Business Calc Homework w answers_Part_79

Business Calc Homework w answers_Part_79 - Section 9.5 391...

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Unformatted text preview: Section 9.5 391 (b) S n 1 3n2n 3 3n2 1 1 3 n n 1 3n23 1 . 53. ln (1 n 1 x) n 1 This series converges by the Direct Comparison Test, since 1 and n2 n ( 1) n ( n 1 1)n n 1 x n , so at x 1, the series is . This series converges by the Alternating Series n12 is convergent as a 1 Test. 54. arctan x At x .... n 0 p-series with p 2. n 0 ( 1 1)n x 2n 1 2n 1 50. (a) From the list of Maclaurin series in Section 9.2, ln (1 (b) 1 x x) 1 3 2 1 2 1, the sequence is 1)n( 1)2n 2n 1 n n 0 x x2 2 x3 3 ... ( 1)n 1x n n ( 2(n 1) 1 , which converges by the n Alternating Series Test. At x n 0 1, the sequence is (c) To estimate ln , we would let x 2(n 1) 1 , which converges by the Alternating Series Test. un 1 The truncation error is less than the magnitude of the sixth nonzero term, or x6 6 1 26 6 1 384 55. (a) It fails to satisfy un (b) The sum is n 13 for all n 1 1 2 N. 1/3 1/3 1 1/2 1/2 1 n n 1 21n 0.002605 1 1 . 2 Thus, a bound for the (absolute) truncation error is 0.002605. (d) 51. lim k 56. Answers will vary. 1 2n n 1 ak ( 1)n 1x 2n 2n 1 k 1 1 ( 1)n 1(x 2)n n ln (k 2) 2k x k 1 ln (1 2 x ) 2 57. (a) lim n n an lim n n n2 2n lim n ( n n) 2 2 1 2 The series converges. (b) lim n n ak 1 2 2 x lim 3) k ln (k 1 . 2 1 : 2 k 1 2x 1 , 2 The series converges absolutely for x or x an lim n n n n 2n 1 n 2n lim n 1 1 2 The series converges. n n Check x ( 1) converges by the Alternating Series Test. 2) k 0 ln (k 1 Check x : 2 1 diverges by the Direct Comparison Test, since 2) k 0 ln (k 1 1 1 for k 2 and diverges. The original k ln (k 2) k 2 k 1 1 series converges for x . 2 2 k (c) lim n , n odd an lim n , n odd n 2n n lim n , n odd n n 2 1 2n 1 2 1 2 lim n , n even n an an lim n lim n , n even Thus, lim n n n 1 , so the series converges. 2 n x 58. (a) lim 52. (a) The series converges by the Direct Comparison Test, since 1 n ln n p n an 1n 4 n x 4 1 x 4 1 1 for n np 3, and n 3 1 converges as a np The series converges absolutely if 3 x 5. 1, or p-series when p 1. 1 , which diverges by the n ln n b (b) For p 1, the series is 2 Check x Check x 3: ( 1)n diverges. n 0 n 2 Integral Test, since (c) For 0 p 1 dx x ln x lim ln (ln x) b 2 . 5: 1n diverges. n 0 The interval of convergence is ( 3, 5). 1 1, we have p n ln n 1 , so n ln n n 2 1 diverges by the Direct Comparison Test with n p ln n 1 from part (b). n ln n 392 Chapter 9 Review x 4 n3 n an 3 x 4n n x 4 The series converges absolutely for 1, 3 an 1 58. continued (b) lim n n n 3 x 2 The series converges absolutely if 3 an lim n n x 2n 2. lim lim n x n 2 x 3 2 x n (n lim 4n 1 1)3 n 1 n 3n 1, or or 7 x 1. 7: 1: ( 1)n converges n 1 diverges. n 1 x 5. 1: 5: n 1 Check x ( 1)n converges. n n 1 Check x Check x Check x (a) 3 (b) [ 7, (c) ( 7, (d) At x n 1 n 1 1 diverges. n 1) 1) 7 The interval of convergence is [ 1, 5). (c) lim n n an lim n n 2n x n 2x 3. This is a geometric series, so it converges absolutely when r r 2 (x 3 The series converges absolutely if 2x 1, or 1 1 x . 2 2 1 : ( 1)n diverges. 2 n 1 1 and diverges for all other values of x. Since 2 (x 3 Check x Check x 1 1), the series converges absolutely when 1, or 1 2 1) 3 2 1 5 , 2 2 1 5 , 2 2 x 1 : 2 n 1 diverges. 1 1 , . 2 2 n 5 . 2 (a) (b) (c) 1, or The interval of convergence is (d) lim n n an lim n ln x n ln x The series converges absolutely if ln x 1 e (d) None 4. lim an an 1 x e. 1 : e n Check: x Check x ln 0 1 n e n n 0 ( n 1)n diverges. e: (ln e) n 0 n 0 1n diverges. 1 ,e . e x 1 2n (2n 1)! 1)! x 1 2n 2 n (2n x 12 lim 0 1)(2n) n (2n lim The series converges absolutely for all x. (a) (b) All real numbers (c) All real numbers (d) None 5. lim an an 1 The interval of convergence is s Chapter 9 Review Exercises (pp. 509511) 1. lim n an an 1 lim n x (n n 1 lim n 1)! n! xn lim n x n 1 0 n 3x (n 1n 1 1)2 n2 3x 1n 3x 1 The series converges absolutely for 3x 1 1 2 . Furthermore, when 3 1 1, we have an and n2 The series converges absolutely for all x. (a) (b) All real numbers (c) All real numbers (d) None 3x 1, or 0 x n 1 1 converges as a p-series with p n2 2, so an also n 1 converges absolutely at the interval endpoints. (a) 1 3 2 3 2 3 (b) 0, (c) 0, (d) None Chapter 9 Review an an (n 2) x 3n 3 (n 1) x 3n n an an xn n 1 393 6. lim n 1 lim x3 1, or 9. lim n 1 lim n 1 n xn x 1, or 1 x 1. The series converges absolutely for x 3 1 x 1. When x The series converges absolutely for x Check x n 1 1, the series diverges by the nth 1: Term Test. (a) 1 (b) ( 1, 1) (c) ( 1, 1) (d) None 7. lim n ( 1)n converges by the Alternating Series Test. n Check x 1 n n 1 1: 1 . 2 diverges as a p-series with p (a) 1 (b) [ 1, 1) lim n an an 1 (n 1 2) 2x 1 n (2n 3)2n 1 1 (2n 1)2n (n 1) 2x 1 n (c) ( 1, 1) (d) At x 10. lim n 2x 2 1 en 1 x n 1 (n 1)e n The series converges absolutely for 3 2 2x 2 1 an an 1 1, or lim ne e xn n ex 1, x 1 2x 1 . When 2 2 1, the series diverges by the The series converges absolutely for e x or 1 e nth-Term Test. (a) 1 (b) (c) 3 1 , 2 2 3 1 , 2 2 x 1 . e Furthermore, when e x 1, we have an n 1 converges as a p-series with p e, so an also converges 1 and ne n 1 1 ne absolutely at the interval endpoints. (a) xn 1 1)n n (n 1 nn xn nn 1)(n 1 e 1 1 , e e 1 1 , e e (d) None 8. lim n an an 1 lim 1 x lim n (b) 1) n (n x lim n (n 1) 1 1 n n x 1 lim e n n 1 0 (c) (d) None 11. lim 3n an (n 1) x 2n 1 n n x2 The series converges absolutely when 1, 3 an 1 The series converges absolutely for all x. Another way to see that the series must converge is to observe that for n 2x, we have xn nn 1 n , so the terms 2 lim (n 2) x 2n 3n 1 1 x2 3 are (eventually) bounded by the terms of a convergent geometric series. A third way to solve this exercise is to use the nth Root Test (see Exercises 5758 in Section 9.5). (a) (b) All real numbers (c) All real numbers (d) None or 3 x 3. 3, the series diverges by the nth Term Test. When x (a) (b) ( (c) ( 3 3, 3, 3) 3) (d) None 394 Chapter 9 Review an an x 1 2n 2n 3 3 12. lim n 1 lim n 2n x 1 1 2n 1 x 12 12 1, 16. This is a geometric series with r absolutely when x2 2 1 x2 2 1 , so it converges x 3. It The series converges absolutely when x or 0 x 2. 0: 2: n 0 1, or 3 diverges for all other values of x. ( 1) ( 1) 2n 1 n 2n 1 Check x n 0 ( 1) converges 2n 1 n (a) (b) ( (c) ( 3 3, 3, 3) 3) conditionally by the Alternating Series Test. Check x n 0 ( 1)n converges conditionally by the 2n 1 (d) None 17. f (x) 1 1 x Alternating Series Test. (a) 1 (b) [0, 2] (c) (0, 2) (d) At x 13. lim n 1 x 1 . Sum 4 x2 1 x2 2 ... 1 1 4 ( 1)nx n 4 . 5 ..., evaluated at x 0 and x lim n 2 1)! x 2n 2n 1 2 18. f (x) 2n n! x 2n ln (1 x) x 2 . Sum 3 x3 3 ... 2 3 ( 1)n ln 5 . 3 1x n n , an an 1 (n evaluated at x 19. f (x) sin x x ln 1 x5 5! lim n (n 1)x 2 2 0, x , x 0 0 0. x3 3! ... sin ( 1)n 0. x 2n 1 (2n 1)! ..., evaluated at x The series converges only at x (a) 0 (b) x (c) x 0 only 0 21. f (x) 10x n 1 ln (n 1) n . Sum 1 3 x2 2! x4 4! 20. f (x) cos x ... cos 3 xn n! ( 1)n 1 . 2 x 2n (2n)! ..., evaluated at x ex 1 . Sum x2 2! (d) None 14. lim n x e ln 2 x 1 3 ... 2. ..., evaluated at an an 1 lim ln n 10x n 10x 1, x 22. f (x) ln 2. Sum tan 1 The series converges absolutely for 10x or 1 10 x x 1 . 10 1 : 10 n x3 3 x5 5 ... tan 1 ( 1)n 1 3 x 2n 1 2n 1 ..., evaluated at x 2 . Sum Check n Series Test. Check n ( 1) converges by the Alternating ln n n 6 . (Note that 1 when n is replaced by n becomes ( 1) n 1 x 2n 1 1, the general term of tan x 2n 1 , which matches the general term 1 : 10 n 1 Test, since ln n 1 10 1 1 , 10 10 1 1 , 10 10 2 1 diverges by the Direct Comparison ln n 1 1 for n 2 and diverges. n n 2 n given in the exercise.) 23. Replace x by 6x in the Maclaurin series for the end of Section 9.2. 1 1 6x 1 1 x given at (a) (b) (c) 1 1 (6x) 6x (6x)2 36x 2 ... ... (6x)n (6x)n ... ... 1 1 x 24. Replace x by x 3 in the Maclaurin series for 1 10 given at (d) At x 15. lim n the end of Section 9.2. 1 an an 1 (n 2)! x n 1 (n 1)! x n n lim lim (n n 2) x (x 0) 1 x3 1 1 (x 3) x3 (x 3) 2 x6 ... ... ( x 3)n ... ... The series converges only at x (a) 0 (b) x (c) x 0 only 0 0. ( 1)nx 3n 25. The Maclaurin series for a polynomial is the polynomial itself: 1 2x 2 x 9. (d) None Chapter 9 Review 4x 1 x 1 1 x 395 26. 4x 33. Use the Maclaurin series for e x given at the end of x x2 4x 3 4x(1 4x ... ... xn 4x n ...) 1 Section 9.2. ... xe x2 4x 2 x1 x x3 ( x 2) x 2! 5 ( x 2)2 2! ... x n! ... ( 1)n 2n 1 ( x 2 )n n! ... ... 1 27. Replace x by x in the Maclaurin series for sin x given at the end of Section 9.2. sin x x ( x)3 3! ( x)5 5! 34. Replace x by 3x in the Maclaurin series for tan ... ( 1)n ( x)2n 1 (2n 1)! x given at ... the end of Section 9.2. tan 1 3x 3x 28. Replace x by 2x in the Maclaurin series for sin x given at 3 (3x)3 3 (3x)5 5 ... ( 1)n (3x)2n 1 2n 1 ... x) 35. Replace x by 2x in the Maclaurin series for ln (1 the end of Section 9.2 2x sin 3 ( 2x 3 2x 3 3 2x 5 3 3! 5! ... ( 2x 2n+1 1)n 3 (2n 1)! ) given at the end of Section 9.2. ln (1 2x) 2x ( 2x)2 ( 2x)3 ... 2 3 n ( 2x) ... ( 1)n 1 n n 8x 3 ... (2x) 2x 2x 2 3 n 2x 3 4x 3 81 4x 5 3645 ... ( 1)n (2n 1 2x 2n 1 3 .... 1)! 29. x sin x x x3 x5 x7 ... 3! 5! 7! x 2n 1 ... ( 1)n (2n 1)! 2n 1 x3 x5 x7 ... ( 1)n x 3! 5! 7! (2n 1)! 36. Use the Maclaurin series for ln (1 Section 9.2. x ln (1 ... x x2 x x) x ln [1 2 x) given at the end of x ( x)] 3 30. ex 2 e x 1 1 2 x 1 1 2 x2 2! x2 2! ... x2 2! xn n! ... ( 1)n ... xn n! x3 2 ( x) 2 x4 3 1 ( x) 3 ... xn n 1 ( 1)n ... n 1 ( x) n ... ... 1 1 x x4 4! ... x 2n (2n)! ... 37. f (2) f (2) f (2) f (2) f (n)(2) (3 (3 2(3 6(3 x) x) x 2 2 x 2 3 x 2 4 x 2 n 1 1 ... x) x) 2, so 31. Replace x by 5x in the Maclaurin series for cos x given at f (2) 2! f (2) 6, so 3! 1 1 1 ... n!(3 1 (x (x x) the end of Section 9.2. cos 5x 1 ( 5x)2 2! 5x)4 ... 4! ( 5x)2n ... ( 1)n (2n)! n 5x (5x)2 ... ( 1)n (5x) 2! 4! (2n)! ( 3 x 2 f (n)(2) n!, so n! 2 1 x 2) 2)n 2x 2 4x) x 4) x 1 (x ... 5) x 1 1 2) (x 2) 3 1 ... 38. f ( 1) f ( 1) f ( 1) f ( 1) f (n)( 1) (x 3 (3x 2 (6x 6x 1 2 7 10, so f ( 1) 2! 32. Replace x by x in the Maclaurin series for e x given at the 2 x 2 2 x n 2 5 end of Section 9.2. e x/2 f ( 1) 6, so 3! 1 0 for n 5 2 4. 7(x 1) 5(x 1)2 (x 1)3 4 is 0. 1 1 x 2 x 2 2! 2 2 ... ... n! 1 n! 2 x n ... ... x3 2x 2 x 8 This is a finite series and the general term for n ...
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This document was uploaded on 10/31/2011 for the course MAC 2311 at University of Florida.

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