is a finite set, then
some ak is repeated infinite number of times and hence some subsequences of
{an}∞n=m converges to ak. By the preceding argument {an}∞n=m also converges
to ak. This completes the proof of this theorem.

Theorem 8.2.2 Let {f (n)}∞n=m be a sequence where f is a differentiable
function defined for all real numbers x ≥ m. Then the sequence {f (n)}∞n=m

is

(a) increasing if f (x) > 0 for all x > m;

(b) decreasing if f (x) < 0 for all x > m;

(c) nondecreasing if f (x) ≥ 0 for all x > m;

(d) nonincreasing if f (x) ≤ 0 for all x > m.

Proof. Suppose that m ≤ a < b. Then by the Mean Value Theorem for
derivatives, there exists some c such that a < c < b and

f (b) − f (a)
b − a = f (c),

f (b) = f (a) + f (c)(b − a).

The theorem follows from the above equation by considering the value of
f (c). In particular, for all natural numbers n ≥ m,

f (n + 1) = f (n) + f (c),

for some c such that n < c < n + 1.

Part (a). If f (c) > 0, then f (n + 1) > f (n) for all n ≥ m.
Part (b). If f (c) < 0, then f (n + 1) < f (n) for all n ≥ m.
Part (c). If f (c) ≥ 0, then f (n + 1) ≥ f (n) for all n ≥ m.
Part (d). If f (c) ≤ 0, then f (n + 1) ≤ f (n) for all n ≤ m.

This completes the proof of this theorem.

8.3. INFINITE SERIES 323

8.3 Infinite Series

Definition 8.3.1 Let {tn}∞n=1 be a given sequence. Let

n

s1 = t1, s2 = t1 + t2, s3 = t1 + t2 + t3, · · · , sn = tk,

k=1

for all natural number n. If the sequence {sn}∞n=1 converges to a finite number
L, then we write

∞

L = t1 + t2 + t3 + · · · = tk.

k=1

n

We call tk an infinite series and write

k=1

∞n

tk = lim tk = L.

k=1 n→∞ k=1

We say that L is the sum of the series and the series converges to L. If a

series does not converge to a finite number, we say that it diverges. The
sequence {sn}n∞=1 is called the sequence of the nth partial sums of the series.

Theorem 8.3.1 Suppose that a and r are real numbers and a = 0. Then
the geometric series

∞ a
,
a + ar + ar2 + · · · = ark = 1−r

k=0

if |r| < 1. The geometric series diverges if |r| ≥ 1.

Proof. For each natural number n, let
sn = a + ar + · · · + arn−1.

On multiplying both sides by r, we get

rsn = ar + ar2 + · · · + arn−1 + arn
sn − rsn = a − arn
(1 − r)sn = a(1 − rn)

sn = 1 a r − a rn.
− 1−r

324 CHAPTER 8. INFINITE SERIES

If |r| < 1, then

lim sn = a − a lim rn = a
− − .
n→∞ 1 r 1 r n→∞ 1−r

If |r| > 1, then lim rn is not finite and so the sequence {sn}n∞=1 of nth partial

n→∞

sums diverges.

If r = 1, then sn = na and lim na is not a finite number.

n→∞

This completes the proof of the theorem.

∞

Theorem 8.3.2 (Divergence Test) If the series tk converges, then lim tn =

k=1 n→∞

0. If lim tn = 0, then the series diverges.

n→∞

Proof. Suppose that the series converges to L. Then

n n−1

lim an = lim ak − ak
n→∞
n→∞ k=1 k=1

n n−1

= lim ak − lim ak
n→∞
n→∞
k=1 k=1

=L−L

= 0.

The rest of the theorem follows from the preceding argument. This completes
the proof of this theorem.

Theorem 8.3.3 (The Integral Test) Let f be a function that is defined,
continuous and decreasing on [1, ∞) such that f (x) > 0 for all x ≥ 1. Then

∞ ∞

f (n) and f (x)dx

n=1 1

either both converge or both diverge.

Proof. Suppose that f is decreasing and continuous on [1, ∞), and f (x) > 0
for all x ≥ 1. Then for all natural numbers n, we get,

n+1 n+1 n

f (k) ≤ f (x)dx ≤ f (k)

k=2 1 k=1

8.3. INFINITE SERIES 325
graph

It follows that, ∞∞ ∞

f (k) ≤ f (x)dx ≤ f (k).

k=2 1 k=1

Since f (1) is a finite number, it follows that

∞ ∞

f (k) and f (x)dx

k=1 1

either both converge or both diverge. This completes the proof of the theo-
rem.

Theorem 8.3.4 Suppose that p > 0. Then the p-series

∞1
np

n=1

converges if p > 1 and diverges if 0 < p ≤ 1. In particular, the harmonic
∞
series n=1 1 diverges.
n

Proof. Suppose that p > 0. Then

∞1 ∞
1 xp dx =
x−pdx

1
x1−p ∞
= 1−p 1

1 lim x1−p − 1 .
= 1−p
x→∞

It follows that the integral converges if p > 1 and diverges if p < 1. If p = 1,

then ∞ 1 dx = ln x ∞ = ∞.

1x 1

Hence, the p-series converges if p > 1 and diverges if 0 < p ≤ 1. This

completes the proof of this theorem.

326 CHAPTER 8. INFINITE SERIES

Exercises 8.1

1. Define the statement that the sequence {an}n∞=1 converges to L.

2. Suppose the sequence {an}n∞=1 converges to L and the sequences {bn}n∞=1
converges to M . Then prove that

(a) {can}n∞=1 converges to cL, where c is constant.

(b) {an + bn}∞n=1 converges to L + M .

(c) {an − bn}n∞=1 converges to L − M .

(d) {anbn}∞n=1 converges to LM .

an ∞ L
converges to , if M = 0.
(e) bn n=1 M

3. Suppose that 0 < an ≤ an+1 < M for each natural number n. Then
prove that

(a) {an}n∞=1 converges.

(b) {−an}∞n=1 converges.

(c) ank ∞ converges for each natural number k.
n=1

4. Prove that xn ∞ converges to 0 for every real number x.

n! n=1

5. Prove that n! ∞ converges to 0.

nn n=1

6. Prove that for each natural number n ≥ 2,

(a) 1 + 1 + ··· + 1 < ln(n) < 1 + 1 + ··· + n 1 .
2 3 n 2 − 1

(b) 1 + 1 + ··· + 1 < n1 dt < 1+ 1 + ··· + 1 for each
2p 3p np 1 tp 2p (n − 1)p

p > 0.

n ∞ ∞1

(c) 1 converges if and only if 1 tp dt converges. De-

kp n=1 ∞

k=1 1

n

termine the numbers p for which kp converges.

n=1 n=1

8.4. SERIES WITH POSITIVE TERMS 327

7. Prove that n∞

rk converges if and only if |r| < 1.

k=0 n=1

8. Prove that n1 ∞
9. Prove that k
diverges.
k=1
n=1

n ∞

1 diverges.

k=2 k ln k n=2

10. Prove that for each natural number m ≥ 2,

m m+1

(a) (ln t)dt < ln(m!) < (ln t)dt

11

(b) m(ln(m) − 1) < ln(m!) < (m + 1)(ln(m + 1) − 1).

mm (m + 1)m+1
(c) em−1 < m! < .
em

(d) lim (m!)1/m = +∞.
m→+∞

(e) lim (m!)1/m 1
=
m→+∞ m e

11. Prove that {(−1)n}n∞=1 does not converge.

12. Prove that sin(1/n) ∞ converges to 1.

(1/n) n=1

13. Prove that sin n ∞ converges to zero.

n n=1

8.4 Series with Positive Terms

Theorem 8.4.1 (Algebraic Properties) Suppose that ∞ ak and ∞ bk
k=1 k=1

are convergent series and c > 0. Then

∞ ∞∞

(i) (ak + bk) = ak + bk

k=1 k=1 k=1

328 CHAPTER 8. INFINITE SERIES

∞ ∞∞

(ii) (ak − bk) = ak − bk

k=1 k=1 k=1

∞∞

(iii) c ak = c ak

k=1 k=1

(iv) If m is any natural number, then the series

∞ and ∞

ck ck

k=1 k=m

either both converge or both diverge.

Proof.
Part (i)

∞n

(ak ± bk) = lim (ak ± bk)

k=1 n→∞ k=1

nn

= lim ak ± lim bk
n→∞
k=1 n→∞ k=1

∞∞

= ak ± bk.

k=1 k=1

Part (ii) This part also follows from the preceding argument.

Part(iii) We see that

∞n

c ak = lim c ak

k=1 n→∞ k=1

n

= c lim ak
n→∞
k=1

∞

= c ak.

k=1

8.4. SERIES WITH POSITIVE TERMS 329

Part (iv) We observe that

∞ m−1 ∞

ak = ak + ak.

k=1 k=1 k=1

Therefore,

∞n

ak = lim ak

k=1 n→∞ k=1

m−1 n

= ak + lim ak.

k=1 n→∞ k=m

It follows that the series

∞ and ∞

ak ak

k=1 k=m

either both converge or both diverge. This completes the proof of this theo-
rem.

Theorem 8.4.2 (Comparison Test) Suppose that 0 < an ≤ bn for all natural
numbers n ≥ 1.

(a) If there exists some M such that n ak ≤ M, for all natural numbers
k=1
∞
n, then k=1 ak converges. If there exists no such M , then the series

diverges.

(b) If ∞ bk converges, then ∞ ak converges.
(c) If k=1 k=1

∞ ak diverges, then ∞ bk diverges.
k=1 k=1

(d) If cn > 0 for all natural numbers n, and

lim cn = L, 0 < L < ∞,
n→∞ an

then the series ∞ ak and ∞ ck either both converge or both diverge.
k=1 k=1

330 CHAPTER 8. INFINITE SERIES

nn

Proof. Let An = ak, Bn = bk, 0 < an ≤ bn for all natural numbers

k=1 k=1
nAup. rpTeephrrebessoeeuqnnutdetnhocefe{sleB{aAnst}nn∞u}=p∞n1=p1eranbdou{nBdn}o∞nf={1Aanr}en∞s=t1riacntldy
increasing sequence. Let
let B represent the least

Part (a) If An ≤ M for all natural numbers, then {An}n∞=1 is a bounded
and strictly increasing sequence. Then A is a finite number and {An}∞n=1

converges to A and

∞

A = ak.

k=1

∞∞

Part (b) If bk converges, then bk = B and An ≤ Bn ≤ B for all

k=1 k=1
∞

natural numbers n. By Part (a), ak converges to A.

k=1

∞

Part (c) If ak diverges, then the sequence {An}∞n=1 diverges. Since {An}n∞=1

k=1

is strictly increasing and divergent, for every M there exists some m such
that

M < An ≤ Bn

for all natural numbers n ≥ m. It follows that {Bn}∞n=1 diverges.

Part (d) Suppose that 0 < an and 0 < cn, 0 < L < ∞, L
= and
2

lim cn = L.
n→∞ an

Then there exists some natural number m such that

cn −L 1
<
an 2

8.4. SERIES WITH POSITIVE TERMS 331

for all natural numbers n ≥ m. Hence, for all n ≥ m, we have

−L < cn −L < L L < cn < 3 L
,
2 an 2 2 an 2

L an ≤ cn ≤ 3
2 L an.
2

L n mck ≤ 3 n
2 L
ak ≤ ak
2
k=m k=m k=m

n∞ Lm n∞

If ak diverges, then ak diverges and, hence ck
2
k=1 n=∞1 k=m k=m n=m
n

and ck both diverge.

k=1 ∞ k=1 3 n∞
n

If ak converges, then L ak converges and, hence,
2
k=1 ∞ k=1 k=m n=m
n∞

ck and ck both converge.

k=m n=m k=1 n=1

This completes the Proof of Theorem 8.4.2.

Theorem 8.4.3 (Ratio Test) Suppose that 0 < an for every natural number

n and lim an+1 = r.
n→∞ an

Then the series ∞ ak
k=1

(a) converges if r < 1;

(b) diverges if r > 1;

(c) may converge or diverge if r = 1; the test fails.

Proof. Suppose that 0 < an for every natural number n and
lim an+1 = r.
n→∞ an

332 CHAPTER 8. INFINITE SERIES

Let > 0 be given. Then there exists some natural number M such that

an+1 − r < , − + r < an+1 < r + (1)
an an

(r − )an < an+1 < (r + )an

for all natural numbers n ≥ M .
Part (a) Suppose that 0 ≤ r < 1 and = (1 − r)/2. Then for each natural
number k, we have

am+k < (r + )kam = 1+r k (2)
2 am . . .

Hence, by (2), we get

∞ m−1 ∞

an = an + am+k

n=1 n=1 k=0

m−1 ∞ 1+r k
2
< an + am

n=1 k=0

= m−1 an + am
n=1
1 − 1+r
2

m−1 2am
1−r
= an +

n=1

< ∞.

∞

It follows that the series an converges.

n=1

Part (b) Suppose that 1 < r, = (r − 1)/2. Then by (1) we get

3r − 1
an < 2 an < an+1

for all n ≥ m. It follows that

0 < am ≤ lim am+k = lim an.

k→∞ n→∞

8.4. SERIES WITH POSITIVE TERMS 333

∞

By the Divergence test, the series an diverges.

n=1

∞1 ∞1
Part (c) For both series and ,
n n2
n=1 n=1

lim an+1 = 1.
n→∞ an

∞1 ∞1
But, by the p-series test, diverges and converges. Thus, the
n n2
n=1 n=1

ratio test fails to test the convergence or divergence of these series when

r = 1.

This completes the proof of Theorem 8.4.3.

Theorem 8.4.4 (Root Test) Suppose that 0 < an for each natural number

n and

nl→im∞(an)1/n = r.

Then the series ∞ ak
k=1

(a) converges if r < 1;

(b) diverges if r > 1;

(c) may converge or diverge if r = 1; the test fails.

Proof. Suppose that 0 < an for each natural number n and

lim (an)1/n = r.

n→∞

Let > 0 be given. Then there exists some natural number m such that

(an)1/n − r < (3)
r − < (an)1/n < r + . . .

for all natural numbers n ≥ m.

Part (a) Suppose r < 1 and = 1+r . Then, by (3), for each natural number

n ≥ m, we have 2

(an)1/n < 1 + r and an < 1−r n
2 .

2

334 CHAPTER 8. INFINITE SERIES

it follows that

∞ m−1 ∞

ak = an + an

n=1 n=1 n=m

m−1 ∞ 1+r n
2
< an +

n=1 n=m

m−1 1+r m 1
2
= an + 1− 1+r
2
n=1

m−1 1+r m 2
2 1−r
= an +

n=1

< ∞.

∞ = (r − 1)/2. Then, by (3), for each natural

Therefore, ak converges.

n=1

Part (b) Suppose r > 1 and
number n ≥ m, we have

1+r < (an)1/n
1< =r+

2
1+r n
1< < an.
2

∞

It follows that lim an = 0 and, by the Divergence test, the series an

n→∞ n=1

diverges.

∞1 ∞1
Part (c) For each of the series and n2 we have r = 1, where
n
n=1 n=1

r = lim (an)1/n.

n→∞

∞1 ∞1
But the series diverges and the series converges by the p-series
n n2
n=1 n=1

test. Therefore, the test fails to determine the convergence or divergence for

these series when r = 1. This completes the proof of Theorem 8.4.4.

8.4. SERIES WITH POSITIVE TERMS 335

Exercises 8.2

∞

1. Define what is meant by ak.

k=1

2. Define what is meant by the sequence of nth partial sums of the series
∞
ak.
k=1

∞ a
1−r
3. Suppose that a = 0. Prove that ark converges to if |r| < 1.

k=0

∞1 3
4. Prove that the series converges to .
k(k + 2) 4
k=1

5. Prove that ∞1 converges to 1 if p > 1 and diverges otherwise.

k=1 kp p − 1

6. Prove that n∞ ∞n
is an increasing sequence and the series ln
n + 1 n=1 n+1

n=1

diverges.

∞ 1 if |x| < 1.

7. Prove that (−1)kxk converges to 1+x

k=0

8. ∞ x2k converges to 1 if |x| < 1.
1 − x2
Prove that

k=0

∞ 1 if |x| < 1.

9. Prove that (−1)kx2k converges to 1 + x2

k=0

∞

10. Prove that if ak converges, then lim ak = 0. Is the converse true?
k→∞
k=0

Explain your answer.

∞∞

11. Suppose that if ak converges to L and bk converges to M . Prove

k=0 k=0

that

336 CHAPTER 8. INFINITE SERIES

∞

(a) (c ak) converges to cL for each constant c.

k=0
∞

(b) (ak + bk) converges to L + M .

k=0
∞

(c) (ak − bk) converges to L − M .

k=0
∞

(d) akbk may or may not converge to LM .

k=0

∞1 ∞1
12. Prove that kp converges if and only if tp dt converges. Deter-
1
k=1

mine the values of p for which the series converges.

13. Suppose that f (x) is continuous and decreasing on the interval [a, +∞).
∞
Let ak = f (k) for each natural number k. Then the series ak con-

∞ k=1

verges if and only if f (x)dx converges.

a

n

14. Suppose that 0 ≤ ak ≤ ak+1 for each natural number k, and sn = ak.

k=1
∞

Prove that if sn ≤ M for some M and all natural numbers n, then ak

converges. k=1

15. Suppose that 0 ≤ ak ≤ bk for each natural number k. Prove that

∞∞

(a) if bk converges, then ak converges.

k=1 k=1

∞∞

(b) if ak diverges, then bk diverges.

k=1 k=1

∞

(c) if lim ak = 0, then ak diverges.
k→∞
k=1

8.4. SERIES WITH POSITIVE TERMS 337

∞

(d) if lim ak = 0, then ak may or may not converge.
k→∞
k=1

16. Suppose that 0 < ak for each natural number k. Prove that if lim (ak+1/ak) <
k→∞
∞
1, then ak converges.
k=1

17. Suppose that 0 < ak for each natural number k. Prove that if lim (ak+1/ak) >
k→∞
∞
1, then ak diverges.
k=1

18. Suppose that 0 < ak for each natural number k. Prove that if kl→im∞(ak+1/ak) =
∞
1, then ak may or may not converge.
k=1

19. Suppose that 0 < ak and 0 < bk for each natural number k. Prove that
∞∞

if 0 < lim (ak/bk) < ∞, then ak converges if and only if bk
k→∞
k=1 k=1
converges.

20. Suppose that 0 < ak for each natural number k. Prove that if lim (ak)1/k <
k→∞
∞
1, then ak converges.
k=1

21. Suppose that 0 < ak for each natural number k. Prove that if lim (ak)1/k >
k→∞
∞
1, then ak diverges.
k=1

22. Suppose that 0 < ak for each natural number k. Prove that if lim (ak)1/k =
k→∞
∞
1, then ak may or may not converge.
k=1

∞∞

23. A series ak is said to converge absolutely if |ak| converges. Sup-

k=1 k=1

pose that lim |ak+1/ak | = p. Prove that

k→∞

338 CHAPTER 8. INFINITE SERIES

∞

(a) ak converges absolutely if p < 1.

k=1
∞

(b) ak does not converge absolutely if p > 1.

k=1
∞

(c) ak may or may not converge absolutely if p = 1.

k=1

∞∞

24. A series ak is said to converge absolutely if |ak| converges. Sup-

k=1 k=1

pose that lim (|ak|)1/k = p. Prove that
k→∞

∞

(a) ak converges absolutely if p < 1.

k=1
∞

(b) ak does not converge absolutely if p > 1.

k=1
∞

(c) ak may or may not converge absolutely if p = 1.

k=1

∞

25. Prove that if ak converges absolutely, then it converges. Is the

k=1

converse true? Justify your answer.

26. Suppose that ak = 0, bk = 0 for any natural number k and lim ak = p.
k→∞ bk

∞

Prove that if 0 < p < 1, then the series ak converges absolutely if

k=1
∞

and only if bk converges absolutely.

k=1

∞∞

27. A series ak is said to converge conditionally if ak converges but

k=1 k=1

8.4. SERIES WITH POSITIVE TERMS 339

∞ ∞ (−1)n+1
converges
|ak| diverges. Determine whether the series
n
k=1 n=1

conditionally or absolutely.

28. Suppose that 0 < ak and |ak+1| < |ak| for every natural number k. Prove
∞∞

that if lim ak = 0, then the series (−1)k+1ak and (−1)kak are

k→+∞ k=1 k=1

both convergent. Furthermore, show that if s denotes the sum of the

series, then s is between the nth partial sum sn and the (n + 1)st partial
sum sn+1 for each natural number n.

∞ n
3n
29. Determine whether the series (−1)n converges absolutely or condi-

n=1

tionally.

∞ (2n)!
n10
30. Determine whether the series (−1)n converges absolutely or con-

n=1

ditionally.

In problems 31–62, test the given series for convergence, conditional conver-
gence or absolute convergence.

∞ n! 32. ∞ (−1)n+1 5n
5n n!
31. (−1)n
n=1
n=1

33. ∞ 4n 34. ∞ 4n

(−1)nn 5 (−1)n+1n2 5

n=1 n=1

∞ (−1)n ∞ (−1)n+1
35. 36.

n3/2 n1/2

n=1 n=1

∞ (−1)n ∞ (−1)n+1
37. , 0 < p < 1 38. , 1 < p

np np

n=1 n=1

∞ (n + 1) ∞ (n + 1)2
n2 +2 3n
39. (−1)n 40. (−1)n+1

n=1 n=1

340 CHAPTER 8. INFINITE SERIES

∞ (n + 2)2 ∞ 3n
(n + 1)3
41. (−1)n+1 42. (−1)n−1 2

n=1 n=1 n2

∞ (−1)n(4/3)n ∞ (−4)n
43. 44.

n4 (n!)n

n=1 n=1

∞ n ∞ (n + 1)!
5· ·· (2n
45. (−3)n (2n)! 46. (−1)n 1 · 3 · + 1)

n=1 n=1

47. ∞ (−1)n (n!)22n ∞ (n − 1)
(2n)! n3/2
48. (−1)n+1
n=1
n=1

49. ∞ (−1)n(n!)2 4n ∞ 2 · 4 · · · (2n + 2)
(2n)! · 4 · 7 · · · (3n + 1)
50. (−1)n 1
n=1
n=1

∞ 5n+1 52. ∞ (−1)n+1 (n + 1)
24n (n + 3)
51. (−1)n−1
n=1
n=1

∞ (n + 2) ∞ (n + 2)
n5/4 n7/4
53. (−1)n+1 54. (−1)n

n=1 n=1

∞ (3n2 + 2n − 1) ∞ (−1)n
2n3 56.
55. (−1)n
n(ln n)
n=1
n=2

57. ∞ (−1)n (ln n) ∞ (ln n)
n n2
58. (−1)n+1
n=2
n=1

∞ n! 60. ∞ (−1)n np , 0 < p < 1
np n!
59. (−1)n , 0 < p < 1
n=1
n=1

∞ n! 62. ∞ (−1)n np , 1 < p
np n!
61. (−1)n , 1 < p
n=1
n=1

8.5. ALTERNATING SERIES 341

∞

63. Suppose that 0 < ak for each natural number k and ak converges.

∞ k=1

Prove that apk converges for every p > 1.

k=1

∞

64. Suppose that 0 < ak for each natural number k and ak diverges.

∞ k=1

Prove that apk, for 0 < p < 1.

k=1

65. Suppose that 0 < r < 1 and |ak+1/ak| < r for all k ≥ N . Prove that
∞
ak converges absolutely.
k=1

∞ an
+ bn
66. Prove that (−1)n 3 converges absolutely if 0< a< b.

k=1

8.5 Alternating Series

Definition 8.5.1 Suppose that for each natural number n, bn is positive or
∞
negative. Then the series k=1 bk is said to converge

(a) absolutely if the series ∞ |bk| converges;
k=1

(b) conditionally if the series ∞ bk converges but ∞ |bk| diverges.
k=1 k=1

Theorem 8.5.1 If a series converges absolutely, then it converges.

∞

Proof. Suppose that |bk| converges. For each natural number k, let

k=1

ak = bk + |bk| and ck = 2|bk|. Then 0 ≤ ak ≤ ck for each k. Since

∞∞ ∞

ck = 2|bk| = 2 |bk|,

k=1 k=1 k=1

342 CHAPTER 8. INFINITE SERIES

∞∞

the series ck converges. by the comparison test ak also converges. It

k=1 k=0

follows that

∞∞

bk = (ak − |bk|)

k=1 k=1 ∞
∞

= ak − |bk|

k=1 k=1

∞

and the series bk converges. This completes the proof of the theorem.

k=1

Definition 8.5.2 Suppose that for each natural number n, an > 0. Then an
alternating series is a series that has one of the following two forms:

n

(a) a1 − a2 + a3 − · · · + (−1)n+1an + · · · = (−1)k+1ak

k=1

∞

(b) −a1 + a2 − a3 + · · · + (−1)nan + · · · = (−1)kak.

k=1

Theorem 8.5.2 Suppose that 0 < an+1 < an for all natural numbers m, and

lim an = 0. Then

n→∞

∞∞

(a) (−1)nan and (−1)n+1an both converge.

n=1 n=1

∞n

(b) (−1)k+1an − (−1)k+1an < an+1, for all n;

k=1 k=1

∞∞

(c) (−1)kak − (−1)kak < an+1, or all n.

k=1 k=1

Proof.

8.5. ALTERNATING SERIES 343

n

Part (a) For each natural number n, let sn = (−1)k+1ak. Then,

k=1

2n+2 2n

s2n+2 − s2n = (−1)k+1ak − (−1)k+1ak

k=1 k=1

= (−1)2n+3a2n+2 + (−1)2n+2a2n+1

= a2n+1 − a2n+2 > 0.

Therefore, s2n+2 > s2n and {s2n}∞n=1 is an increasing sequence. Similarly,

s2n+3 − s2n+1 = (−1)2n+4a2n+3 − (−1)2n+2a2n+1 = a2n+3 − a2n+1 < 0.

Therefore, s2n+3 < s2n+1 and {s2n+1}∞n=0 is a decreasing sequence. Further-
more,

s2n = a1 − a2 + a3 − a4 + · · · + (−1)2n+1a2n
= a1 − (a2 − a3) − (a4 − a5) − · · · − (a2n−2 − a2n−1) − a2n < a1.

TThhuerse, fo{rse2,n{}sn∞=2n1}n∞is=1acnoninvcerregaessintog sequence which is bounded above by a1.
some number s ≤ a1. Then

lim s2n+1 = lim s2n + lim a2n+1

n→∞ n→∞ n→∞

= lim s2n

n→∞

= s.

It follows that

lim sn = s

n→∞

∞∞

and the series (−1)n+1ak converges to s and the series (−1)nak con-

n=1 n=1

verges to −s.

Part (b) In the proof of Part (a) we showed that

s2n < s < s2n+1 < s2n−1 . . . (1)
for each natural number n. It follows that

0 < s − s2n < s2n+1 − s2n = a2n+1

344 CHAPTER 8. INFINITE SERIES

and ∞ 2n
Similarly,
(−1)k+1ak − (−1)k+1ak < a2n+1.

k=1 k=1

s2n − s2n−1 < s − s2n−1
s2n−1 − s2n > s2n−1 − s
s − s2n−1 < s2n−1 − s2n = a2n

∞ 2n−1

(−1)k+1ak − (−1)k+1ak < a2n.

k=1 k=1

It follows that for all natural numbers n,

∞n

(−1)k+1ak − (−1)k+1ak < an+1.

k=1 k=1

∞n

Part (c) (−1)kak − (−1)kak

k=1 k=1

∞n

= (−1) (−1)k+1ak − (−1)k+1ak

k=1 k=1

∞n

= (−1)k+1ak − (−1)k+1ak < a2n+1.

k=1 k=1

This concludes the proof of this theorem.

Theorem 8.5.3 Consider a series ∞ ak. Let
k=1

lim an+1 =L , lim |an|1/n = M
an
n→∞ n→∞

(a) If L < 1, then the series ∞ ak converges absolutely.
(b) If L > 1, then the series k=1
(c) If M < 1, then the series
∞ ak does not converge absolutely.
k=1

∞ ak converges absolutely.
k=1

8.5. ALTERNATING SERIES 345

(d) If M > 1, then the series ∞ ak does not converge absolutely.
k=1

(e) If L = 1 or M = 1, then the series ∞ ak may or may not converge
k=1

absolutely.

∞

Proof. Suppose that for a series ak,

k=1

lim an+1 = L and lim |an|1/n = M.
n→∞ an
n→∞

∞

Part (a) If L < 1, then the series |ak| converges to the ratio test, since

k=1

lim |an+1| = lim an+1 = L < 1.
|an| an
n→∞ n→∞

∞

Hence, the series ak converges absolutely.

k=1

∞

Part (b) As in Part (a), the series |ak| diverges by the ratio test if L > 1,

k=1

since |an+1| an+1
|an| an
lim = lim = L > 1.

n→∞ n→∞

∞

Part (c) If M < 1, then the series |ak| converges by the root test, since

k=1

lim |ak |1/n = M < 1.

n→∞

∞

Part (d) If M > 1, then the series |ak| diverges by the root test as in

k=1

Part (c).

∞1 ∞1 ∞1
Part (e) For the series and k2 , L = M = 1, but diverges
k k
k=1 k=1 k=1
∞1
and k2 converges by the p-series test. Thus, L = 1 and M = 1 fail to
k=1

determine convergence or divergence.

346 CHAPTER 8. INFINITE SERIES
This completes the proof of Theorem 8.5.3.

Exercises 8.3 Determine the region of convergence of the following series.

∞ (−1)nxn ∞ (−1)n(x + 2)n
71. 72.

2n 3nn2

n=1 n=1

∞ (−1)n(x − 1)n ∞ (−1)nn!(x − 1)n
73. 74.

n! 5n

n=1 n=1

∞ ∞ (x + 2)n
76.
75. (−2)nxn
2nn2
n=0
n=1
∞ (x + 1)n
3nn3 ∞ (−1)n(x − 3)n
77. (−1)n 78. n3/2

n=1 n=1

∞ (2x)n ∞ (−1)nxn
79. 80.

n! (2n)!

n=1 n=1

∞ (n + 1)!(x − 1)n ∞ (−1)n(2n)!xn
81. 4n 82.

n=1 n!

n=1

∞ ∞ (−1)nn!(x − 1)n
84. 1 · 3 · · · 5 · · · (2n + 1)
83. n2(x + 1)n
n=1
n=1
∞ (−1)n3nxn
∞ (−1)n(n!)2(x − 1)n 86.
85.
23n
3n(2n)!
n=1
n=1
∞ ln(n + 1)2n(x + 1)n
∞ (−1)n(x + 1)n 88.
87.
n+2
(n + 1) ln(n + 1)
n=1
n=1
90. ∞ (−1)n1 · 3 · 5 · · · (2n + 1) xn
∞ (−1)n(ln n)3nxn 2 · 4 · 6 · · · (2n + 2)
89.
n=1
4nn2

n=1

8.6. POWER SERIES 347

8.6 Power Series

Definition 8.6.1 If a0, a1, a2, . . . is a sequence of real numbers, then the
∞
series k=1 ak xk is called a power series in x. A positive number r is called

the radius of convergence and the interval (−r, r) is called the interval of

convergence of the power series if the power series converges absolutely for

all x in (−r, r) and diverges for all x such that |x| > r. The end point x = r is
∞
included in the interval of convergence if k=1 ak rk converges. The end point

x = −r is included in the interval of convergence if the series ∞ (−1)kak rk
k=1

converges. If the power series converges only for x = 0, then the radius of

convergence is defined to be zero. If the power series converges absolutely

for all real x, then the radius of convergence is defined to be ∞.

∞

Theorem 8.6.1 If the series cnxn converges for x = r = 0, then the

n=1
∞

series cnxn converges absolutely for all numbers x such that |x| < |r|.

n=0

∞

Proof. Suppose that cnrn converges. Then, by the Divergence Test,

n=0

lim cnrn = 0.

n→∞

For = 1, there exists some natural number m such that for all n ≥ m,

|cnrn| < = 1.

Let
M = max{|cnrn| + 1 : 1 ≤ n ≤ m}.

Then, for each x such that |x| < |r|, we get |x/r| < 1 and

∞ ∞ xn
r
|cnxn| = |cnrn| ·

n=0 n=0

∞ xn
≤M

r

n=0

M < ∞.
=
1− x
r

348 CHAPTER 8. INFINITE SERIES

∞

By the comparison test the series |cnxn| converges for x such that |x| <

n=0

|r|. This completes the proof of Theorem 8.6.1.

∞

Theorem 8.6.2 If the series cn(x − a)n converges for some x − a =

n=0
∞

r = 0, then the series cn(x − a)n converges absolutely for all x such that

n=0

|x − a| < |r|.

∞

Proof. Let x−a = u. Suppose that cnun converges for some u = r. Then

n=0
∞

by Theorem 8.6.1, the series cnun converges absolutely for all u such that

n=0 ∞

|u| < |r|. It follows that the series cn(x − a)n converges absolutely for all

n=0

x such that |x − a| < |r|. This completes the proof of the theorem.

∞

Theorem 8.6.3 Let cnxn be any power series. Then exactly one of the

n=0

following three cases is true.

(i) The series converges only for x = 0.

(ii) The series converges for all x.

(iii) There exists a number R such that the series converges for all x with
|x| < R and diverges for all x with |x| > R.

Proof. Suppose that cases (i) and (ii) are false. Then there exist two

∞∞

nonzero numbers p and q such that cnpn converges and cnqn diverges.

n=0 n=0

By Theorem 8.6.1, the series converges absolutely for all x such that |x| < |p|.

Let
∞

A = {p : cnpn converges}.

n=0