Solution manual calculus 8th edition varberg, purcell, rigdon ch09

9

CHAPTER

Infinite Series
7
26
63
124
215
5. a1 = , a2 =
, a3 = , a4 =
, a5 =
8
27
64
125
216

9.1 Concepts Review
1. a sequence

lim

n3 + 3n 2 + 3n
(n + 1)3

n →∞

2.

lim an exists (finite sense)

n →∞

4. –1; 1
6. a1 =

Problem Set 9.1
1
2
3
4
5
1. a1 = , a2 = , a3 = , a4 = , a5 =
2
5
8
11
14
n
1
1
lim
= lim
= ;
3
n →∞ 3n –1 n →∞ 3 – 1
n

converges

5
8
11
14
17
2. a1 = , a2 = , a3 = , a4 = , a5 =
2
3
4
5
6
3 + n2
3n + 2
= lim
= 3;
n →∞ n + 1
n →∞ 1 + 1
lim

n

converges

a4 =

+ 32 + 13
n

+ 3n 2 + 3n + 1

=1

n

n→∞ 1 + 3
n

n3 + 3n 2 + 3n

n→∞ n3

1 + n3 + 32

= lim

3. bounded above

= lim

n

5
14
29
, a2 =
, a3 =
,
3
5
7

50 5 2
77
=
, a5 =
9
9
11

3 + 22
3n 2 + 2
3
n
= lim
=
;
2
n →∞ 2n + 1
n→∞ 2 + 1
n
converges
lim

1
2 1
3
4 2
7. a1 = – , a2 = = , a3 = – , a4 = = ,
3
4 2
5

6 3
5
a5 = –
7
n
1
= lim
= 1, but since it alternates
lim
n →∞ n + 2 n→∞ 1 + 2
n

6
18
38
3. a1 = = 2, a2 =
= 2, a3 = ,
3
9
17
66 22
102 34
a4 =
= , a5 =
=
27 9
39 13
lim

4n 2 + 2

n →∞ n 2

+ 3n – 1

= lim

4+

n →∞ 1 + 3
n

2
n2

–

1
n2

between positive and negative, the sequence
diverges.

= 4;

converges

2
3
4

5
8. a1 = –1, a2 = , a3 = – , a4 = , a5 = –
3
5
7
9
⎧−1 for n odd
cos(nπ) = ⎨
⎩ 1 for n even
lim

n

n →∞ 2n – 1

4.

a1 = 5, a2 =

14
29
50
77
, a3 =
, a4 = , a5 =
3
5
7
9

3n + n
3n 2 + 2
= lim
= ∞;
n →∞ 2n –1
n→∞ 2 – 1
2

= lim

1

n→∞ 2 – 1
n

=

1
, but since cos(n π )
2

alternates between 1 and –1, the sequence
diverges.

lim

n

diverges

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material may be reproduced, in any form or by any means, without permission in writing from the publisher.


1
1
1
1
9. a1 = –1, a2 = , a3 = – , a4 = , a5 = –
2
3
4
5
n
1 cos(nπ) 1
cos ( nπ ) = ( −1) , so – ≤
≤ .
n
n
n
1
1
lim – = lim = 0, so by the Squeeze
n →∞ n n→∞ n

Theorem, the sequence converges to 0.
10. a1 = e –1 sin1 ≈ 0.3096, a2 = e –2 sin 2 ≈ 0.1231,
a3 = e

–3

sin 3 ≈ 0.0070, a4 = e

–4

sin 4 ≈ –0.0139,

a5 = e –5 sin 5 ≈ –0.0065
–1 ≤ sin n ≤ 1 for all n, so

– e – n ≤ e – n sin n ≤ e – n .
lim – e – n = lim e – n = 0, so by the Squeeze

n →∞

n →∞

Theorem, the sequence converges to 0.
11. a1 =
a3 =

e2
e4
≈ 2.4630, a2 =
≈ 6.0665,

3
9
e6
e8
≈ 23.7311, a4 =
≈ 110.4059,
17
27

e10
≈ 564.7812
39
Consider
e2 x
2e2 x
4e2 x
= lim
= lim
=∞
lim
x →∞ x 2 + 3 x –1 x →∞ 2 x + 3 x →∞ 2
by using l’Hôpital’s Rule twice. The sequence
diverges.
a5 =

e2
e4
12. a1 =
≈ 1.8473, a2 =
≈ 3.4124,

4
16
6

8

e
e
a3 =
≈ 6.3036, a4 =
≈ 11.6444,
64
256
a5 =
e

2n

4n

10

e
≈ 21.510
1024
⎛ e2
=⎜
⎜ 4
⎝

n

⎞ e2
> 1 so the sequence diverges.
⎟ ,
⎟ 4
⎠

π
π2
13. a1 = – ≈ –0.6283, a2 =
≈ 0.3948,
5
25
a3 = –

π3
π4
≈ –0.2481, a4 =
≈ 0.1559,
125
625

a5 = –

π5
≈ –0.0979
3125

(– π)n

n

1
1
+ 3 ≈ 1.9821, a2 = + 3 = 3.0625,
4
16
1
1
a3 =
+ 3 3 ≈ 5.2118, a4 =
+ 9 ≈ 9.0039,
64
256
1
a5 =
+ 9 3 ≈ 15.589
1024

14. a1 =

n

1
⎛1⎞
⎜ ⎟ converges to 0 since –1 < < 1.
4
4
⎝ ⎠

3n / 2 =

( 3)

n

3 ≈ 1.732 > 1 .

diverges since

Thus, the sum diverges.
15. a1 = 2.99, a2 = 2.9801, a3 ≈ 2.9703,
a4 ≈ 2.9606, a5 ≈ 2.9510
(0.99)n converges to 0 since –1 < 0.99 < 1, thus
2 + (0.99)n converges to 2.

16. a1 =
a3 =

a5 =

1
2100
≈ 0.3679, a2 =
≈ 1.72 × 1029 ,
2
e
e
3100

≈ 2.57 × 1046 , a4 =

3

e
5100

4100

e

4

≈ 2.94 × 1058 ,

≈ 5.32 × 1067

e5

Consider lim

x →∞

x100
ex

. By Example 2 of

x100

Section 8.2, lim

ex

x →∞

= 0 . Thus, lim

n100
en

n →∞

= 0;

converges
17. a1 =
a3 =
a5 =

ln1
1
ln 3
3
ln 5
5

= 0, a2 =

ln 2

2

≈ 0.4901,

≈ 0.6343, a4 =

ln 4
≈ 0.6931 ,
2

≈ 0.7198

Consider lim

x →∞

ln x
x

= lim

x →∞

1
x
1

= lim

2 x

x →∞

using l’Hôpital’s Rule. Thus, lim

n →∞

ln n
n

2
x

= 0 by

= 0;

converges.

π
⎛ π⎞
= ⎜ – ⎟ , – 1 < – < 1, thus the sequence
n
5
⎝ 5⎠
5
converges to 0.

514

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material may be reproduced, in any form or by any means, without permission in writing from the publisher.


18. a1 =
a3 =
a5 =

ln1
2

= 0, a2 =

ln 13
6
ln 15
10

lim

ln 1n

n →∞

2n

2

ln 14
2 2

a3 = 61/ 6 ≈ 1.3480, a4 = 81/ 8 = 23 / 8 ≈ 1.2968,

≈ –0.4901,

a5 = 101/10 ≈ 1.2589

Consider lim (2 x)1/ 2 x . This limit is of the form
x →∞

≈ –0.5089
ln 1x

x →∞

x →∞

20. a1 = 21/ 2 ≈ 1.4142, a2 = 41/ 4 = 21/ 2 ≈ 1.4142,

≈ –0.3466,

≈ –0.4485, a4 =

Consider lim
= lim −

ln 12

2

2x

∞ 0 . Let y = (2 x)1/ 2 x , then ln y =

= lim

− ln x

x →∞

2x

= lim

x →∞

− 1x
1
2x

= 0 by using l’Hôpital’s Rule. Thus,

x

ln 2 x
2x

∞
This limit is of the form .
∞
lim

1

1
ln 2 x.
2x

x →∞ 2 x

ln 2 x = lim

x →∞

1

ln 2 x
1
= lim x = lim
=0
x →∞ 2 x
x →∞ 2 x →∞ 2 x
lim

= 0; converges

lim (2 x)1/ 2 x = lim eln y = 1

x →∞

1/ 2

⎛ 2⎞
19. a1 = ⎜1 + ⎟
⎝ 1⎠

= 3 ≈ 1.7321,

⎛ 2⎞
a2 = ⎜ 1 + ⎟
⎝ 2⎠
⎛ 2⎞
a3 = ⎜ 1 + ⎟
⎝ 3⎠

= 1; converges

Thus lim (2n)
n →∞

2/ 2

= 2,

3/ 2

⎛ 2⎞

a4 = ⎜1 + ⎟
⎝ 4⎠

x →∞
1/ 2 n

4/2

5/ 2

⎛5⎞
=⎜ ⎟
⎝3⎠

n
1
or an = 1 −
;
n +1
n +1
1 ⎞
1
⎛
lim ⎜ 1 −
= 1; converges
⎟ = 1 − lim
n +1⎠
n →∞ ⎝
n→∞ n + 1

21. an =
3/ 2

≈ 2.1517,
2

9
⎛3⎞
=⎜ ⎟ = ,
2
4
⎝ ⎠

22. an =

5/ 2

⎛ 2⎞
⎛7⎞
a5 = ⎜ 1 + ⎟
=⎜ ⎟
≈ 2.3191
5
⎝
⎠
⎝5⎠
2
Let = h, then as n → ∞, h → 0 and
n
n/2

⎛ 2⎞
lim ⎜1 + ⎟
= lim (1 + h)1/ h = e by
n⎠
h→0
Theorem 6.5A; converges
n →∞ ⎝

n

2

n +1

x

Consider

2

x

. Now, lim

x

= lim

x →∞ 2 x

1

x →∞ 2 x

by l’Hôpital’s Rule. Thus, lim

n

n →∞ 2n +1

=0

ln 2

= 0;

converges
n
n
; lim
2n − 1 n→∞ 2n − 1
1
1
= lim
= , but due to (−1)n , the terms of
2
n →∞ 2 − 1

23. an = (−1)n

n

the sequence alternate between positive and
negative, so the sequence diverges.
24. an =

1
1 − nn−1

= n;

lim n = ∞ ; diverges

n →∞

25. an =

n

=

n

=

n – (n –1)
n – (n – 2n + 1)
n
1

1
lim
= lim
= ; converges
2
n →∞ 2n –1 n→∞ 2 – 1
2

2

2

2

n
;
2n –1

n

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material may be reproduced, in any form or by any means, without permission in writing from the publisher.

26. an =

n
n(n + 1)
n2 + n
=
=
;
(n + 1) − n1+1 (n + 1)2 − 1 n 2 + 2n
n +n
2

lim

n →∞ n 2

+ 2n

= lim

1 + 1n

n→∞ 1 + 2
n

= 1; converges

1

n

sin x
= 1; converges
lim
x →0 x

n2

n2
3n

;

n →∞ 3n

29. an =
lim

2n
n2
2n

n →∞ n 2

= lim

2n

n →∞ 3n

= lim

2

=0
ln 3
by using l’Hôpital’s Rule twice; converges
lim

n→∞ 3n (ln 3) 2

;
2n ln 2
2n (ln 2) 2
= lim
= ∞;
2
n →∞ 2n
n→∞

= lim

diverges
30. an =

an =

2n 2 − 1
n2 + n

1
1
n +1 – n
1
–
;
=
=
n n + 1 n(n + 1) n(n + 1)

to a limit L ≤ 2.

1
5
9
13
31. a1 = , a2 = , a3 = , a4 =
2
4
8
16
an is positive for all n, and an +1 < an for all
4n − 7

converges to a limit L ≥ 0.

2n +1

, so {an }

n + 2n

, so {an } converges

⎛
1 ⎞
1
< 1, so
an +1 = an ⎜ 1 −
and 1 −
⎜ (n + 1) 2 ⎟⎟
(n + 1)2
⎝
⎠
{an } converges to a limit L ≥ 0.

3
5
41
34. a1 = 1; a2 = ; a3 = ; a4 =
2
3
24
an < 2 for all n since
1
1
1
1
1
+ + +

1+ + + ≤
0
1
n
n! 2
2!
2
2 +1
<

1
= 0; converges
n →∞ n( n + 1)

2
2

3
⎛ 3 ⎞⎛ 8 ⎞ 2
33. a2 = ; a3 = ⎜ ⎟ ⎜ ⎟ = ;
4
⎝ 4 ⎠⎝ 9 ⎠ 3
⎛ 3 ⎞ ⎛ 8 ⎞⎛ 15 ⎞ 5
a4 = ⎜ ⎟ ⎜ ⎟⎜ ⎟ = ;
⎝ 4 ⎠ ⎝ 9 ⎠⎝ 16 ⎠ 8
⎛ 3 ⎞⎛ 8 ⎞ ⎛ 15 ⎞ ⎛ 24 ⎞ 3
a5 = ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ =
⎝ 4 ⎠⎝ 9 ⎠ ⎝ 16 ⎠ ⎝ 25 ⎠ 5
an > 0 for all n and an +1 < an since

lim

n ≥ 2 since an +1 − an = −

< 2 for all n, and an < an +1 for all

n since an +1 − an =

sin n
1
1
27. an = n sin ; lim n sin = lim
= 1 since
n n→∞
n n→∞ 1

28. an = (–1)n

1
7
17
31
32. a1 = ; a2 = ; a3 = ; a4 =
2
6
12
20

∞

k

⎛1⎞
∑ ⎜⎝ 2 ⎟⎠ = 2
k =0

the sum never reaches 2. an < an +1 since each
term is the previous term plus a positive quantity,
so {an } converges to a limit L ≤ 2.
1
3
1⎛3⎞ 7
35. a1 = 1, a2 = 1 + (1) = , a3 = 1 + ⎜ ⎟ = ,
2
2
2⎝2⎠ 4
1 ⎛ 7 ⎞ 15
a4 = 1 + ⎜ ⎟ =
2⎝4⎠ 8
1 1
Suppose that 1 < an < 2, then < an < 1, so
2 2
3
1
3
< 1 + an < 2, or < an +1 < 2. Thus, since
2
2
2
1 < a2 < 2, every subsequent term is between

3
2

and 2.
1
1
an < 1, so an < 1 + an = an +1
2
2
and the sequence is nondecreasing, so {an }
converges to a limit L ≤ 2.
an < 2 thus

516

Section 9.1

Instructor’s Resource Manual

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material may be reproduced, in any form or by any means, without permission in writing from the publisher.


1⎛
2⎞
⎜2+ ⎟ =
2⎝
2⎠
1 ⎛ 3 4 ⎞ 17

a3 = ⎜ + ⎟ = , a4
2 ⎝ 2 3 ⎠ 12

36. a1 = 2, a2 =

un +1 = 3 + un > un if 3 + un > un 2 or

3
,
2
1 ⎛ 17 24 ⎞ 577
= ⎜ + ⎟=
2 ⎝ 12 17 ⎠ 408

un 2 – un – 3 < 0. un 2 – un – 3 = 0 when

Suppose an > 2 , and consider

(

1⎛
2 ⎞
2
>2 2⇔
⎜ an + ⎟ > 2 ⇔ an +
an ⎠
an
2⎝
2

> 0 , which is always true. Hence,

an > 2 for all n. Also,

n
1
2
3
4
5
6
7
8
9
10
11
lim un ≈ 2.3028

u=

41.

(

)

1
1 + 13 , then
2

(

(

)

)

(

)

Instructor’s Resource Manual

)

)

)

)

(

n
1
2
3
4
5

6
7
8
lim un ≈ 1.1118

un
0
1
1.1
1.11053
1.11165
1.11177
1.11178
1.11178

n →∞

)

(

(

(

1
1 ± 13 so
2

1⎛

2 ⎞
40. If a = lim an where an +1 = ⎜ an + ⎟ , then
an ⎠
2⎝
n →∞
1⎛
2⎞
a = ⎜ a + ⎟ or 2a 2 = a 2 + 2; a 2 = 2 when
a⎠
2⎝
a = ± 2, so a = 2, since a > 0.

1
3 < 3 + un < 7 + 13 and
2
1
1
3 < 3 + un = un +1 <
7 + 13 = 1 + 13
2
2
⎛ 1
1
7 + 13 = 1 + 13 can be seen by
⎜⎜
2
⎝ 2
squaring both sides of the equality and noting
that both sides are positive.) Hence, since
1

0 < u1 = 3 ≈ 1.73 < 1 + 13 ≈ 2.3028,
2
1
3 < un < 1 + 13 for all n; {un } is bounded
2
above.

)

)

1
1 + 13 ≈ 2.3028 since u > 0 and
2

(

n →∞

(

)

1
1 – 13 < 0.
2

un
1.73205
2.17533

2.27493
2.29672
2.30146
2.30249
2.30271
2.30276
2.30277
2.30278
2.30278

(

) (

u 2 – u – 3 = 0 when u =

2 < an +1 ≤ an and the

38. Suppose that 0 < un <

(

n →∞

series converges to a limit L ≥ 2.
37.

)

39. If u = lim un , then u = 3 + u or u 2 = 3 + u;

1⎛
2 ⎞
an +1 ≤ an ⇔ ⎜ an + ⎟ ≤ an
⎜
2⎝
an ⎟⎠
1 1
⇔
≤ an ⇔ 2 ≤ an
an 2

which is true. Hence,

)

(

an 2 + 2 > 2 2an ⇔ an 2 − 2 2an + 2 > 0 ⇔

( an − 2 )

(

1
1 ± 13 , thus un +1 > un if
2
1
1
1

1 – 13 < un < 1 + 13 ,
1 – 13 < 0
2
2
2
1
and 0 < un < 1 + 13 for all n, as shown
2
above, so {un } is increasing. Hence, by Theorem
D, {un } converges.
un =

)

42. Since 1.1 > 1, 1.1a > 1.1b if a > b. Thus, since
u3 = 1.1 > 1 = u2 , u4 = 1.11.1 > 1.11 = u3 .
Suppose that un < un +1 for all n ≤ N. Then
u N +1 = 1.1u N > 1.1u N –1 = u N , since u N > u N –1
by the induction hypothesis. Thus, un is
increasing.

1.1un < 2 if and only if un ln1.1 < ln 2;
un <

ln 2
≈ 7.3. Thus, unless un > 7.3,
ln1.1

un +1 = 1.1un < 2. This means that {un } is
bounded above by 2, since u1 = 0.

Section 9.1

517

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material may be reproduced, in any form or by any means, without permission in writing from the publisher.


1
k
→ 0; using Δx = , an equivalent
n
n
definite integral is

43. As n → ∞,
1

∫0 sin x dx = [– cos x]0 = – cos1 + cos 0 = 1 – cos1
1

≈ 0.4597

1
k
→ 0; using Δx = , an equivalent
n
n
definite integral is

1 1
π
–1 1
–1
–1
∫0 1 + x2 dx = [tan x]0 = tan 1 – tan 0 = 4

44. As n → ∞,

45.

n
n − (n + 1)
−1
1
−1 =
=
=
;
n +1
n +1
n +1 n +1
1
1
< ε is the same as < n + 1. For any given
ε
n +1
1
ε > 0 , choose N > − 1 then

ε

n≥N ⇒

46. For n > 0,

n
n +1

=

n

.

n
2

< ε is the

n2 + 1
1 1
= n+ > .
n
n ε
1
1
Since n + > n , it suffices to take n > . So for
n
ε

1
any given ε > 0 , choose N > , then

ε

n
n +1
2

and L is a lower bound for {an }. Then {– an } is
a nondecreasing sequence and –L is an upper
bound for {– an }. By what was just proven,
{– an } converges to a limit A ≤ –L, so {an }
converges to a limit B = –A ≥ L.
49. If {bn } is bounded, there are numbers N and M

with N ≤ bn ≤ M for all n. Then
an N ≤ an bn ≤ an M .
lim an N = N lim an = 0 and
n→∞

n→∞

n →∞

by the Squeeze Theorem, and by Theorem C,
lim an bn = 0.
n →∞

50. Suppose {an + bn } converges. Then, by

Theorem A
lim [(an + bn ) – an ] = lim (an + bn ) – lim an .
n →∞

n→∞

n→∞

But since (an + bn ) – an = bn , this would mean
that {bn } converges. Thus {an + bn } diverges.

{an } and {bn }

48. Suppose that {an } is a nondecreasing sequence,
and U is an upper bound for {an }, so
S = {an : n ∈ } is bounded above. By the
completeness property, S has a least upper bound,
which we call A. Then A ≤ U by definition and
an ≤ A for all n. Suppose that lim an ≠ A, i.e.,
n →∞

that {an } either does not converge, or does not
converge to A. Then there is some ε > 0 such that

Section 9.1

< A for all n,
2
which contradicts A being the least upper bound
for the set S. For the second part of Theorem D,

suppose that {an } is a nonincreasing sequence,

51. No. Consider an = (−1)n and bn = (−1)n +1 . Both

< ε.

47. Recall that every rational number can be written
as either a terminating or a repeating decimal.
Thus if the sequence 1, 1.4, 1.41, 1.414, … has a
limit within the rational numbers, the terms of the
sequence would eventually either repeat or
terminate, which they do not since they are the
decimal approximations to 2, which is
irrational. Within the real numbers, the least
upper bound is 2.

518

ε

lim an M = M lim an = 0, so lim an bn = 0

same as

n≥N ⇒

A – an > ε for all n, an < A –

n →∞

n +1 n +1
2

A – an ≤ ε for n ≥ N since {an } is
nondecreasing and an ≤ A for all n. However, if

n →∞

n
−1 < ε.
n +1

2

A – an > ε for all n, since if A – a N ≤ ε ,

diverge, but

an + bn = (−1) + (−1)n +1 = (−1) n (1 + (−1)) = 0 so
n

{an + bn }
52. a.

converges.

f3 = 2, f 4 = 3, f5 = 5, f 6 = 8,
f 7 = 13, f8 = 21, f9 = 34, f10 = 55

b. Using the formula,

1 ⎡1 + 5 1 − 5 ⎤ 1 ⎡ 2 5 ⎤
f1 =
−
⎢
⎥=
⎢
⎥ =1
2 ⎦
5⎣ 2
5⎣ 2 ⎦
2
2⎤
⎡
1 ⎢⎛ 1 + 5 ⎞ ⎛ 1 − 5 ⎞ ⎥
f2 =
⎜⎜
⎟⎟ − ⎜⎜
⎟⎟
5 ⎢⎝ 2 ⎠ ⎝ 2 ⎠ ⎥
⎣
⎦
⎡
⎤
1 1 + 2 5 + 5 − (1 − 2 5 + 5)
=
⎢
⎥
4
5⎣
⎦

1 ⎡4 5 ⎤
=
⎢
⎥ = 1.
5⎣ 4 ⎦

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material may be reproduced, in any form or by any means, without permission in writing from the publisher.


f
φ n +1 − (−1)n +1φ − n −1
lim n +1 = lim
n →∞ f n
n→∞
φ n − (−1) n φ − n
n +1

= lim

φ n +1 − ( −1)n +1
φ

n→∞

φn −

φ

n +1

= lim

( −1)n

φ − ( −1)
2 n +1
φ

n→∞

n

(

x

⎛ 1⎞
54. Let f ( x) = ⎜ 1 + ⎟ .
⎝ x⎠

)

1−

2

( −1) n

φ

x

=φ

n

⎛ 1⎞
lim ⎜1 + ⎟ = e .
n⎠
n →∞ ⎝

2n

(

)

1
⎡1
⎤
1+ 5 ⎥ − 1+ 5 −1
2
⎣2
⎦
⎛3
5⎞ ⎛1
5⎞

= ⎜⎜ +
⎟⎟ − ⎜⎜ +
⎟⎟ − 1 = 0
⎝2 2 ⎠ ⎝2 2 ⎠

φ 2 − φ −1 = ⎢

c.

⎛ 1⎞
lim ⎜1 + ⎟ = lim (1 + x )1/ x = e, so
x⎠
x →∞ ⎝
x →0 +

Therefore φ satisfies x 2 − x − 1 = 0 .
Using the Quadratic Formula on
x 2 − x − 1 = 0 yields
1± 1+ 4 1± 5
=
.
2
2
1+ 5
;
φ=
2
1
2
2(1 − 5) 1 − 5

− =−
=−
=
1− 5
2
φ
1+ 5
x=

53.

x

1 ⎞
⎛
55. Let f ( x) = ⎜ 1 + ⎟ .
⎝ 2x ⎠
x

1/ x

1 ⎞
⎛
⎛ x⎞
lim ⎜1 + ⎟ = lim ⎜ 1 + ⎟
+
2x ⎠
2⎠
x →∞ ⎝
x →0 ⎝

1/ 2

⎡⎛ x ⎞ 2 / x ⎤
= lim ⎢⎜ 1 + ⎟ ⎥
2⎠ ⎥
x →0+ ⎢⎝
⎣
⎦

= e1/ 2 , so

n

1 ⎞
⎛
lim ⎜ 1 + ⎟ = e1/ 2 .
2n ⎠
n →∞ ⎝
x

1 ⎞
⎛
56. Let f ( x) = ⎜1 + ⎟ .
⎝ x2 ⎠
1

x
⎛ ⎛ 1 ⎞ 2 ⎞1/ x
1 ⎞
⎛

lim ⎜ 1 + 2 ⎟ = lim ⎜ 1 + ⎜ ⎟ ⎟
x →∞ ⎝
x →∞ ⎜
x ⎟
x ⎠
⎝ ⎝ ⎠ ⎠

⎛1⎞
Using the fact that lim f ( x ) = lim+ f ⎜ ⎟ , we
x →∞
x →0
⎝ x⎠
can write
1

⎛ ⎛ 1 ⎞ 2 ⎞1/ x
lim ⎜ 1 + ⎜ ⎟ ⎟ = lim+ 1 + x 2
x →∞ ⎜
x →0
x ⎟
⎝ ⎝ ⎠ ⎠

(

From the figure shown, the sides of the triangle
have length n – 1 + 2x. The small right triangles
3
marked are 30-60-90 right triangles, so x =
;
2

thus the sides of the large triangle have lengths
2
3
n − 1 + 3 and Bn =
n −1+ 3
4
3 2
=
n + 2 3n − 2n − 2 3 + 4 while
4

(

)

n(n + 1) ⎛ 1 ⎞
π
π ⎜ ⎟ = ( n 2 + n)
2
8
⎝2⎠

An
= lim
n →∞ Bn
n →∞
lim

= lim

n→∞ 2

(

π (n2
8

+ 2 3n – 2n – 2 3 + 4)

π 1+

1
n

3 1 + 2 n 3 – n2 – 2

3
n2

+

4
n2

)

=

π
2 3

which leads

1/ x

. Then,

(

ln y = ln 1 + x 2
ln y =

)

1/ x

(

1
ln 1 + x 2
x

lim+ ln y = lim+

)

(

ln 1 + x 2

x→0

x→0

+ n)

)

)

= lim+

3 2
(n
4

(

(

Let y = 1 + x 2

x →0

2

An =

1/ x

to the indeterminate form 1∞ .

)

(

)

1 + x2

2

x →0

x
2x

) = lim 1+2xx
+

1

=0

This gives us
lim+ ln y = 0
x →0

ln ⎛⎜ lim+ y ⎞⎟ = 0
⎝ x →0 ⎠

lim+ y = e0 = 1 or

x →0

(

lim+ 1 + x 2

x →0

)

1/ x

=1

n

1 ⎞
⎛
Thus, lim ⎜ 1 + 2 ⎟ = 1 .
n →∞ ⎝
n ⎠

Instructor’s Resource Manual

Section 9.1

519

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material may be reproduced, in any form or by any means, without permission in writing from the publisher.


x

x

⎛ x −1 ⎞
57. Let f ( x ) = ⎜
⎟ .
⎝ x +1⎠
⎛1⎞
Using the fact that lim f ( x ) = lim+ f ⎜ ⎟ , we
x →∞
x →0
⎝ x⎠
can write
1/ x

x
⎛ 1x − 1 ⎞
⎛ x −1 ⎞
=
lim ⎜
lim
⎜
⎟
⎟
x →∞ ⎝ x + 1 ⎠

x → 0+ ⎜ 1 + 1 ⎟
⎝x ⎠

1/ x

⎛ 1− x ⎞
= lim+ ⎜ 1+xx ⎟
⎟
x →0 ⎜
⎝ x ⎠

1/ x

⎛ 1− x ⎞
= lim+ ⎜
⎟ which leads to the
x →0 ⎝ 1 + x ⎠
indeterminate form 1∞ .
1/ x

⎛ 1− x ⎞
Let y = ⎜
⎟
⎝ 1+ x ⎠

lim+ ln y = lim+
x →0

y ⎤⎥ = lim+
⎦ x →0

= lim+
x →0

1 ⎛ 1− x ⎞
ln ⎜
⎟
x ⎝ 1+ x ⎠
ln

x →0

1/ x

⎛ 2 x2 + 1 ⎞
= lim+ ⎜ 2
⎟
x →0 ⎜ 3 x + 1 ⎟
⎝
⎠

which leads

. Then,

1− x
1+ x

1 − x2

⎛ n −1 ⎞

−2
Thus, lim ⎜
⎟ =e .
n →∞ ⎝ n + 1 ⎠

ln 2 x2 +1
3 x +1
⎡
⎤
ln ⎢ lim+ y ⎥ = lim+
x
⎣ x →0 ⎦ x →0
6x ⎤
⎡ 4x
= lim+ ⎢ 2
−
⎥ (l'Hopital's Rule)
x →0 ⎣ 2 x + 1 3 x 2 + 1 ⎦
=0
This gives us,
ln ⎡⎢ lim+ y ⎤⎥ = 0
⎣ x →0 ⎦
2

(l'Hopital's Rule)

⎛ 1− x ⎞
lim+ ⎜
⎟
x →0 ⎝ 1 + x ⎠

n

x→0

1 ⎛ 2 x2 + 1 ⎞
ln ⎜
⎟
x ⎜⎝ 3 x 2 + 1 ⎟⎠

( )

( )

x
−2

or

Section 9.1

1/ x

lim+ ln y = lim+

1/ x

520

⎞
⎟

⎟
⎠

2 x 2 +1
x2
3 x 2 +1
x2

1/ x

−2
This gives us,
ln ⎡⎢ lim+ y ⎤⎥ = −2
⎣ x →0 ⎦
x →0

⎛
= lim+ ⎜
x →0 ⎜
⎝

1/ x

⎞
⎟
⎟
⎠

⎛ 2 x2 + 1 ⎞
ln y = ln ⎜ 2

⎜ 3 x + 1 ⎟⎟
⎝
⎠

1 ⎛ 1− x ⎞
ln ⎜
⎟
x ⎝ 1+ x ⎠

lim+ y = e −2

x
⎛ 2 + 12
⎞
x
⎟ = lim+ ⎜
⎟
⎜ 3 + 12
0
x
→
⎠
x
⎝

1/ x

1/ x

ln ⎡⎢ lim+

⎣ x →0

⎛ 2 + x2
lim ⎜
x →∞ ⎜⎝ 3 + x 2

⎛ 2 x2 + 1 ⎞
Let y = ⎜ 2
⎜ 3 x + 1 ⎟⎟
⎝
⎠

⎛ 1− x ⎞
ln y = ln ⎜
⎟
⎝ 1+ x ⎠

x →0

⎛1⎞
Using the fact that lim f ( x ) = lim+ f ⎜ ⎟ , we
x →∞
x →0
⎝ x⎠
can write

to the indeterminate form 1∞ .

. Then,
1/ x

ln y =

⎛ 2 + x2 ⎞
58. Let f ( x) = ⎜
⎟ .
⎜ 3 + x2 ⎟
⎝
⎠

= e−2

1/ x

lim+ y = e0 = 1 or

x →0

⎛ 1− x ⎞
lim+ ⎜
⎟
x →0 ⎝ 1 + x ⎠

=1

Thus,
⎛ 2 + n2
lim ⎜
n →∞ ⎜ 3 + n 2
⎝

n

⎞
⎟⎟ = 1 .
⎠

Instructor’s Resource Manual

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material may be reproduced, in any form or by any means, without permission in writing from the publisher.


⎛ 2 + x2
59. Let f ( x ) = ⎜
⎜ 3 + x2
⎝

Problem Set 9.2

x2

⎞
⎟⎟
⎠

∞

⎛1⎞
Using the fact that lim f ( x ) = lim+ f ⎜ ⎟ , we

x →∞
x →0
⎝ x⎠
can write
⎛ 2 + x2
lim ⎜
x →∞ ⎜⎝ 3 + x 2

⎛
= lim+ ⎜
x →0 ⎜
⎝

⎞
⎟
⎟
⎠

2 x 2 +1
x2
3 x 2 +1
x2

x2

1/ x

⎞
⎟
⎟

⎠

2

∞

1/ x 2

∞

. Then,

3.

1/ x 2

)

⎤
⎥ (l'Hopital's Rule)
⎥
⎦⎥

= −1
This gives us,
ln ⎡⎢ lim+ y ⎤⎥ = −1
⎣ x →0 ⎦
⎛ 1− x ⎞
lim ⎜
⎟

⎝1+ x ⎠

x → 0+

Thus,
⎛ 2 + n2
lim ⎜
n →∞ ⎜ 3 + n 2
⎝

⎞
⎟⎟
⎠

n2

= e −1 .

9.2 Concepts Review
1. an infinite series
2. a1 + a2 +…+ an
3.

r < 1;

k

⎛ 1⎞
+⎜– ⎟
⎝ 4⎠

–5

+…

2

k

∞

4.

1/ x 2

or

–4

a
1– r

2

1
series with a = 3, r = – ;
5
3
3 5
S=

= =
⎛ 1⎞ 6 2
1– ⎜ – ⎟ 5
⎝ 5⎠
Thus, by Theorem B,
k
k
∞ ⎡
⎛1⎞
⎛ 1 ⎞ ⎤ 8 5 31
∑ ⎢⎢2 ⎜⎝ 4 ⎟⎠ + 3 ⎜⎝ – 5 ⎟⎠ ⎥⎥ = 3 + 2 = 6
k =0 ⎣
⎦

2 x 2 +1
3 x 2 +1
2

)(

lim y = e −1

⎛ 1⎞
+⎜– ⎟
⎝ 4⎠

1 ⎛1⎞
⎛ 1⎞
∑ 3 ⎜⎝ – 5 ⎟⎠ = 3 – 3 ⋅ 5 + 3 ⎜⎝ 5 ⎟⎠ −…; a geometric
k =0

( )

x
⎡
−1
= lim+ ⎢
⎢
2
x →0
2 x + 1 3x 2 + 1
⎣⎢

x → 0+

–3

1
⎛1⎞
⎛1⎞
∑ 2 ⎜⎝ 4 ⎟⎠ = 2 + 2 ⋅ 4 + 2 ⎜⎝ 4 ⎟⎠ +…; a geometric
k =0

∞

⎛ 2 x2 + 1 ⎞
lim+ ln y = lim+ 2 ln ⎜ 2
⎜ 3x + 1 ⎟⎟
x →0
x →0 x

⎝
⎠
1

(

⎛ 1⎞
=⎜– ⎟
⎝ 4⎠

1
6

1
2
2 8
= = .
series with a = 2, r = ; S =
4
1 – 14 34 3

⎛ 2 x2 + 1 ⎞
ln y = ln ⎜ 2
⎜ 3 x + 1 ⎟⎟
⎝
⎠

y ⎤⎥ = lim+
⎦ x →0

– k –2

=

a = (–4)3 , r = –4; r = 4 > 1 so the series diverges.

1/ x 2

ln ⎡⎢ lim+
⎣ x →0

⎛ 1⎞
∑ ⎜⎝ – 4 ⎟⎠
k =1

1
7
6
7

= (–4)3 + (–4)4 + (–4)5 +…; a geometric series
with

which

leads to the indeterminate form 1∞ .

ln

2

1

2.

⎛ 2x2 + 1 ⎞
= lim+ ⎜ 2
⎟
x →0 ⎜ 3x + 1 ⎟
⎝
⎠

⎛ 2 x2 + 1 ⎞
Let y = ⎜ 2
⎜ 3x + 1 ⎟⎟
⎝
⎠

k

1 1 1 1⎛1⎞
⎛1⎞
∑ ⎜⎝ 7 ⎟⎠ = 7 + 7 ⋅ 7 + 7 ⎜⎝ 7 ⎟⎠ +…; a geometric
k =1

1
1
series with a = , r = ; S = 7 =
7
7

1 – 17

1/ x 2

⎛ 2 + 12
x
= lim ⎜
x →0+ ⎜ 3 + 12
x
⎝

⎞
⎟
⎟
⎠

1.

= e −1

k

2

5 5 1 5⎛1⎞
⎛1⎞
∑ 5 ⎜⎝ 2 ⎟⎠ = 2 + 2 ⋅ 2 + 2 ⎜⎝ 2 ⎟⎠ +…; a geometric
k =1
5

5
1
series with a = , r = ; S = 2 =
2
2
1 – 12
∞

⎛1⎞
∑ 3 ⎜⎝ 7 ⎟⎠
k =1

k +1

3
49
1 – 17

=

= 5.

2

=

3
3 1 3 ⎛1⎞
+ ⋅ + ⎜ ⎟ +…; a
49 49 7 49 ⎝ 7 ⎠

geometric series with a =
S=

5
2
1
2

3
49
6
7

=

3
1
,r = ;
49
7

1
14

Thus, by Theorem B,
k
k +1
∞ ⎡
1 69

⎛1⎞
⎛1⎞ ⎤
2
–
3
⎢
∑ ⎢ ⎜⎝ 4 ⎟⎠ ⎜⎝ 7 ⎟⎠ ⎥⎥ = 5 – 14 = 14 .
k =1 ⎣
⎦

4. diverges

Instructor’s Resource Manual

Section 9.2

521

© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this
material may be reproduced, in any form or by any means, without permission in writing from the publisher.


∞

5.

∞

k –5
4 3 2 1

1 2
∑ k + 2 = – 3 – 4 – 5 – 6 + 0 + 8 + 9 +…;
k =1

6.

1 – k5
k –5
= lim
= 1 ≠ 0; the series
k →∞ k + 2 k →∞ 1 + 2
k

diverges.

∞

⎛1

1 ⎞

⎛ 1 1⎞ ⎛ 1

1⎞ ⎛1

2

9
9 9
series with a = , r = ; > 1, so the series

8
8 8
diverges.

lim

7.

k

9 9 9 9⎛9⎞
⎛9⎞
∑ ⎜⎝ 8 ⎟⎠ = 8 + 8 ⋅ 8 + 8 ⎜⎝ 8 ⎟⎠ +…; a geometric
k =1

1⎞

∑ ⎜⎝ k – k – 1 ⎟⎠ = ⎜⎝ 2 – 1 ⎟⎠ + ⎜⎝ 3 – 2 ⎟⎠ + ⎜⎝ 4 – 3 ⎟⎠ +…;

k =2

1 ⎞ ⎛1
1 ⎞
1
⎛1 ⎞ ⎛1 1⎞
⎛ 1
–
Sn = ⎜ – 1⎟ + ⎜ – ⎟ +…+ ⎜
⎟+⎜ –
⎟ = –1 + ;

2
3
2
–
1
–
2
–
1
n
n
n
n
n
⎝
⎠ ⎝
⎠
⎝
⎠ ⎝
⎠
∞
1
1 ⎞
⎛1
= –1, so ∑ ⎜ –
⎟ = –1
n
k =2 ⎝ k k – 1 ⎠

lim Sn = lim –1 +

n →∞
∞

8.

n→∞

∞

3

k =1

∞

which diverges since

k =1

∞

9.

1

∑ k = 3∑ k
k!

∑ 100k

1

∑k

diverges.

k =1

1
2
6
+
+
+…
100 10, 000 1, 000, 000

=

k =1

n +1
1
an , a1 =
. an > 0 for all n, and for n >99, an +1 > an , so the
100
100
sequence is eventually an increasing sequence, hence lim an ≠ 0. The sequence can also be described by

Consider {an }, where an +1 =

n →∞

an =
∞

10.

n!
100

n

∞

2

k!

∑ 100k

, hence

diverges.

k =1

2

2

∞

2

⎛1

1 ⎞

⎛1 1 ⎞ ⎛ 1

1⎞ ⎛1

1⎞ ⎛1

1⎞

∑ (k + 2)k = 3 + 8 + 15 +… = ∑ ⎜⎝ k – k + 2 ⎟⎠ = ⎜⎝ 1 – 3 ⎟⎠ + ⎜⎝ 2 – 4 ⎟⎠ + ⎜⎝ 3 – 5 ⎟⎠ + ⎜⎝ 4 – 6 ⎟⎠ +…

k =1

k =1

1 ⎞ ⎛1
1 ⎞
⎛ 1⎞ ⎛1 1⎞ ⎛1 1⎞
⎛ 1
–
Sn = ⎜1 – ⎟ + ⎜ – ⎟ + ⎜ – ⎟ +…+ ⎜
⎟+⎜ –

⎟
3
2
4
3
5
–
1
1
2⎠
n
n
+
n
n
+
⎝
⎠ ⎝
⎠ ⎝
⎠
⎝
⎠ ⎝
1
1
1
3
2n + 3
3
2n + 3
–

= 1+ –
= –
= –
2
2 n + 1 n + 2 2 (n + 1)(n + 2) 2 n + 3n + 2
2+ 3
∞
2
3
3
2n + 3
3
3
n n2
= – lim
= , so ∑
lim Sn = – lim
= .
2
3
2
2 n→∞ n + 3n + 2 2 n→∞ 1 + + 2 2
2
n →∞
k =1 ( k + 2) k
n n

∞

11.

⎛e⎞

∑ ⎜⎝ π ⎟⎠

k +1

k =1

2

e
e
(
(
π)
π)
S=
=
2

∞

12.

∑

4k +1

k =1 7

522

k –1

2

π–e
π

1 – πe

2

2

2

2

e
⎛e⎞ ⎛e⎞ e ⎛e⎞ ⎛e⎞
⎛e⎞
= ⎜ ⎟ + ⎜ ⎟ ⋅ + ⎜ ⎟ ⎜ ⎟ +…; a geometric series with a = ⎜ ⎟ , r = < 1;
π
⎝π⎠ ⎝π⎠ π ⎝π⎠ ⎝π⎠
⎝π⎠
=

e2

≈ 5.5562
π(π – e)
2

=

16
4
4
16
16 112
⎛4⎞
+ 16 ⋅ + 16 ⎜ ⎟ +…; a geometric series with a = 16, r = < 1; S =
=
=
3
1
7
7
3
1 – 74
⎝7⎠
7

Section 9.2

Instructor’s Resource Manual

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material may be reproduced, in any form or by any means, without permission in writing from the publisher.

∞

13.

⎛

3 ⎞

3

⎛3

3⎞ ⎛3

3⎞ ⎛3

3⎞

∑ ⎜⎜ (k – 1)2 – k 2 ⎟⎟ = ⎜⎝ 1 – 4 ⎟⎠ + ⎜⎝ 4 – 9 ⎟⎠ + ⎜⎝ 9 – 16 ⎟⎠ +…;

k =2 ⎝

⎠

⎛
3⎞ ⎛ 3 1⎞ ⎛1 3 ⎞
3
3

⎛
–
Sn = ⎜ 3 – ⎟ + ⎜ – ⎟ + ⎜ – ⎟ +…+ ⎜
⎜ (n – 2)2 (n – 1)2
4 ⎠ ⎝ 4 3 ⎠ ⎝ 3 16 ⎠
⎝
⎝
= 3–
∞

3

; lim Sn
n 2 n→∞

⎛

= 3 – lim

3

n →∞ n 2

⎞ ⎛ 3
3
–
⎟⎟ + ⎜⎜
2
n2
⎠ ⎝ (n – 1)

⎞
⎟⎟
⎠

= 3, so

3 ⎞

3

∑ ⎜⎜ (k – 1)2 – k 2 ⎟⎟ = 3.

k =2 ⎝

⎠

∞

14.

2
2 2 2 2
∑ k – 5 = 1 + 2 + 3 + 4 +…
k =6

20. 0.36717171... =

∞

∞
1
1
which diverges since ∑ diverges.
k
k =1
k =1 k

15. 0.22222 … =

∞

2⎛1⎞

∑ 10 ⎜⎝ 10 ⎟⎠

=

1
1 – 10

=

100

s < 1, and

2
=
9

∞

16. 0.21212121… =

=
21 ⎛ 1 ⎞

∑ 100 ⎜⎝ 100 ⎟⎠

k –1

=

22.

21 7
=
99 33

17. 0.013013013... =

∞

k =0

k =0

∞

1− s

∑ (1 – s)s k –1 = 1 − s = 1

∞

∞

∞

k =0

k =0

k =1

∑ (−1)k xk = ∑ (− x)k = ∑ (– x)k –1;

13 ⎛ 1 ⎞

∑ 1000 ⎜⎝ 1000 ⎟⎠

k –1

∞

1

1

∑ (– x)k –1 = 1 − (− x) = 1 + x

k =1

13
=
999

18. 0.125125125... =
=

∞

if –1 < x < 1 then
–1 < –x < 1 so x < 1 ;

k =1
13
= 1000
1
1 − 1000

∞

∑ r (1 − r )k = ∑ (1 − s)s k

k =1

k =1
21
= 100

1
1 – 100

36 10,000
727
+
=
100 1 − 1
1980

21. Let s = 1 – r, so r = 1 – s. Since 0 < r < 2,
–1 < 1 – r < 1, so

k =1
2
10

k –1

71

= 2∑

k –1

36 ∞ 71 ⎛ 1 ⎞
+∑
⎜
⎟
100 k =1 10, 000 ⎝ 100 ⎠

125
1000
1
1 − 1000

=

19. 0.4999... =

∞

125 ⎛ 1 ⎞
∑ 1000 ⎜⎝ 1000 ⎟⎠
k =1

k –1

125
999

4 ∞ 9 ⎛1⎞
+∑
⎜ ⎟
10 k =1 100 ⎝ 10 ⎠

k –1

9

=

4
1
+ 100 =
10 1 − 1 2
10

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Section 9.2

523

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material may be reproduced, in any form or by any means, without permission in writing from the publisher.


k
= ln k – ln(k + 1)
k +1
Sn = (ln1 – ln 2) + (ln 2 – ln 3) + (ln 3 – ln 4) +…+ (ln(n – 1) – ln n) + (ln n – ln(n + 1)) = ln1 – ln(n + 1) = – ln(n + 1)

23. ln

lim Sn = lim – ln(n + 1) = – ∞, thus

n →∞

n →∞

∞

k

∑ ln k + 1 diverges.

k =1

⎛
1 ⎞
k2 –1
= ln
= ln(k 2 – 1) – ln k 2 = ln[(k + 1)(k – 1)] – ln k 2 = ln(k + 1) + ln(k – 1) – 2 ln k
24. ln ⎜ 1 –
⎟
2
2
k
⎝ k ⎠
Sn = (ln 3 + ln1 – 2 ln 2) + (ln 4 + ln 2 – 2 ln 3) + (ln 5 + ln 3 – 2 ln 4) +…
+(ln n + ln(n – 2) – 2 ln(n – 1)) + (ln(n + 1) + ln(n – 1) – 2 ln n)

= –ln 2 + ln(n + 1) – ln n = – ln 2 + ln
lim Sn = – ln 2 + lim ln

n →∞

n →∞

n +1
n

n +1
n +1⎞
⎛
= – ln 2 + ln ⎜ lim
⎟ = – ln 2 + ln1 = – ln 2
n
⎝ n→∞ n ⎠
2

⎛2⎞
⎛2⎞
⎛2⎞
25. The ball drops 100 feet, rebounds up 100 ⎜ ⎟ feet, drops 100 ⎜ ⎟ feet, rebounds up 100 ⎜ ⎟ feet, drops
⎝3⎠
⎝3⎠
⎝3⎠
2

⎛2⎞
100 ⎜ ⎟ , etc. The total distance it travels is
⎝3⎠
2

3

2

3

⎛2⎞
⎛2⎞
⎛2⎞
⎛2⎞
⎛2⎞
⎛2⎞
100 + 200 ⎜ ⎟ + 200 ⎜ ⎟ + 200 ⎜ ⎟ + ... = −100 + 200 + 200 ⎜ ⎟ + 200 ⎜ ⎟ + 200 ⎜ ⎟ + ...
⎝3⎠
⎝3⎠
⎝3⎠
⎝3⎠
⎝3⎠
⎝3⎠
∞

⎛2⎞
= −100 + ∑ 200 ⎜ ⎟
⎝3⎠
k =1

26. Each gets

k –1

= −100 +

200
= 500 feet

1 − 23

∞
1 1 1 1⎛1 1⎞
1⎛1⎞
+ ⋅ + ⎜ ⋅ ⎟ + ... = ∑ ⎜ ⎟
4 4 4 4⎝4 4⎠
k =1 4 ⎝ 4 ⎠

1
4

k –1

=

1−

1
4

=

1
3

(This can be seen intuitively, since the size of the leftover piece is approaching 0, and each person gets the same
amount.)
∞

27. $1 billion + 75% of $1 billion + 75% of 75% of $1 billion + ... =

∑ ($1 billion)0.75k –1 =

k =1
∞

28.

∑ $1 billion (0.90)k –1

k =1

=

$1 billion
= $4 billion
1 − 0.75

$1 billion
= $10 billion
1 − 0.90

29. As the midpoints of the sides of a square are connected, a new square is formed. The new square has sides

1

2
1
times the sides of the old square. Thus, the new square has area the area of the old square. Then in the next step,

2
1
of each new square is shaded.
8

Area =

∞
1
1 1 1 1
1⎛1⎞
⋅1 + ⋅ + ⋅ + ... = ∑ ⎜ ⎟
8
8 2 8 4
k =1 8 ⎝ 2 ⎠

The area will be

524

Section 9.2

k –1

=

1
8

1−

1
2

=

1
4

1
.
4

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material may be reproduced, in any form or by any means, without permission in writing from the publisher.


∞

1 1⎛8⎞ 1⎛8 8⎞
+ ⎜ ⎟+ ⎜ ⋅ ⎟+
30.
9 9⎝9⎠ 9⎝9 9⎠

31.

1⎛8⎞
=∑ ⎜ ⎟

k =1 9 ⎝ 9 ⎠

k –1

=

1
9

= 1; the whole square will be painted.

1 − 89

∞
3 3 ⎛ 1 1 ⎞ 3 ⎛ 1 1 ⎞⎛ 1 1 ⎞
3⎛ 1 ⎞
+ ⎜ ⋅ ⎟ + ⎜ ⋅ ⎟⎜ ⋅ ⎟ + ... = ∑ ⎜ ⎟
4 4 ⎝ 4 4 ⎠ 4 ⎝ 4 4 ⎠⎝ 4 4 ⎠
k =1 4 ⎝ 16 ⎠

k –1

=

3
4
1
1 − 16

=

4
5

The original does not need to be equilateral since each smaller triangle will have

1
area of the previous larger
4

triangle.
∞

π

32. Ratio of inscribed circle to triangle is

3 3

, so

π

3⎛1⎞
∑ 3 3 ⋅ 4 ⎜⎝ 4 ⎟⎠
k =1

k –1

=

( )=
π
4 3
1 − 14

π
3 3

(This can be seen intuitively, since every small triangle has a circle inscribed in it.)
33. a. We first note that, at each stage, the number of sides is four times the number in the previous stage and the
length of each side is one-third the length in the previous stage. Summarizing:
Stage

0

# of sides

length/
side (in.)

perimeter

9

27

3

pn

()

36

( )

⎛4⎞
27 ⎜ ⎟
⎝3⎠

1

3(4)

9

n

3(4n )

9

1
3

1
3n

n

n

4
⎛4⎞
The perimeter of the Koch snowflake is lim pn = lim 27 ⎜ ⎟ which is infinite since > 1 .
3
n →∞
n →∞
⎝3⎠

b. We note the following:
3 2
s
4
2. The number of new triangles added at each stage is equal to the number of sides the figure had at the
previous stage and

1. The area of an equilateral triangle of side s is

3. the area of each new triangle at a given stage is

3
(side length at that stage)2 . Using results from part a. we
4

can summarize:
Stage Additional triangles Area of each new Δ Additional area, An
(col 2, part a.)
(see col 3, part a.)

( )

3 2
9
4

( )

36

0

original

3 2
9
4

1

3

3 2
3
4

n

3 4n −1

(

)

Instructor’s Resource Manual

3⎛ 9 ⎞
⎜ ⎟
4 ⎝ 3n ⎠

2

( )

⎛4⎞
3 3⎜ ⎟
⎝9⎠

n−2

Section 9.2

525

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material may be reproduced, in any form or by any means, without permission in writing from the publisher.


Thus the area of the Koch snowflake is

∞

∑ An =

n=0

81 3 27 3 ∞
⎛4⎞
+
+ ∑3 3 ⎜ ⎟
4
4
⎝9⎠
n =1

n −1

=

81 3 27 3 ⎛ 3 3 ⎞
⎟
+
+⎜
⎜ (1 − 4 ) ⎟
4
4
9 ⎠
⎝

=

81 3 1 ⎛ 81 3 ⎞ 4 ⎛ 81 3 ⎞ 8 ⎛ 81 3 ⎞
+ ⎜⎜
⎟+ ⎜
⎟= ⎜
⎟
4
3 ⎝ 4 ⎟⎠ 15 ⎜⎝ 4 ⎟⎠ 5 ⎜⎝ 4 ⎟⎠

Note: By generalizing the above argument it can be shown that, no matter what the size of the original
equilateral triangle, the area of the Koch snowflake constructed from it will be

8
5

times the area of the original

triangle.
34. We note the following:

1. Each triangle contains the angles 90,θ ,90 − θ
succeeding triangle. Summarizing:
# triangle

base

2. The height of each triangle will be the hypotenuse of the

height

area An
1 2
h sin θ cos θ
2
1 2 3
h sin θ cos θ
2

1

h cos θ

h sin θ

2

h sin θ cos θ

h sin 2 θ

n

h (sin n −1 θ ) cos θ

h sin n θ

1 2 2 n −1
θ cos θ
h sin
2

∞

Thus the total area of the small triangles is A = ∑ An =
n =1

h 2 ⎛ cos θ ⎞ ∞
2 n −1
⎜
⎟ ∑ (sin θ )
2 ⎝ sin θ ⎠n = 2

∞

Now consider the infinite geometric series S = ∑ (sin 2 θ )n −1 =
n =1

∞

then:

1

sin 2 θ

∑ (sin 2 θ )n−1 = S − 1 = cos2 θ − 1 = cos2 θ

1
1 − sin θ

ABC , height = h and base = h tan θ ; thus the area of

=

1
cos 2 θ

A=

Therefore:

n=2

In

2

ABC =

h 2 ⎛ cos θ
⎜
2 ⎝ sin θ

⎞ ⎛ sin θ
⎟ ⎜⎜
⎠ ⎝ cos 2 θ
2

⎞ h2
tan θ
⎟=

⎟ 2
⎠

1
h2
(h tan θ )h =
tan θ , the same as A .
2
2

35. Both Achilles and the tortoise will have moved.
100 + 10 + 1 +
=

∞
1
1
⎛1⎞
+
+ ... = ∑ 100 ⎜ ⎟
10 100
⎝ 10 ⎠
k =1

k –1

100
1
= 111 yards
1

9
1 − 10

Also, one can see this by the following reasoning. In the time it takes the tortoise to run
run d yards. Solve d = 100 +

526

Section 9.2

d
yards, Achilles will
10

d
1000
1
.d =
= 111 yards
10
9
9

Instructor’s Resource Manual

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material may be reproduced, in any form or by any means, without permission in writing from the publisher.


36. a.

Say Trot and Tom start from the left, Joel
from the right. Trot and Joel run towards
each other at 30 mph. Since they are 60
miles apart they will meet in 2 hours. Trot
will have run 40 miles and Tom will have
run 20 miles, so they will be 20 miles apart.
Trot and Tom will now be approaching each
other at 30 mph, so they will meet after
2/3 hour. Trot will have run another
40/3 miles and will be 80/3 miles from the
left. Joel will have run another 20/3 miles
and will be at 100/3 miles from the left, so
they will be 20/3 miles apart. They will meet
after 2/9 hour, during which Trot will have
run 40/9 miles, etc. So Trot runs
40 +

40 40
+
+
3
9

=

∞

⎛1⎞

∑ 40 ⎜⎝ 3 ⎟⎠

k –1

=

k =1

40
1 − 13

39. Let X = number of rolls needed to get first 6 For
X to equal n , two things must occur:

1
6

2. Mary rolls a 6 (probability =

) on the nth

roll. Thus,
⎛5⎞
Pr( X = n) = ⎜ ⎟
⎝6⎠

n −1

⎛1⎞
⎜ ⎟ and

⎝6⎠
1
n
−
1
∞
⎛5⎞ ⎛1⎞
=
∑ ⎜⎝ 6 ⎟⎠ ⎜⎝ 6 ⎟⎠ (1 − 65 ) = 1
6
n =1
n −1

1⎛ 1 ⎞ ∞
1 ⎛ 6 ⎞⎛ 5/ 6
= ⎜ ⎟ ∑ n ⋅ pn = ⎜ ⎟ ⎜
6 ⎝ p ⎠ n =1
6 ⎝ 5 ⎠ ⎜ (1 − 5 / 6 )2
⎝
1 ⎛ 6 ⎞ ⎛ 5 ⎞ ⎛ 36 ⎞
= ⎜ ⎟⎜ ⎟⎜ ⎟ = 6
6 ⎝ 5 ⎠⎝ 6 ⎠⎝ 1 ⎠

b. Tom and Joel are approaching each other at
20 mph. They are 60 miles apart, so they will
meet in 3 hours. Trot is running at 20 mph
during that entire time, so he runs 60 miles.
37. Note that:

⎞

⎟
⎟
⎠

∞

41. (Proof by contradiction) Assume

1. If we let tn be the probability that Peter wins
on his nth flip, then the total probability that
Peter wins is T =

k =1

∞

∑ tn

n −1

∞

⎛ 1 ⎞⎛ 4 ⎞
and T = ∑ ⎜ ⎟ ⎜ ⎟
n =1 ⎝ 3 ⎠ ⎝ 9 ⎠

⎛ 13 ⎞ 1 9 3
⎜
⎟= ⋅ =
⎜ (1 − 4 ) ⎟ 3 5 5

9
⎝
⎠

In this case (see problem 37),
2
tn = p ⎡(1 − p ) ⎤
⎣
⎦

∞

n −1

so

2
T = ∑ p ⎡(1 − p ) ⎤
⎣
⎦
n =1

p

(2 p − p )
2

=

n −1

1
2− p

Instructor’s Resource Manual

=

p

(1 − (1 − p ) )
2

∞

k =1

k =1

1 ∞

1

would also

k =1

converge, by Theorem B(i).
42.

⎛ 1 ⎞⎛ 4 ⎞
tn = ⎜ ⎟ ⎜ ⎟
⎝ 3 ⎠⎝ 9 ⎠

∞

∑ ak = ∑ c cak = c ∑ cak

2. The probability that neither man wins in their
k
k
⎛2 2⎞
⎛4 ⎞
first k flips is ⎜ ⋅ ⎟ = ⎜ ⎟ .
⎝3 3⎠
⎝9 ⎠
3. The probability that Peter wins on his nth flip
requires that (i) he gets a head on the nth
flip, and (ii) neither he nor Paul gets a head
on their previous n-1 flips. Thus:

∑ cak

1
converges, and c ≠ 0. Then is defined, so
c

n =1

=

) on

each of her first n-1 rolls, and

∞
∞
1 ⎛5⎞
40. EV ( X ) = ∑ n ⋅ Pr ( X = n ) = ∑ n ⋅ ⋅ ⎜ ⎟
6 ⎝6⎠
n =1
n =1

= 60 miles .

38.

5
6

1. Mary must get a non-6 (probability =

1 1 1 1
+ + + +
2 4 6 8
∞

since

1

∑k

=

∞

1⎛1⎞

1 ∞ 1

∑ 2 ⎜⎝ k ⎟⎠ = 2 ∑ k

k =1

diverges

k =1

diverges.

k =1

n −1

=

43. a.

The top block is supported exactly at its

center of mass. The location of the center of
mass of the top n blocks is the average of the
locations of their individual centers of mass,
so the nth block moves the center of mass
1
of the location of its center of
left by
n
1 1
1
to the left. But
mass, that is, ⋅ or
n 2
2n
this is exactly how far the (n + 1)st block
underneath it is offset.
1 1 1
1 ∞ 1
+ + + ... = ∑ , which
2 4 6
2 k =1 k
diverges, there is no limit to how far the top
block can protrude.

b. Since

Section 9.2

527

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material may be reproduced, in any form or by any means, without permission in writing from the publisher.


44. N = 31; S31 ≈ 4.0272 and S30 ≈ 3.9950.

48. If

∞

45. (Proof by contradiction) Assume

∑ (ak + bk )

converges, so would

∞

∞

∞

k =1

k =1

k =1

∑ ak = ∑ (ak + bk ) + (−1) ∑ bk ,

S=

∞

by

∞

=r+

⎛1⎞

S=

49. a.

k –1

∑

k =1

2

k

=

∞

⎛1⎞
∑ ⎜⎝ 2 ⎟⎠
k =1

k –1

=

so

∞

∞

k =2

k =2

∑ kr k – ∑ (k – 1)r k

∞

∑ [k – (k – 1)]r k = r +

∞

∑ rk =

k =2

∞

∑ rk

k =1

r
, thus
1– r

∞

1
r
∑ r k = (1 – r )2 .
1 – r k =1
A=

∞

∞

=

⎛ 1 ⎞

∑ Ce−nkt = ∑ C ⎜ ekt ⎟

n =0

.

k =1

while

k =2

k =1

Taking horizontal strips, the area is
∞
1
1
1
1
k
⋅1 + ⋅ 2 + ⋅ 3 + ⋅ 4 + = ∑
.
k
2
4
8
16
k =1 2
k

k =2

Since r < 1, ∑ r k =

47. Taking vertical strips, the area is
∞

k =1
∞

∞

∞
⎛1 1⎞
+
=
(
)
a
b
∑ n n ∑ ⎜⎝ n − n ⎟⎠ converges to 0.
n =1
n =1

∑ ⎜⎝ 2 ⎟⎠

∞

k =2

∞

=

∞

∑ kr k +1 = ∑ (k – 1)r k

∑ kr k = r + ∑ kr k

S – rS = r +

1
46. (Answers may vary). ∑ an = ∑ and
n =1
n =1 n
∞
∞
1
∑ bn = ∑ (−1) n both diverge, but
n =1
n =1

1
1
1
1 ⋅1 + 1 ⋅ + 1 ⋅ + 1 ⋅ +
2
4
8

k =1
∞

k =1

Theorem B(ii).

∞

k =1

rS = r ∑ kr k =

k =1

a.

k =1

∞

k =1

∑ bk

∞

∑ kr k converges, so will r ∑ kr k , by

Theorem B.

∞

converges. Since

∞

n =1

⎝

n –1

⎠

kt

C
Ce
=
1
1 − kt ekt − 1
e

b.

1
=2
1 − 12

1
ln 2

4
= e− kt = e−6k ⇒ k =
⇒ A = C;
2
6
3
8
if C = 2 mg, then A = mg.
3

b. The moment about x = 0 is
∞

∞
∞
k
k
⎛1⎞
∑ ⎜⎝ 2 ⎟⎠ ⋅ (1)k = ∑ 2k = ∑ 2k = 2.
k =0
k =0
k =1
moment 2
x=
= =1
area
2
k

2k

50. Using partial fractions,
(2

k +1

k

– 1)(2 – 1)

=

1
k

–

2 –1 2

1
k +1

–1

1 ⎞ ⎛ 1
1 ⎞
1 ⎞ ⎛ 1
1 ⎞
⎛ 1
⎛ 1

–
–
–
–
Sn = ⎜
⎟+⎜ 2
⎟ +…+ ⎜ n –1
⎟+⎜ n
⎟
1
2
3
n
n +1
⎝ 2 –1 2 –1⎠ ⎝ 2 –1 2 –1⎠
⎝2
–1 2 –1⎠ ⎝ 2 –1 2
–1⎠
1
1
1
=
= 1–
–
n +1
2 – 1 2n +1 – 1
2
–1
1
lim Sn = 1 – lim

=1– 0 =1
n →∞
n →∞ 2n +1 – 1

528

Section 9.2

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material may be reproduced, in any form or by any means, without permission in writing from the publisher.


51.

f
– fk
1
1
1
–
= k +2
=
f k f k +1 f k +1 f k + 2
f k f k +1 f k + 2
fk fk +2
since f k + 2 = f k +1 + f k . Thus,
∞

∞ ⎛
⎞
1
1
1
–
= ∑⎜
⎟ and
f
f
f
f
f
f
k +1 k + 2 ⎠
k =1 k k + 2
k =1 ⎝ k k +1
⎛ 1
⎛ 1
⎞
1 ⎞ ⎛ 1
1 ⎞
1 ⎞ ⎛ 1
1
–
–
–
–
Sn = ⎜
⎟+⎜

⎟+ +⎜
⎟+⎜
⎟
⎝ f1 f 2 f 2 f3 ⎠ ⎝ f 2 f3 f3 f 4 ⎠
⎝ f n –1 f n f n f n +1 ⎠ ⎝ f n f n +1 f n +1 f n + 2 ⎠
1
1
1
1
1
–
–
=
=
=1–
f1 f 2 f n +1 f n + 2 1 ⋅1 f n +1 f n + 2
f n +1 f n + 2
The terms of the Fibonacci sequence increase without bound, so
1
lim Sn = 1 – lim
=1– 0 =1
n →∞
n →∞ f n +1 f n + 2

∑

9.3 Concepts Review

4.

1. bounded above
2. f(k); continuous; positive; nonincreasing

is continuous, positive, and
2x +1
nonincreasing on [1, ∞) .
3

3 ⎛π
⎞
−1
2⎟< ∞
⎜ − tan
2⎝2
⎠
The series converges.
=

4. p > 1

Problem Set 9.3
1
is continuous, positive, and nonincreasing
x+3
on [0, ∞) .

5.

∞

1
∞
dx = ⎡⎣ ln x + 3 ⎤⎦ = ∞ – ln 3 = ∞
0
x+3
The series diverges.

2

is continuous, positive, and
x+2
nonincreasing on [1, ∞) .
∞

∫0

x+2
∞

3
2.
is continuous, positive, and nonincreasing
2x – 3
on [2, ∞) .

x

is continuous, positive, and nonincreasing
x +3
on [ 2, ∞ ) .

2

∞

1
⎡1
⎤
2
∫2 x2 + 3 dx = ⎢⎣ 2 ln x + 3 ⎥⎦ 2 = ∞ − 2 ln 7 = ∞
The series diverges.
∞

x

Instructor’s Resource Manual

∑

k =1
∞

–2

∑

k =1 k + 2

∞

3

3
⎡3
⎤
dx = ⎢ ln 2 x – 3 ⎥ = ∞ – ln1 = ∞
2x – 3
2
⎣2
⎦2
The series diverges.
∞

∫2

∞

2

∫1

Thus

3.

∞

⎡ 3
⎤
–1
∫1 2 x 2 + 1 dx = ⎢⎣ 2 tan 2 x ⎥⎦
1

∞

3. convergence or divergence

1.

3

2

6.

dx = ⎡⎣ 4 x + 2 ⎤⎦ = ∞ – 4 3 = ∞
1

2
k+2
∞

diverges, hence

= –∑

2

k =1 k + 2

also diverges.

3

is continuous, positive, and
( x + 2) 2
nonincreasing on [100, ∞) .
∞

3 ⎤
3
3
⎡
∫100 ( x + 2)2 dx = ⎢⎣ – x + 2 ⎥⎦100 = 0 + 102 = 102 < ∞
The series converges.
∞

3

Section 9.3

529

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7.

7
is continuous, positive, and
4x + 2
nonincreasing

on. [2, ∞)

12.

8.

1000
<∞
ln 5
The series converges.

13.

–x

∞ 2 –x

∫2

2

∞ –x ⎞
+ 2 ⎛⎜ [– xe – x ]∞
2 + ∫2 e dx ⎟
⎝
⎠

−2

= 0 + 4e + 4e + 2e

The series converges.

9.

−

2

= 10e

<∞
∞

15.

10.

1000 x

2

is continuous, positive, and

1+ x
nonincreasing on [2, ∞) .
3

∞ 1000 x 2

⎡1000

3
∫2 1 + x3 dx = ⎢⎣ 3 ln 1 + x
1000
ln 9 = ∞
=∞–
3
The series diverges.

∞

⎤
⎥
⎦2

2

11. xe –3x is continuous, positive, and
nonincreasing on [1, ∞) .

= 1 ≠ 0, so the series

∞

k –1

; a geometric series with

3
3 3
,r = ;

< 1 so the series converges.
π
π π

⎛1⎞

∑ ⎜⎝ 2 ⎟⎠

k

1 1
is a geometric series with r = ; < 1
2 2

so the series converges.
1
1–
k –1
k –1
k = 1 ≠ 0, so
, lim
= lim
In ∑
1 2
k =1 2k + 1 k →∞ 2k + 1 k →∞ 2 +
k
the series diverges. Thus, the sum of the series
diverges.
∞

∞

⎡
⎤
6
∫1 (4 + 3x)7 / 6 dx = ⎢⎢ – (4 + 3x)1/ 6 ⎥⎥
⎣
⎦1
6
= 0+
= 6 ⋅ 7 –1/ 6 < ∞
71/ 6
The series converges.

k
k →∞ 1 + 52
k

k

k =1

is continuous, positive, and
(4 + 3x)7 / 6
nonincreasing on [1, ∞) .
3

+5

1 + 12

= lim

3⎛3⎞
⎛3⎞
∑ ⎜⎝ π ⎟⎠ = ∑ π ⎜⎝ π ⎟⎠
k =1
k =1
a=

–2

3

∞

k →∞ k 2

∞

14.

= [– x 2 e – x – 2 xe – x – 2e – x ]∞
2
–2

k 2 +1

lim

diverges.

∞

–x
x e dx = [– x 2 e – x ]∞
2 + 2 ∫ xe dx

= [– x 2 e – x ]∞
2

1000

=

u = x , i = 1, 2 and dv = e dx,
i

∞

1000
⎡ 1000 ⎤
∫5 x(ln x)2 dx = ⎢⎣ – ln x ⎥⎦5 = 0 + ln 5

x2

is continuous, positive, and nonincreasing
ex
[2, ∞) . Using integration by parts twice, with

is continuous, positive, and
x(ln x)2
nonincreasing on [5, ∞) .
∞

∞

7
7
⎡7
⎤
∫2 4 x + 2 dx = ⎢⎣ 4 ln 4 x + 2 ⎥⎦ 2 = ∞ – 4 ln10 = ∞
The series diverges.
∞

1000

16.

1
x2

is continuous, positive, and nonincreasing on

[1, ∞) .
∞

1

∑ k2

k =1
∞

1

1

converges.
∞

⎛1⎞

∑ 2k = ∑ ⎜⎝ 2 ⎟⎠

k =1

∞

⎡ 1⎤
∫1 x 2 dx = ⎢⎣ – x ⎥⎦1 = 0 + 1 = 1 < ∞, so
∞

k

; a geometric series with

k =1

1 1

r = ; < 1, so the series converges. Thus, the
2 2
sum of the series converges.

∞

1 –3
⎡ 1 –3 x 2 ⎤
–3 x 2
∫1 xe dx = ⎢⎣ – 6 e ⎥⎦1 = 0 + 6 e
1
=
<∞
6e3
The series converges.
∞

530

Section 9.3

Instructor’s Resource Manual

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material may be reproduced, in any form or by any means, without permission in writing from the publisher.


k = 4 j +1
⎧1
⎛ kπ ⎞ ⎪

17. sin ⎜ ⎟ = ⎨ –1 k = 4 j + 3,
⎝ 2 ⎠ ⎪
⎩0 k is even

1
1
→ 0. Let y = , then
k
k
1
1
sin y
lim k sin = lim sin y = lim
= 1 ≠ 0, so
k y →0 y
k →∞
y →0 y
the series diverges.

18. As k → ∞,

where j is any nonnegative integer.
⎛ kπ ⎞
Thus lim sin ⎜ ⎟ does not exist, hence
k →∞
⎝ 2 ⎠
⎛ kπ ⎞
lim sin ⎜ ⎟ ≠ 0 and the series diverges.
⎝ 2 ⎠

k →∞

3

19. x 2 e – x is continuous, positive, and
nonincreasing on [1, ∞) .
∞

3⎤
1
⎡ 1
dx = ⎢ – e – x ⎥ = 0 + e –1 < ∞, so
3
3
⎣
⎦1
the series converges.

∞ 2 – x3

∫1

x e

1
1⎞ ⎛1
1 ⎞
⎛1 1 ⎞ ⎛ 1 1 ⎞
⎛ 1
20. Sn = ⎜ – ⎟ + ⎜ – ⎟ +…+ ⎜

– ⎟+⎜ –
⎟ =1–
n –1
⎝1 2 ⎠ ⎝ 2 3 ⎠
⎝ n –1 n ⎠ ⎝ n n –1⎠
1
lim Sn = 1 – lim
=1– 0 =1
n →∞
n →∞ n –1
The series converges to 1.

21.

tan –1 x

24.

is continuous, positive, and

1 + x2
nonincreasing on [1, ∞) .

∞

tan –1 x

⎡1
–1 2 ⎤
∫1 1 + x 2 dx = ⎢⎣ 2 (tan x) ⎥⎦1

∞

2

x
x

26.

∞

∞

1
x3 / 2

5

2
⎡ 2 ⎤
dx = ⎢ –
⎥ = 0+
x ⎦5
5
⎣

∞

1

∞

∑ 1 + k 2 ≤ ∫5

1
1+ x

2

dx = [tan –1 x]5∞

π
– tan –1 5 ≈ 0.1974
2

1
is continuous, positive, and
x( x + 1)

nonincreasing on [5, ∞) .

is continuous, positive, and nonincreasing on

E=

∞

1

∞

∑ k (k + 1) ≤ ∫5

k =6

∞⎛ 1
1
1 ⎞
dx = ∫ ⎜ –
dx
5 ⎝ x x + 1 ⎟⎠
x ( x + 1)
∞

k

∞

∑ ek ≤ ∫5

k =6
–x

≤∫

is continuous, positive, and nonincreasing
1 + x2
on [5, ∞) .

=

e
[5, ∞) .
E=

k

k =6

∞

2

∞

1

1

E=

⎡1
⎤
dx = ⎢ tan –1 (2 x ) ⎥
⎣2
⎦1
1 + 4x
1⎛π⎞ 1
= ⎜ ⎟ – tan –1 2 < ∞,
2⎝ 2⎠ 2

so the series converges.

23.

25.

is continuous, positive, and
1 + 4x 2
nonincreasing on [1, ∞) .
1

∞

≈ 0.8944

2

1

∞

1

3/ 2

∑k

k =6

1⎛π⎞

1⎛π⎞
3π2
< ∞, so the series
⎜ ⎟ – ⎜ ⎟ =
2⎝ 2⎠
2⎝ 4⎠
32
converges.

∫1

=

is continuous, positive, and
x x x
nonincreasing on [5, ∞) .
E=

=

22.

1

= [– xe
≈ 0.0404

x
e

– e – x ]5∞

x

∞ –x
e dx
5

dx = [– xe – x ]5∞ + ∫

= 0 + 5e –5 + e –5 = 6e –5

Instructor’s Resource Manual

x ⎤
5
∞ ⎡
= ⎡⎣ln x – ln x + 1 ⎤⎦ = ⎢ ln
= 0 – ln
⎥
5
6
⎣ x + 1 ⎦5
6
= ln ≈ 0.1823
5

Section 9.3

531

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material may be reproduced, in any form or by any means, without permission in writing from the publisher.


27. En =

∞

1

∑

k = n +1 k

2

∞

1

n

2

<∫

x

A

∫
A→∞ n

dx = lim

1
x

2

32. En =

dx =

⎡1 1 ⎤ 1
lim ⎢ − ⎥ =
A→∞ ⎣ n A ⎦ n
1
< 0.0002 ⇒ n > 5000
n

28. En =

∞

1

∑

k = n +1 k

3

∞

1

n

3

<∫

x

29. En =

dx = lim

∫

A

A→∞ n

1
x3

dx =

∑

k = n +1 1 + k

= lim

∫

2

1

A

1

n

1 + x2

dx

dx = lim ⎡ tan
A→∞ ⎣

1 + x2

A→∞ n

33. Consider

∞

<∫

−1

A − tan

−1

du =

n⎤
⎦

∞

∫2

= π2 − tan −1 n

(2

2

)

⇒ n > tan π − 0.0002 ≈ 5000

30. En =

lim

A→∞

∞

∑

k

k = n +1 e k

2

∞

x

n

x2

<∫

e

34.

n2

31. En =

∑

∞

k = n +1 1 + k

4

∞

1

ln 2 u p

du which converges for

1
dx
x ln x ln(ln x)

= ∞ – ln(ln(ln 3)) = ∞
The series diverges.

1
0.0004

35.

<∫

∞

n

x
1+ x

4

u = x2
du = 2 xdx

1 A du
∫2
A→∞ 2 n 1 + u 2

dx = lim

1
1 ⎡π
⎤
lim ⎡ tan −1 A − tan −1 n 2 ⎤ = ⎢ − tan −2 n 2 ⎥
⎣
⎦
2 A→∞

2⎣2
⎦
1 ⎡π
⎤
− tan −2 n 2 ⎥ < 0.0002
2 ⎢⎣ 2
⎦

( )

=

( )

⇒

π
2

( )

( )

− tan −2 n 2 < 0.0004 ⇒ tan −1 n 2 > 1.5703963

⇒ n > tan (1.5703963) ≈ 50

532

dx = ∫

dx. Let u = ln x,

1
dx.
x ln x
∞
∞
1
1
∞
∫3 x ln x ln(ln x) dx = ∫ln(ln 3) u du = [ln u ]ln(ln 3)

∴n > 2
k

p

x(ln x) p

Let u = ln(ln x), du =

< 0.0002 ⇒ n > ln

∞

1
dx.
x
1

1

1
is continuous, positive, and
x ln x ln(ln x )

∫3

u = x2
du = 2 x dx

1 A 1
du =
2
2 ∫n eu

2e
≈ 2.797

∞

nonincreasing on [3, ∞) .

dx =

⎡ 1
1 ⎤
1
⎛ 1⎞

⎜ − ⎟ lim ⎢ A − 2 ⎥ =
2
2
n
→∞
A
⎝
⎠
e ⎥⎦ 2e n
⎣⎢ e

1

1 ⎞

∫2

x(ln x)
p > 1.

π − tan −1 n < 0.0002 ⇒ tan −1 n > π − 0.0002
2

A⎛ 1

⎡ ⎛ A ⎞
⎛ n ⎞⎤
⎛ n ⎞
lim ⎢ ln ⎜
⎟ − ln ⎜

⎟ ⎥ = 0 − ln ⎜
⎟=
A→∞ ⎣ ⎝ A + 1 ⎠
⎝ n + 1 ⎠⎦
⎝ n +1⎠
⎛ n +1⎞
ln ⎜
⎟
⎝ n ⎠
1
⎛ n +1⎞
0.0002
ln ⎜
≈ 1.0002
⎟ < 0.0002 ⇒ 1 + < e
n
⎝ n ⎠
1
⇒n>
= 5000
0.0002

1 ⎤
1
⎡ 1
−
=
lim ⎢
2⎥
2 A ⎦ 2n 2

1
1
< 0.0002 ⇒ n >
= 50
2
0.0004
2n
1

∞
1
1
<∫
dx =
n x( x + 1)
(
1)
k
k
+
k = n +1

∑

∫ ⎜ − ⎟ dx =
A→∞ n ⎝ x x + 1 ⎠
lim

A→∞ ⎣ 2n 2

∞

∞

Section 9.3

The upper rectangles, which extend to n + 1 on
1 1
1
the right, have area 1 + + +…+ . These
2 3
n
1
rectangles are above the curve y = from x = 1
x
to x = n + 1. Thus,
n +1 1
n +1
∫1 x dx = [ln x]1 = ln(n + 1) – ln1 = ln(n + 1)
Instructor’s Resource Manual

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material may be reproduced, in any form or by any means, without permission in writing from the publisher.


1 1
1
+ +…+ .
n
2 3

The lower (shaded) rectangles have area
1 1
1
+ +…+ . These rectangles lie below the
2 3
n
1
curve y = from x = 1 to x = n. Thus
x
n1
1 1
1
+ +…+ < ∫ dx = ln n, so
2 3
n 1 x
1 1
1
1 + + +…+ < 1 + ln n.
2 3
n
< 1+

b. The leftmost rectangle has area
1 ⋅ f (1) = f (1). If each shaded region to the
right of x = 2 is shifted until it is in the
leftmost rectangle, there will be no overlap
of the shaded area, since the top of each
rectangle is at the bottom of the shaded
region to the left. Thus, the total shaded area
is less than or equal to the area of the

leftmost rectangle, or Bn ≤ f (1).
c.

By parts a and b, {Bn } is a nondecreasing
sequence that is bounded above, so lim Bn
n →∞

exists.
36. From Problem 35, Bn is the area of the region
within the upper rectangles but above the curve
1
y = . Each time n is incremented by 1, the
x
added area is a positive amount, thus Bn is
increasing.
From the inequalities in Problem 35,
1 1
1
0 < 1 + + +…+ – ln(n + 1) < 1 + ln n – ln(n + 1)
2 3
n
n
= 1 + ln
n +1
n
n
< 1, ln
< 0, thus Bn < 1 for all n,
Since
n +1

n +1
and Bn is bounded by 1.
37. {Bn } is a nondecreasing sequence that is
bounded above, thus by the Monotonic Sequence
Theorem (Theorem D of Section 9.1), lim Bn
n →∞

exists. The rationality of γ is a famous unsolved
problem.
38. From Problem 35, ln(n + 1) <
ln(10, 000, 001) ≈ 16.1181 <

n

1

∑ k < 1 + ln n,

k =1
10,000,000

∑

k =1

thus

1
k

< 1 + ln(10, 000, 000) ≈ 17.1181

39. γ + ln(n + 1) > 20 ⇒ ln(n + 1) > 20 – γ ≈ 19.4228

1
, then
x
n +1
n +1 1
∫1 f ( x)dx = ∫1 x dx = ln(n + 1) and
lim Bn = γ as defined in Problem 37.

d. Let f ( x) =

n →∞

41. Every time n is incremented by 1, a positive
amount of area is added, thus { An } is an
increasing sequence. Each curved region has
horizontal width 1, and can be moved into the
heavily outlined triangle without any overlap.
This can be done by shifting the nth shaded
region, which goes from (n, f(n)) to
(n + 1, f(n + 1)), as follows:
shift (n + 1, f(n + 1)) to (2, f(2)) and (n, f(n)) to
(1, f(2)–[f(n + 1) – f(n)]).
The slope of the line forming the bottom of the
shaded region between x = n and x = n + 1 is
f (n + 1) – f (n)
= f (n + 1) – f (n) > 0

(n + 1) – n
since f is increasing.
By the Mean Value Theorem,
f (n + 1) – f (n) = f ′(c) for some c in (n, n + 1).
Since f is concave down, n < c < n + 1 means that
f ′(c) < f ′(b) for all b in [1, n]. Thus, the nth
shaded region will not overlap any other shaded
region when shifted into the heavily outlined
triangle. Thus, the area of all of the shaded
regions is less than or equal to the area of the
heavily outlined triangle, so lim An exists.
n →∞

⇒ n +1 > e
≈ 272, 404,867
⇒ n > 272, 404,866
19.4228

40. a.

Each time n is incremented by 1, a positive
amount of area is added.

Instructor’s Resource Manual

Section 9.3

533

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material may be reproduced, in any form or by any means, without permission in writing from the publisher.


42. ln x is continuous, increasing, and concave down on [1, ∞) , so the conditions of Problem 41 are met.
a.

See the figure in the text for Problem 41. The area under the curve from x = 1 to x = n is

n

∫1 ln x dx

and the

n
ln n + ln(n + 1)
ln(n –1) + ln(n) ⎤
⎡ ln1 + ln 2
, thus An = ∫ ln x dx – ⎢
+…+
⎥.
1
2
2
2
⎣
⎦
1
Using integration by parts with u = ln x, du = dx, dv = dx, v = x
x

area of the nth trapezoid is

n

n

∫1 ln x dx = [ x ln x]1 – ∫1 dx = [ x ln x – x]1
n

n

= n ln n – n – (ln1 –1) = n ln n – n + 1

The sum of the areas of the n trapezoids is
ln1 + ln 2 ln 2 + ln 3
ln(n – 2) + ln(n –1) ln(n –1) + ln(n) 2 ln 2 + 2 ln 3 +…+ 2 ln(n –1) + ln n
+
+…+
+
=
2
2
2
2
2
ln n
ln n
= ln 2 + ln 3 +…+ ln n –
= ln(2 ⋅ 3 ⋅…⋅ n) –

= ln n !– ln n
2
2

(

Thus, An = n ln n – n + 1 – ln n !– ln n

)

= n ln n – n + 1 – ln n !+ ln n = ln n n – ln en + 1 – ln n !+ ln n

n
⎡⎛ n ⎞ n n ⎤
n
⎛n⎞
= ln ⎜ ⎟ + 1 + ln
= 1 + ln ⎢⎜ ⎟
⎥
n!
⎝e⎠
⎢⎣⎝ e ⎠ n ! ⎥⎦

⎡
⎡⎛ n ⎞ n n ⎤ ⎤
b. By Problem 41, lim An exists, hence part a says that lim ⎢1 + ln ⎢⎜ ⎟
⎥ ⎥ exists.
n →∞
n →∞ ⎢
⎢⎣⎝ e ⎠ n ! ⎥⎦ ⎥⎦

⎣
n
⎡
⎡⎛ n ⎞ n n ⎤ ⎤
⎡⎛ n ⎞ n n ⎤
⎡
⎛n⎞ n⎤
⎥ ⎥ = 1 + lim ln ⎢⎜ ⎟
⎥ = 1 + ln ⎢ lim ⎜ ⎟
⎥
lim ⎢1 + ln ⎢⎜ ⎟
e ⎠ n! ⎥ ⎥
n →∞ ⎢
n→∞ ⎢⎝ e ⎠ n ! ⎥
n→∞ ⎝ e ⎠ n ! ⎥
⎝
⎢
⎢
⎣
⎦⎦
⎣
⎦
⎣
⎦
⎣
⎛n⎞
Since the limit exists, lim ⎜ ⎟
n →∞ ⎝ e ⎠
n!

Thus, lim

(e)

n →∞ n n

= lim
n

n

n
= m. m cannot be 0 since lim ln x = – ∞.
n!
x →0 +

1

( )

n
n →∞ n
n
e

1

=

( )

n n
lim ne
n!
n →∞

n!

=

15

n

c.

43.

⎛n⎞
⎛ 15 ⎞
From part b, n ! ≈ 2πn ⎜ ⎟ , thus, 15! ≈ 30π ⎜ ⎟
e
⎝ ⎠
⎝ e ⎠
The exact value is 15! = 1,307, 674,368, 000 .

(Refer to fig 2 in the text). Let bk = ∫
t

∑

k = n +1

En =

534

k +1

k

Therefore
∞

∑

k = n +1

Section 9.3

ak ≥

ak = lim

t

∑

bk = ∫

∑

ak ≥ lim ∫

k = n +1
t

t →∞ k = n +1

1
, i.e., the limit exists.
m

t

n +1

≈ 1.3004 × 1012

f ( x) dx ; then from fig 2, it is clear that ak ≥ bk for k = 1, 2,… , n,…

f ( x) dx so that
t

t →∞ n +1

f ( x) dx = ∫

∞

n +1

f ( x) dx .

Instructor’s Resource Manual

© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this
material may be reproduced, in any form or by any means, without permission in writing from the publisher.


9.4 Concepts Review

5.

1. 0 ≤ ak ≤ bk
2.

an +1
8n +1 n !
8
= lim
= lim
= 0 <1
n →∞ an
n →∞ ( n + 1)!8n
n→∞ n + 1
lim

The series converges.

ak
k →∞ bk
lim

6.

3. ρ < 1; ρ > 1; ρ = 1

an +1
5n +1 n5
5n 5
= lim
= lim
n →∞ an
n→∞ (n + 1)5 5n
n →∞ ( n + 1)5
lim

5n5

= lim

+ 5n 4 + 10n3 + 10n 2 + 5n + 1
5
= lim
= 5 >1
n→∞ 1 + 5 + 102 + 103 + 54 + 15
n
n→∞ n5

4. Ratio; Limit Comparison

Problem Set 9.4

n

n

1. an =

; bn =

1
n

n + 2n + 3
an
n2
1
lim
= lim
= lim
2
2
n →∞ bn
n→∞ n + 2n + 3 n→∞ 1 + +
n
2

3

n2

= 1;

7.

2. an =

n3 – 4

(1)
(3)

(n + 1) 3
a
8. lim n +1 = lim
n
n →∞ an
n→∞
n 1

1

; bn =

n2

3 + 1n
an
3n3 + n 2

lim
= lim
= lim
= 3;
n →∞ bn
n→∞ n3 – 4
n→∞ 1 – 43

0<3< ∞
∞

∞

converges ⇒ ∑ an converges

n =1

3. an =

n =1

1
n n +1

=

1
n +n
3

2

9.
; bn =

1
n

∞

∑ bn

n =1

10.

n =1

n

; bn =

a
n
lim n = lim
n →∞ bn
n→∞
= lim
n→∞
∞

2 + 1n
1

2n + 1
n

2

= lim

2n + n

n→∞

= 2;0 < 2 < ∞
∞

+

3
n3

4+

+

3 + 1
n 4 n5
6+ 2

n n2

n3 + 3n 2 + 3n + 1

= lim

n →∞

4 n5 + 6 n 4 + 2 n3

= 0 <1

an +1
(3n +1 + n + 1)n !
= lim
n →∞ an
n →∞ ( n + 1)!(3n + n)
3n +1 + n + 1

n →∞ (3n

∑ bn converges ⇒ ∑ an converges

n =1

1
n2

= lim

1
n

n +1
3n

lim

3/ 2

3/ 2

n →∞

The series converges.

converges ⇒ ∑ an converges

2

(n + 1)3

n→∞

∞

2n + 1

= lim

an +1
(n + 1)3 (2n)!
= lim
n →∞ an
n→∞ (2n + 2)! n3

= lim

1
= 1; 0 < 1 < ∞
1 + 1n

4. an =

n +1

=

n→∞ (2n + 2)(2n + 1) n3

an
n
n
= lim
= lim
n →∞ bn
n→∞ n3 + n 2
n→∞ n3 + n 2
n→∞

=1

lim

3

lim

= lim

1 + 1n

= lim

3/ 2

3/ 2

99

1
<1
3
n→∞ 3
The series converges.
= lim

n

∑ bn

( )

⎛ n +1⎞
= ∞ since lim ⎜
⎟
n →∞ ⎝ n ⎠

The series diverges.

n =1

3n + 1

n

n→∞ n +1 99
n

diverges ⇒ ∑ an diverges.

n =1

n

an +1
(n + 1)!n100
n100
= lim
= lim

n →∞ an
n→∞ ( n + 1)100 n ! n→∞ ( n + 1)99

∞

∑ bn

n

lim

= lim

0<1< ∞
∞

n

The series diverges.

4

n

4

3

=

+ n)(n + 1)

3+

n + 1
3n 3n
lim
2
n→∞ n + 1 + n + n
3n 3n
2

and lim

n

n →∞ 3n

= lim

3n +1 + n + 1

n→∞ n3n

+ 3n + n 2 + n

= 0 < ∞ since lim

n

n →∞ 3n

=0

= 0 which can be seen by using

l’Hôpital’s Rule. The series converges.

n =1

Instructor’s Resource Manual

Section 9.4

535

© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this
material may be reproduced, in any form or by any means, without permission in writing from the publisher.


11.

n
1
= lim
=1≠ 0
n →∞ n + 200 n→∞ 1 + 200

17. an =

lim

n

lim

n →∞

an +1
(n + 1)!(5 + n)
= lim
an
n →∞ (6 + n) n !

∞

∑ bn converges ⇒

n =1

18.

The series diverges; Ratio Test
n+3

13. an =
lim

n2 n
an

n →∞ bn

; bn =

= lim

n

1
+ 3n

1 + n3

= lim

n→∞

∞

∑ bn converges ⇒

0<1< ∞.

n =1

1

= 1;

∞

∑ an

n =1

14. an =

n2 + 1

; bn =

∞

∑ bn

n =1

1 + 1n
1+

n 2 + 2n + 2
3n + 3
2

n3 / 2

19. an =

1

n2

∑ bn

∞

converges ⇒ ∑ an converges; Limit
n =1

an +1
n + 2n + 1
(n + 1) n !
= lim
= lim
2
n →∞ an
n→∞ (n + 1)!n
n →∞ ( n + 1) n 2
2

n + 2n + 1
2

n3 + n 2

= lim

n →∞

1

n

+

2
n2

+

1
n3

1 + 1n

= 0 <1

n =1

∞

converges ⇒ ∑ an converges;
n =1

n

(n + 1)

an +1
ln(n + 1)2n
ln(n + 1)

= lim
= lim
+
1
n
n →∞ an
n→∞ 2
n →∞ 2 ln n
ln n
Using l’Hôpital’s Rule,
ln(n + 1)
=
2 ln n

1
lim n +1
n →∞ 2
n

2

=

n
n + 2n + 1
2

; bn =

1

n

an
1
n2
= lim
= lim
2
2
n →∞ bn
n →∞ n + 2n + 1 n→∞ 1 + +
lim

n

= 1;

1
n2

= lim

∑ bn diverges ⇒

n =1

∞

∑ an

diverges;

n =1

Limit Comparison Test
21. an =

n +1
n +1
1
; bn =
=
n(n + 2)(n + 3) n3 + 5n 2 + 6n
n2

1+ n
an
n3 + n 2
= lim
= lim
= 1;
3
2
n →∞ bn
n→∞ n + 5n + 6n n→∞ 1 + 5 + 6
2
1

lim

lim

1
<1
3

0<1< ∞

lim

n →∞

=

Limit Comparison Test
20. an =

= 1; 0 < 1 < ∞.

The series converges; Ratio Test
16.

3
n2

1
1
1
; bn =
=

2
n(n + 1) n + n
n2

∞

n→∞

3+

n →∞

2
n2

The series converges; Ratio Test

∞

2

= lim

= lim

1 + n2 +

0<1< ∞

Comparison Test

15.

converges; Limit

n

1

an
n3 / 2 n + 1
n 4 + n3
= lim
= lim
n →∞ bn
n→∞
n→∞ n 2 + 1
n2 + 1
n→∞

∑ an

n =1

lim

lim

= lim

∞

an
n2
1
= lim
= lim
= 1;
2
n →∞ bn
n →∞ n + n n→∞ 1 + 1

converges; Limit Comparison Test
n +1

n

an +1
[(n + 1) 2 + 1]3n
= lim
n →∞ an
n→∞ 3n +1 ( n 2 + 1)
n→∞

3/ 2

n5 / 2

n→∞

n2

lim

= lim

n3 / 2

5/ 2

1

Comparison Test

= ∞ >1

+1

6
n

n→∞

; bn =

n

(n + 1)(5 + n)
n 2 + 6n + 5
= lim
6+n

6+n
n→∞
n →∞
n + 6 + 5n

2

0<4< ∞

= lim
= lim

n – 4n + 1
5

4 + 32
an
4n5 + 3n3
n
lim
= lim
= lim
= 4;
n →∞ bn
n→∞ n5 – 4n 2 + 1 n→∞ 1 – 43 + 15

The series diverges; nth-Term Test
12.

4n3 + 3n

n

n →∞ 2( n + 1)

lim

n

0<1< ∞
∞

∑ bn

n =1

n

∞

converges ⇒ ∑ an converges;
n =1

Limit Comparison Test

1
1
= <1.
2
2

n→∞ 2 +
n

= lim

The series converges; Ratio Test

536

Section 9.4

Instructor’s Resource Manual

© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this
material may be reproduced, in any form or by any means, without permission in writing from the publisher.


n

22. an =

; bn =

1
n

n +1
a
n2
1

= lim
= 1; 0 < 1 < ∞
lim n = lim
n →∞ bn
n→∞ n 2 + 1 n→∞ 1 + 12
2

28.

3

1
= <1
3
The series converges; Ratio Test

n

∞

∞

∑ bn

diverges ⇒ ∑ an diverges;

n =1

n =1

Limit Comparison Test
23. an =

n

; lim

3n n→∞

an +1
(n + 1)3n
= lim
an
n →∞ 3n +1 n

1+ n 1
n +1
= lim
= <1
3
n→∞ 3n
n→∞ 3
The series converges; Ratio Test
1

= lim

29. –1 ≤ cos n ≤ 1 for all n, so
3 4 + cos n 5
for all n.

3 ≤ 4 + cos n ≤ 5 ⇒
≤
≤
n3
n3
n3
∞
∞
5
4 + cos n
∑ n3 converges ⇒ ∑ n3 converges;
n =1
n =1
Comparison Test
30.

24. an =

1 + 1n
an +1
5(3n + 1)
3
= lim
= lim
n →∞ an
n→∞ (3n +1 + 1)5 n→∞ 3 + 1n
lim

3n
;

n!

an +1
52 n + 2 n !
25
= lim
= lim
= 0 <1
n →∞ an
n→∞ ( n + 1)!52 n
n →∞ n + 1
lim

The series converges; Ratio Test

an +1
3n +1 n !
3
= lim
= lim
= 0 <1
n
n →∞ an
n →∞ ( n + 1)!3
n →∞ n + 1
lim

31.

an +1

(n + 1)n +1 (2n)!
= lim
n →∞ an
n→∞ (2n + 2)!n n
lim

The series converges; Ratio Test
1

25. an =

=

1

;

1

is continuous, positive,

n n n3 / 2 x 3 / 2
and nonincreasing on [1, ∞ ).
∞

⎡ 2 ⎤
∫1 x3 / 2 dx = ⎢⎣ – x ⎥⎦ = 0 + 2 = 2 < ∞
1
The series converges; Integral Test
∞

1

ln n ln x
is continuous, positive, and
;
n2 x2
nonincreasing on [2, ∞) . Use integration by parts

26. an =

with u = ln x and dv =

1
x2

dx for

∞

∞ 1
⎡ ln x ⎤
dx = ⎢ –
⎥ + ∫2 2 dx
x
⎣
⎦2
x
x

∞ ln x

∫2

(n + 1)n +1

= lim

2

∞

ln 2 1
⎡ ln x 1 ⎤
– ⎥ = 0+
+ <∞
= ⎢–
x ⎦2
2
2
⎣ x
ln x
⎛
⎞
= 0 by l'Hôpital's Rule. ⎟
⎜ lim
⎝ x →∞ x
⎠
The series converges; Integral Test

n→∞ (2n + 2)(2n + 1) n n

(n + 1)n +1

= lim

n →∞ 2( n + 1)(2n + 1) n n

⎡ 1 ⎛ n + 1 ⎞n ⎤
= lim ⎢
⎜
⎟ ⎥
n→∞ 2(2n + 1) n n
n →∞ ⎢ 4n + 2 ⎝ n ⎠ ⎥
⎣
⎦

(n + 1)n

= lim

n
1 ⎤⎡
⎡
⎛ n +1⎞ ⎤
= ⎢ lim
⎢
lim
⎜
⎟ ⎥ = 0⋅e = 0 <1

⎥
⎣ n→∞ 4n + 2 ⎦ ⎢⎣ n→∞ ⎝ n ⎠ ⎥⎦

(The limits can be separated since both limits
exist.) The series converges; Ratio Test
x

⎛ 1⎞
⎛ 1⎞
32. Let y = ⎜ 1 − ⎟ ; ln y = x ln ⎜ 1 − ⎟
⎝ x⎠
⎝ x⎠

(

ln 1 − 1x
⎛ 1⎞
lim x ln ⎜ 1 − ⎟ = lim
1
x →∞
⎝ x ⎠ x →∞
x
= lim

x →∞

1/ x 2
1− 1x

( )

−

1
x2

= lim −
x →∞

1

(1 − 1x )

)

= −1
n

⎛ 1⎞
Thus lim y = e −1 , so lim ⎜ 1 − ⎟ = e−1.
n⎠
x →∞
n→∞ ⎝
The series diverges; nth-Term Test

27. 0 ≤ sin 2 n ≤ 1 for all n, so
1
1
1
2 ≤ 2 + sin 2 n ≤ 3 ⇒ ≥
≥ for all n.

2
2 2 + sin n 3
1
≠ 0 and the series diverges;
Thus, lim
n →∞ 2 + sin 2 n
nth-Term Test

Instructor’s Resource Manual

Section 9.4

537

© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this
material may be reproduced, in any form or by any means, without permission in writing from the publisher.