Chapter 11 CSM Calculus 9th metric 11 solution PDF

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NOT FOR SALE11 INFINITE SEQUENCES AND SERIES11 Sequences1. (a) A sequence is an ordered list of numbers. It can also be defined as a function whose domain is the set of positive integers.(b) The termsapproach 8 asbecomes large. In fact, we can makeas close to 8 as we like by takingsufficiently...

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11

INFINITE SEQUENCES AND SERIES

11.1 Sequences 1. (a) A sequence is an ordered list of numbers. It can also be defined as a function whose domain is the set of positive integers.

(b) The terms  approach 8 as  becomes large. In fact, we can make  as close to 8 as we like by taking  sufficiently large.

SA LE

(c) The terms  become large as  becomes large. In fact, we can make  as large as we like by taking  sufficiently large.  2. (a) From Definition 1, a convergent sequence is a sequence for which lim  exists. Examples: {1}, {12 } →∞

(b) A divergent sequence is a sequence for which lim  does not exist. Examples: {}, {sin } →∞

3.  =

2 , so the sequence is 2 + 1



4.  =

2 − 1 , so the sequence is 2 + 1



5.  =

(−1)−1 , so the sequence is 5

6.  = cos

7.  =

22 23 24 25 2      2(1) + 1 2(2) + 1 2(3) + 1 2(4) + 1 2(5) + 1 1 − 1 4 − 1 9 − 1 16 − 1 25 − 1      1 + 1 4 + 1 9 + 1 16 + 1 25 + 1

1





FO R



 , so the sequence is 2



1 , so the sequence is ( + 1)!

1 −1 1 −1 1      51 52 53 54 55

cos



=





=





=



 2 4 8 16 32      . 3 5 7 9 11

 3 8 15 24 0        . 5 10 17 26

 1 1 1 1 1  .  − −  625 3125 5 25 125

 5 3   cos 2 cos  cos  cos     = {0 −1 0 1 0   }. 2 2 2

1 1 1 1 1      2! 3! 4! 5! 6!



=



 1 1 1 1 1  .     2 6 24 120 720

O T

(−1)  (−1)1 1 −1 , and the sequence is , so 1 = = 2 ! + 1 1! + 1     5 2 −3 4 −5 1 2 3 4 −1  .  = −  −  −     2 3 7 25 121 2 2 + 1 6 + 1 24 + 1 120 + 1

8.  =

9. 1 = 1, +1 = 5 − 3.

Each term is defined in terms of the preceding term. 2 = 51 − 3 = 5(1) − 3 = 2.

N

3 = 52 − 3 = 5(2) − 3 = 7. 4 = 53 − 3 = 5(7) − 3 = 32. 5 = 54 − 3 = 5(32) − 3 = 157. The sequence is {1 2 7 32 157   }.

 1 2 3 4 6 6 3 1 . 2 = = = 6. 3 = = = 3. 4 = = = 1. 5 = = . 4 1 2 3  1 2 3 4   The sequence is 6 6 3 1 14     .

10. 1 = 6, +1 =

11. 1 = 2, +1 =

5 =

 25 1 23 2 2 2 2 3 2 = . 2 = = = = . = . 4 = = . 3 = 1 + 3 1 +  1 + 2 1 + 1 7 1 + 25 1 + 23 1+2 5 3

  4 27 2 = = . The sequence is 2 23  52 27  29     . 1 + 4 9 1 + 27 c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. °

FOR INSTRUCTOR USE ONLY

957

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958

CHAPTER 11

INFINITE SEQUENCES AND SERIES

12. 1 = 2, 2 = 1, +1 =  − −1 .

Each term is defined in term of the two preceding terms.

3 = 2 − 1 = 1 − 2 = −1. 4 = 3 − 2 = −1 − 1 = −2. 5 = 4 − 3 = −2 − (−1) = −1. 6 = 5 − 4 = −1 − (−2) = 1. The sequence is {2 1 −1 −2 −1 1   }.

14. 15.

1

1 1 1 1 , 2 , 4 , 6 , 8 , 10





 1 .    . The denominator is two times the number of the term, , so  = 2

1 4, − 1, 41 , − 16 ,

−1      . The first term is 4 and each term is − 14 times the preceding one, so  = 4 − 14 .

1 , 64

−1   16 .     . The first term is −3 and each term is − 23 times the preceding one, so  = −3 − 23 −3 2 −34 89  − 27

16. {5 8 11 14 17   }. 17.

1 2

Each term is larger than the preceding term by 3, so  = 1 + ( − 1) = 5 + 3( − 1) = 3 + 2.

SA LE

13.

  − 34  94  − 516 256     . The numerator of the nth term is 2 and its denominator is  + 1. Including the alternating signs,

we get  = (−1)+1

2 . +1

18. {1 0 −1 0 1 0 −1 0   }.



 =

3 1 + 6

1

04286

2

04615

3

04737

4

04800

5

04839

6

04865 04884 04898

9

04909

10

04918



→∞

lim

→∞

3 (3) 3 1 3 = lim = lim = = 2 1 + 6 →∞ (1 + 6) →∞ 1 + 6 6

 = 2 +

(−1) 

10000

N

1

It appears that lim  = 05.

O T

7 8

20.

 ( − 1) . and  = cos 2 2

FO

19.

Two possibilities are  = sin

2

25000

3

16667

4

22500

5

18000

6

21667

7

18571

8

21250

9

18889

10

21000

It appears that lim  = 2. →∞

  (−1) 1 (−1) lim 2 + = lim 2 + lim = 2 + 0 = 2 since lim =0 →∞ →∞  →∞ →∞   and by Theorem 6, lim

→∞

(−1) = 0. 

° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

FOR INSTRUCTOR USE ONLY

SECTION 11.1

1

   = 1 + − 12

2

12500

05000

3

08750

4

10625

5

09688

6

10156

7

09922

8 9

10039 09980

10

10010

22.



→∞

    1   = lim 1 + lim − 21 = 1 + 0 = 1 since lim 1 + − 2 →∞ →∞ →∞

  lim −21 = 0 by (9).

→∞

10 9

1

21111

2

22346

3

23717

4

25242

5

26935

6

28817

7

30908

8

33231

9 10

35812 38680

It appears that the sequence does not have a limit.   10 10 , which diverges by (9) since lim  = lim →∞ 9 →∞ 9

10 9

 1.

3 + 52 (3 + 52 )2 5 + 32 5+0 = 5 as  → ∞. Converges , so  → = = 2 2 2 1+0 1 + 1 + ( +  )

O T

23.  =

 = 1 +

It appears that lim  = 1.

AL E



FO R

21.

SEQUENCES

3 + 52 3 + 5 (3 + 52 ) , so  → ∞ as  → ∞ since lim = = →∞ 1 + 1 (1 + ) 1+   1 + 1 = 0 + 1 = 1. Diverges lim →∞ 



 3 + 5 = ∞ and 

N

24.  =

4  4 3 , so  → ∞ as  → ∞ since lim  = ∞ and = = 3 →∞ 1 − 2 / 2 ( − 2)3 3 − 2   2 lim 1 − 2 = 1 − 0 = 1. Diverges →∞ 

25.  =

26.  = 2 + (086) → 2 + 0 = 2 as  → ∞ since lim (086) = 0 by (9) with  = 086. →∞

27.  = 3 7− =

3 = 7

Converges

  3 3 , so lim  = 0 by (9) with  = . Converges 7 →∞ 7

° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

FOR INSTRUCTOR USE ONLY

¤

959

960

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CHAPTER 11

INFINITE SEQUENCES AND SERIES

√ √ √ 3   3  3 3 = √ √ → √ = 28.  = √ = 3 as  → ∞. Converges +2 1+0 1 + 2  (  + 2)  29. Because the natural exponential function is continuous at 0, Theorem 7 enables us to write √

lim  = lim −1

→∞

→∞



√ lim (−1 )

=  →∞

= 0 = 1. Converges

(49) 0 49 4 = → = = 0 as  → ∞ since lim   →∞ (1 + 9 )9 (19) + 1 0+1 1 + 9   1 = 0 by (9). Converges lim →∞ 9

31.  =



1 + 42 = 1 + 2

32.  = cos



 +1





(1 + 42 )2 = (1 + 2 )2

= cos



 ( + 1)





  4 = 0 and 9

SA LE

30.  =

√ (12 ) + 4 → 4 = 2 as  → ∞ since lim (12 ) = 0. Converges →∞ (12 ) + 1

= cos



  , so  → cos  = −1 as  → ∞ since lim 1 = 0 →∞ 1 + 1

Converges

√ √ √ 2  2  3 √ =  , so  → ∞ as  → ∞ since lim  = ∞ and = √ 3 2 →∞ 3 3 1 + 4  + 4   + 4

lim

→∞

 1 + 42 = 1. Diverges

34. If  =

FO R

33.  = √

2 (2) 2 2 = = 2. Since the natural exponential function is , then lim  = lim = lim →∞ →∞ ( + 2) →∞ 1 + 2 1 +2

continuous at 2, by Theorem 7, lim 2(+2) = lim→∞  = 2 . Converges →∞

  (−1)  1 1 1   lim 12 = (0) = 0, so lim  = 0 by (6). Converges | | = lim √ 35. lim  = →∞ →∞ →∞ 2 2 →∞  2    →∞

  1 (−1)+1  1 √ = lim √ √ has odd-numbered terms = lim = 1. Thus,  = √ = →∞ ( + +   ) →∞ 1 + 1  +  1+0

O T

36. lim

that approach 1 and even-numbered terms that approach −1 as  → ∞, and hence, the sequence { } is divergent. (2 − 1)! 1 (2 − 1)! → 0 as  → ∞. Converges = = (2 + 1)(2) (2 + 1)(2)(2 − 1)! (2 + 1)!

N

37.  =

38.  =

ln  ln  = = ln 2 + ln  ln 2

1 1 → = 1 as  → ∞. Converges 0+1 +1

ln 2 ln 

39.  = sin . This sequence diverges since the terms don’t approach any particular real number as  → ∞. The terms take on

values between −1 and 1. Diverges 40.  =

tan−1  . 

41.  = 2 − =

lim tan−1  = lim tan−1  =

→∞

→∞

 by (3), so lim  = 0. Converges →∞ 2

2 H 2 2 H 2 . Since lim  = lim  = lim  = 0, it follows from Theorem 3 that lim  = 0. Converges  →∞ →∞  →∞  →∞  

° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

FOR INSTRUCTOR USE ONLY

SECTION 11.1

42.  = ln( + 1) − ln  = ln 43. 0 ≤

1 cos2  ≤  2 2

44.  =



SEQUENCES

¤

961

   +1 1 = ln 1 + → ln (1) = 0 as  → ∞ because ln is continuous. Converges  

[since 0 ≤ cos2  ≤ 1], so since lim

→∞

1 = 0, 2



cos2  2



converges to 0 by the Squeeze Theorem.

√  1+3 2 = (21+3 )1 = (2 1 23 )1 = 2 123 = 8 · 21, so

lim  = 8 lim 21 = 8 · 2lim→∞(1) = 8 · 20 = 8 by Theorem 7, since the function  () = 2



→∞

→∞

is continuous at 0.

Converges sin  sin(1) sin(1) [where  = 1] = 1, it follows from Theorem 3 = lim . Since lim →∞  1 1 →0+

SA LE

45.  =  sin(1) =

that { } converges to 1.

    1 1 cos   = 0 ≤  ≤ , so lim | | = 0 by (9), and lim  = 0 by (6) Converges  →∞ →∞ 2 2 2

46.  = 2− cos .

  2 1+ 

48.  = 1

⇒ ln  =

lim 1 = lim ln 

→∞

→∞

FO R

  2 ⇒ ln  =  ln 1 + , so     2 1 − 2 1 + 2  2 ln(1 + 2) H = lim lim ln  = lim = lim =2 ⇒ →∞ 1 + 2 →∞ →∞ →∞ −12 1     2 2 = 2 . Converges = lim ln  = 2 , so by Theorem 3, lim 1 + lim 1 + →∞ →∞ →∞  

47.  =

1 1 ln  H 1 = lim =0 ⇒ = lim ln , so lim ln  = lim →∞ →∞  →∞ 1 →∞   √ = 0 = 1, so by Theorem 3, lim   = 1. Converges →∞

O T

49.  = ln(22 + 1) − ln(2 + 1) = ln



22 + 1 2 + 1



= ln



2 + 12 1 + 12



→ ln 2 as  → ∞. Converges

(ln )2 H (ln )2 2(ln )(1) 1 ln  H = 2 lim = 0. Converges = 0, so by Theorem 3, lim = lim = 2 lim →∞ →∞ →∞ 1 →∞  →∞ 1  

50. lim

N

51.  = arctan(ln ). Let  () = arctan(ln ). Then lim  () =

Thus, lim  = lim  () = →∞

52.  =  −

=

→∞

→∞

 . 2

 2

since ln  → ∞ as  → ∞ and arctan is continuous.

Converges

√ √ √ −  + 1  + 3 =  − 2 + 4 + 3 =

√ √ 2 + 4 + 3  + 2 + 4 + 3 · √ 1  + 2 + 4 + 3

−4 − 3 (−4 − 3) −4 − 3 2 − (2 + 4 + 3)  √ √ = √ =  ,  =  + 2 + 4 + 3   + 2 + 4 + 3  + 2 + 4 + 3 1 + 1 + 4 + 32

so lim  = →∞

1+

−4 − 0 −4 √ = = −2. Converges 2 1+0+0

° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

FOR INSTRUCTOR USE ONLY

962

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CHAPTER 11

INFINITE SEQUENCES AND SERIES

53. {0 1 0 0 1 0 0 0 1   } diverges since the sequence takes on only two values, 0 and 1, and never stays arbitrarily close to

either one (or any other value) for  sufficiently large. 1

1 1 1 1 1 1 1 1 3 24 3 5 4 6

lim 2−1 = lim

→∞

→∞

as we like. 55.  =



. 2−1 =

1 1 for all positive integers . lim  = 0 since and 2 = →∞ +2 

1 1 = 0 and lim 2 = lim = 0. For  sufficiently large,  can be made as close to 0 →∞  + 2 →∞ 

Converges

! 1 2 3 ( − 1)  1  = · · · · ·· · · ≥ · 2 2 2 2 2 2 2 2

56. 0  | | =

[for   1] =

 → ∞ as  → ∞, so { } diverges. 4

SA LE

54.

3 3 3 3 3 3 3 3 3 · ≤ · · = · · · · ·· · 1 2  ( − 1)  1 2 3 !

[for   2] =

Theorem and Theorem 6, {(−3)!} converges to 0.

27 → 0 as  → ∞, so by the Squeeze 2

From the graph, it appears that the sequence { } =

57.

 (−1)

 +1



is

divergent, since it oscillates between 1 and −1 (approximately). To prove this,  , then { } converges to 1, suppose that { } converges to . If  = +1     does not exist. This = (−1) , so lim = = . But →∞  1  

FO R

and lim

→∞

contradiction shows that { } diverges.

58.

From the graph, it appears that the sequence converges to 0.    sin   |sin | 1 = | | =  ≤ , so lim | | = 0. By (6), it follows that →∞    || lim  = 0.

59.

O T

→∞

From the graph, it appears that the sequence converges to a number between 07 and 08.  = arctan arctan 1 =

60.



2 +4

2



= arctan



22 + 4)2

(2

 [≈ 0785] as  → ∞. 4



= arctan



1 1 + 42



→

From the graph, it appears that the sequence converges to 5 √ √ √ √ √  5 =  5 ≤  3 + 5 ≤  5 + 5 =  2 5   √  = 2 · 5 → 5 as  → ∞ lim 21 = 2 0 = 1 →∞

Hence,  → 5 by the Squeeze Theorem.

[continued] ° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

FOR INSTRUCTOR USE ONLY

SECTION 11.1

¤

SEQUENCES

963

Alternate solution: Let  = (3  + 5 )1 . Then ln (3 + 5  ) H 3 ln 3 + 5 ln 5 = lim = lim →∞ →∞ →∞ →∞  3 + 5 √   so lim  = ln 5 = 5, and so  3 + 5 converges to 5. lim ln  = lim

 3  ln 3 + ln 5 5   = ln 5, 3  +1 5

→∞

From the graph, it appears that the sequence { } =

61.



2 cos  1 + 2



is

SA LE

divergent, since it oscillates between 1 and −1 (approximately). To prove this, suppose that { } converges to . If  = { } converges to 1, and lim

→∞

lim

→∞

2 , then 1 + 2

   = cos , so = = . But   1

 does not exist. This contradiction shows that { } diverges. 

FO

62.

From the graphs, it seems that the sequence diverges.  =

1 · 3 · 5 · · · · · (2 − 1) . We first prove by induction that !

 −1 3 for all . This is clearly true for  = 1, so let  () be the statement that the above is true for . We must 2  −1 2 + 1 2 + 1 3 2 + 1 3 · show it is then true for  + 1. +1 =  · ≥ (induction hypothesis). But ≥ 2 +1 +1 +1 2

O T

 ≥

N

 −1 3  3  [since 2 (2 + 1) ≥ 3 ( + 1) ⇔ 4 + 2 ≥ 3 + 3 ⇔  ≥ 1], and so we get that +1 ≥ 3 2 · 2 = 2 which    3 −1 is  ( + 1). Thus, we have proved our first assertion, so since diverges [by (9)], so does the given sequence { }. 2

63.

From the graph, it appears that the sequence approaches 0. 0   =

1 2 − 1 5 3 1 · 3 · 5 · · · · · (2 − 1) · ·· · · = · · (2) 2 2 2 2

1 1 → 0 as  → ∞ · (1) · (1) · · · · · (1) = 2 2   1 · 3 · 5 · · · · · (2 − 1) So by the Squeeze Theorem, converges to 0. (2) ≤

° c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

FOR INSTRUCTOR USE ONLY

964

¤

CHAPTER 11

INFINITE SEQUENCES AND SERIES

64. (a) 1 = 1, +1 = 4 −  for  ≥ 1.

1 = 1, 2 = 4 − 1 = 4 − 1 = 3, 3 = 4 − 2 = 4 − 3 = 1,

4 = 4 − 3 = 4 − 1 = 3, 5 = 4 − 4 = 4 − 3 = 1. Since the terms of the sequence alternate between 1 and 3, the sequence is divergent. (b) 1 = 2, 2 = 4 − 1 = 4 − 2 = 2, 3 = 4 − 2 = 4 − 2 = 2. Since all of the terms are 2, lim  = 2 and hence, the →∞

sequence is convergent. 65. (a)  = 1000(106)

⇒ 1 = 1060, 2 = 112360, 3 = 119102, 4 = 126248, and 5 = 133823.

→∞

→∞

66. (a) Substitute 1 to 6 for  in  = 100



SA LE

(b) lim  = 1000 lim (106) , so the sequence diverges by (9) with  = 106  1.  10025 − 1 −  to get 1 = $0, 2 = $025,...