EVNREV08 - 414 66. a2n Chapter 8 Infinite Series 68....

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 414 66. a2n Chapter 8 Infinite Series 68. Answers will vary. 1 0 (odd coefficients are zero) 1 ,g 16 2 70. y 60 , v0 3x 3x 3x 64, k 32 1 16 32 x3 3 64 3 1 2 3 1 16 2 32 x 4 4 64 4 1 2 4 2 32x 2 2 64 2 1 2 32 32 n ... 22x 2 2 64 2 23x3 3 64 316 n 2 24x 4 4 64 4 16 3x 32 ... xn 16 2 n 32 n 2n xn n 16 2 n 64 n n 2 72. (a) f x ln x 2 1 . x2 1.5 From Exercise 8, you obtain P P8 (c) F x 0 x (b) 0 1 x 2n 1 x 0 1 n x 2n n1 x4 3 2 n 1 n x 2n n1 0 − 0.5 2 x2 2 x6 4 x8 5 ln t2 1 dt t2 P8 t d t Gx 0 x Fx Gx 0.25 0.2475 0.2475 0.50 0.4810 0.4810 0.75 0.6920 0.6920 1.00 0.8776 0.8805 1.50 1.1798 5.3064 2.00 1.4096 652.21 (d) The curves are nearly identical for 0 < x < 1. Hence, the integrals nearly agree on that interval. 74. Assume e e Set a a 1 N! e N! 1 N 1! 1 N 1 N N p q is rational. Let N > q and form the following. 1 1 1 2! 1 ... 1 N! 1 N! 1 N 1! N 1 2! ... ... 1 2! , a positive integer. But, 1 N 1 1 N 1 1 N 1 1N 1 1 N 1 ...< 1 N 1 N 1 1 2 ... ... 2 1 N 1 1 1 1 2 ... 1 , a contradiction. N Review Exercises for Chapter 8 2. an n n2 1 4. an n : 3.5, 3, . . . 2 Matches (c) 4 6. an 2 3 Matches (b) 6 n 1 : 6, 4, . . . R eview Exercises for Chapter 8 n 2 1 0 n Converges n→ 415 8. an 2 sin 10. lim 0 12 −2 The sequence seems to diverge (oscillates). sin n : 1, 0, 2 1, 0, 1, 0, . . . 12. lim n→ n ln n n→ lim 1 1n 14. lim 1 n→ 1 2n n k→ lim 1 1 k k 12 e1 2 Diverges 16. Let y ln y bn ln bn n n→ Converges; k cn 1n 2n 1, 2, 3, 4, 5 $20,168.40 18. (a) Vn (b) V5 1 120,000 0.70 n, n 120,000 0.70 5 cn n→ lim ln y lim bn cn bn ln b cn ln c Assume b ≥ c and note that the terms bn ln b bn cn ln c cn bn ln b bn cn cn ln c bn cn converge as n → 20. (a) . Hence an converges. (b) 1 k Sk 5 0.3917 10 0.3228 15 0.3627 20 0.3344 25 0.3564 (c) The series converges by the Alternating Series Test. 0 0 12 22. (a) k Sk 5 0.8333 10 0.9091 15 0.9375 20 0.9524 25 0.9615 1 . n2 (b) 1 (c) The series converges, by the limit comparison test with 0 0 12 24. Diverges. Geometric series, r 2n 2 n 03 2 3 n 1.82 > 1. 26. Diverges. nth Term Test, lim an n→ 2 . 3 28. n 4 n 0 43 12 See Exercise 27. 2 3 n 30. n 0 n 1 1n 2 n 0 2 3 n n 0 1 n 1 1 1 2 n 1 2 1 2 1 3 1 3 1 4 ... 3 1 2 1 1 23 416 Chapter 8 Infinite Series 32. 0.923076 0.923076 1 0.000001 n 0.000001 1 2 ... 923,076 999,999 12 76,923 13 76,923 12 13 0.923076 0.000001 n 0 0.923076 0.000001 39 34. S n 0 32,000 1.055 $4,371,379.65 n 32,000 1 1.05540 1 1.055 36. See Exercise 86 in Section 8.2. A P 12 r 1 r 12 1 12t 1 0.065 12 120 100 12 0.065 1 $16,840.32 1 n3 1 n3 4 3 4 38. n 1 4 40. n 1 n 1 1 n2 1 2n n 1 n2 1 n 1 2n 1 Divergent p-series, p <1 The first series is a convergent p-series and the second series is a convergent geometric series. Therefore, their difference converges. 1 n 13 n 42. n 1 n nn n 1 2 1 nn 1n 2 n→ 44. Since lim n n 1 2 1 converges, n 1 13 n 5 converges by the Limit Comparison Test. n→ lim By a limit comparison test with n 1 , the series diverges. 1n 46. n 1 1n n n1 1 48. Converges by the Alternating Series Test. an an 1 an n1 n ≤ n2 n1 n n 1 0 3 ln n 1 3 ln n < n1 n an, lim n→ 3 ln n n 0 n→ lim By the Alternating Series Test, the series converges. n! n 1e an 1 an n→ 50. n n→ lim lim lim n 1! en 1 n e 1 en n! n→ By the Ratio Test, the series diverges. 52. n 1 12 3 5 an 1 an 5 . . . 2n 8 . . . 3n n→ 1 1 3 . . . 2n 5 . . . 3n 1 2n 1 3n 1 2 2 1 5 . . . 3n 3 . . . 2n 1 1 n→ n→ lim lim 1 2 lim 2n 3n 1 2 2 3 By the Ratio Test, the series converges. Review Exercises for Chapter 8 54. (a) The series converges by the Alternating Series Test. (b) x Sn (c) 0.3 417 5 0.0871 10 0.0669 15 0.0734 20 0.0702 25 0.0721 (d) The sum is approximately 0.0714. 0 0 12 56. No. Let an 3937.5 , then a75 n2 0.7. The series n 1 3937.5 is a convergent p-series. n2 0.75 2! 2 58. f x fx fx fx P3 x tan x sec2 x 2 sec2 x tan x 4 sec2 x tan2 x 1 2x 2 sec4 x 2x f f f f 2 4 4 4 4 8 x 3 0.25 4! 4 1 2 4 16 3 60. cos 0.75 1 0.75 4! 4 0.75 6! 6 0.7317 4 0.25 2! 2 4 0.25 3! 3 4 62. e 0.25 1 0.25 0.779 64. fx P4 x P6 x P10 x cos x 1 1 1 x2 2! x2 2! x2 2! 2 n un 1 un n→ n 10 66. P4 n 10 2x 0 n x4 4! x4 4! x4 4! x6 6! x6 6! − 10 f P6 − 10 P10 Geometric series which converges only if 2x < 1 or 1 1 2 < x < 2. x8 8! x10 10! x n 0 68. n 1 3n x 70. lim 3 n 1 2 2n n n 0 x 2 2 n n→ lim x n 2 1 n 1 n 3n x 2 n Geometric series which converges only if x 2 2 < 1 or 0 < x < 4. 3x 1 3 Center: 2 R 2 Since the series converges at 3 and diverges at 7 , the 3 interval of convergence is 5 ≤ x < 7 . 3 3 5 418 Chapter 8 Infinite Series 3 nx 2n 2nn! 3 n 1 n 72. y n 0 y y n 0 2n x2n 2nn! 1 1 n 0 3 n 1 2n 1 x2n 1 x2n 1 2n n 2 x2n 1! 1 3 3 n 0 n 2n n 1 2 n 1 2 2n n 1! 2 2n n 1! 2x y 3xy 3y 2n n 1 1 1 n 0 n 1n 3 2 1 2n 2 2 1 n 2 n 1n n n 2 x2n 1! 2 n nn 0 1 2n 1 n3n 1x2n 2nn! 1n 3 n 0 2n 2nn! 2n 2n 2n n 0 1 3x 2nn! 1 n 2 2n n 2 0 13 x 2nn! n 0 1 n3n 1x2n 2nn! 13 x 2nn! 1 n 1n nn 1 2n 1 1 n 0 3 2x 2n 2nn! 2 1n 1 n 0 n n 0 3 2x 2n 2nn! 1n 3 1 2n x n 1 2nn! 1 n3n 1x2n 2nn! 32 x2 2n n 1 2n 0 1 n3n 1x 2n 1n 1! 2n 2n 2n 2n n 1 74. 3 2 x 3 02 1 x 2 n 32 x2 a 1 r 1 76. Integral: n 0 n 1 n3xn 1 2n 1 1 n n 0 1 n3xn 2n 1 1 x 8 x 4 32 3 1 x , 3 n 78. 8 2x 3 1 x 2 3 2 3 3 ... n 0 8 32 x 1 8 x 34 4 1<x<7 80. fx fx fx fx cos x cos x sin x cos x sin x f n 0 n 4x n! x 4 4 n 2 2 1 nn 1 2 2 x 2 x 1! n 4 4 n 2 1 2 x 2! 2 4 2 2 x 3! 3 4 2 2 x 4! 4 ... 4 2 1 2 82. fx fx fx fx f 4 n 1 csc x csc x cot x csc3 x csc x cot2 x csc x cot3 x csc x cot4 x 1 1 x 2! 2 5 csc3 x cot x 5 csc5 x f n 0 n x 15 csc3 x cot2 x 2x n! 2 n csc x 2 5 x 4! 4 ... 2 Review Exercises for Chapter 8 84. fx fx fx fx f 4 419 x1 1 x 2 2 12 1 2 1 2 1 2 f n 0 n 1 x 2 1 2 1 2 32 3 x 2 3 2 52 x x 2 2 5 x 2 4 n 7 2, ... 4x n! 4 22 4 22 x x x 4 252! 2 1 n 11 3x 4 283! 3 3 1 3 5x 2114! 4 4 n 4 ... 1 n 2 5 . . . 2n 23n 1n! 3x 86. hx hx hx hx h4 x h5 x 1 1 x 3 1 31 12 1 x 3 x x x x 5 4 60 1 360 1 2520 1 1 3x 6 7 x 12x2 2! x n 8 60x3 3! 1 n 360x4 4! 2520x5 5! ... n 0 1 n n 2 !x n 2n! xn , n! 0 23 n! n n 1 n 0 n n 2n 2 1x 88. ln ln x n 1 1 1 n 1 n 1 , 1 n 0<x≤2 90. ex n <x< 2n n 0 3 n! 1.9477 6 5 n 1 65 n 1 5nn e2 3 n 0 1 n 1 n 1 0.1823 92. sin x n 0 1 1 n 0 n x2n 1 , 2n 1 ! 32n 1 <x< 0.3272 sin 1 3 n 1 2n 1! 94. ex n xn 0 n! xn 1 0 n! 1 x xex x x2 x2 2! x3 2! 1 ... ... xex n 1 xex dx 0 1 0n ex 0 e xn 2 e 1 0 0 1 1 1 1 xn 1 dx 0 n! n 0 n 2 n! n 0 n 2 n! 420 Chapter 8 fx fx fx fx f f f f 4 5 6 7 Infinite Series f0 f0 f0 f0 f f f f 8x3 3! 4 5 6 7 96. (a) sin 2x 2 cos 2x 4 sin 2x 8 cos 2x 16 sin 2x 32 cos 2x 64 sin 2x 128 cos 2x 0 2x 0x2 2! 0 2 0 8 0 32 0 128 32 x5 5! 0x6 6! 128x7 7! ... 2x 43 x 3 45 x 15 87 x 315 ... x x x x 0 0 0 0 0x4 4! sin 2x (b) sin x n 0 1 nx2n 1 2n 1 ! 1 n 2x 2n 2n 1 ! 8x3 6 32x5 120 1 sin 2x n 0 2x 128x7 5040 2x 3 3! ... 2x 5 5! 2x 2x 7 7! 43 x 3 ... 45 x 15 87 x 315 ... 2x (c) sin 2x 2 sin x cos x 2x 2x 2x 2 x3 3 x3 6 x3 2 x5 120 x3 6 2 x5 15 x7 5040 x5 24 4 x7 315 ... x5 12 ... 1 x2 2 x4 24 x6 720 x7 720 43 x 3 45 x 15 ... x7 144 x7 240 x7 5040 ... ... x5 120 2x 87 x 315 98. cos t n 0 1 n t 2n 2n ! 1t 2 2n ! 2n nn 100. et ntn 1 x et n tn n! 0 tn n! 1 tn 1 1 n! tn n! 1n x 0 n cos x t 2 t dt n n 0 1 n cos 0 2 0 2 2n 1 2n ! n nn et t x 1 n 1 1 0 n 0 22n x3 2 x2 2 1x 2n ! n et t 1 1 1 3 5x7 2467 1 3 5x6 2467 dt n 0 xn n! 1n 102. arcsin x arcsin x x x→0 x 1 1 3 3 1 3x5 245 1 3x4 245 ... ... lim arcsin x x 1 arcsin x By L’Hôpital’s Rule, lim x→0 x x→0 lim 1 1 x2 1. Problem Solving for Chapter 8 421 Problem Solving for Chapter 8 2. Let S n 2 1 1 2n 1 1 22 1 22 1 1 22 2 1 . 2 26 2 1 12 1 32 1 42 1 22 1 42 1 32 1 52 ... ... Then 6 1 12 S S S 2 ... 1 32 ... Thus, S 6 12 46 3 64 2 2 8 . 4. (a) Position the three blocks as indicated in the figure. The bottom block extends 1 6 over the edge of the table, the middle block extends 1 4 over the edge of the bottom block, and the top block extends 1 2 over the edge of the middle block. The centers of gravity are located at bottom block: middle block: top block: 1 6 1 6 1 6 1 4 1 2 1 4 1 2 1 2 1 2 1 3 1 12 5 . 12 1 6 1 2 1 4 01 5 6 12 11 12 The center of gravity of the top 2 blocks is 1 12 5 12 2 1 , 6 which lies over the bottom block. The center of gravity of the 3 blocks is 1 3 1 12 5 12 3 0 which lies over the table. Hence, the far edge of the top block lies 1 1 1 11 6 4 2 12 beyond the edge of the table. n (b) Yes. If there are n blocks, then the edge of the top block lies c 1 1 from the 2i edge of the table. Using 4 blocks, 4 c 1 1 2i 1 2 1 4 1 6 1 8 25 24 which shows that the top block extends beyond the table. (c) The blocks can extend any distance beyond the table because the series diverges: 1 2i 1 2c 1 i . c 1 1 422 Chapter 8 b 2 b, n 1 Infinite Series 1 n 1 n 1 6. a If a If a a 3 b 4 1 1 n ... 1 a 1 b 2n a b 2a a b 2n n 1 a n 1 1n n a 2n converges conditionally. b diverges. b implies conditional convergence. b, n 1 2n n 1 No values of a and b give absolute convergence. a x2 2! x4 2! 12 8. ex ex f 12 2 1 1 x x2 ... ... 12! 6! x12 6! ... 0 12! 1 ⇒f 6! 0 665,280 10. (a) If p 1, 2 1 dx x ln x 1 dx x ln x p ln ln x 2 diverges. ln b 1 p 1p ln 2 1 p converges. 1p If p > 1, 2 b→ lim If p < 1, diverges. (b) n 4 1 n ln n2 1 2n 1 diverges by part (a). n ln n 4 1 0.99 1 0.01 12. Let bn bn an r n. 1n 14. (a) an r n 1. bn converges ⇒ an r n converges. (b) 1n 1 0.01 n 0 n an1 n r → Lr as n → . 1 0.01 0.01 2 ... 1 Lr < r r 1.010101 . . . 1 0.98 1 1 0.02 0.02 n 0 n By the Root Test, 1 1 0.02 0.02 0.02 2 ... ... 0.0004 1.0204081632 . . . 1 2 1 n1 2 1 1 2 1 23 2 1 3 1 3 16. (a) Height 21 2 n ... 1 <1 2 p-series, p ...4 n (b) S (c) W 4 4 3 4 3 1 1 1 1n 1 33 2 ... n 1 1 converges. n3 2 ...
View Full Document

This note was uploaded on 05/18/2011 for the course MAC 2311 taught by Professor All during the Fall '08 term at University of Florida.

Ask a homework question - tutors are online