# 1 nx 2n 2 2n 2 0 n 0 x 1 x 1 3n 3 0 3 1

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Unformatted text preview: 80 1 95 3 18033! 95 180 0.75 2 2 0.75 3 3 zn x 1! 95 5 18055! 0.75 4 4 95 7 18077! 0.75 5 (a) Rn x ≤ 0 0 x3 3! 5 0.99594 0.75 6 6 0.75 14 ... 0.5 n 1 < 0.001 n 1! This inequality is true for n 1 1n 1 < 0.001 n 1! This inequality is true for n 14 0.559062 (c) Rn x ≤ 0.5 n 1 < 0.0001 n 1! This inequality is true for n 5. 2n 1 < 0.0001 n 1! This inequality is true for n 4. 10. (d) Rn x ≤ (b) Rn x ≤ xn 1 ≤ 1 ⇒ Rn x ≤ n 1! 6. n un 1 un 10 < x < 10. n 69. n! x n 1 lim n 1 n n→ x R 1 x2 4 2! 12 x 8 f < 0.001, use four terms. 9! 1nx 2 n 12 0 f0 1 8 Geometric series which converges only if x 10 < 1 or 67. x2 2! 1 x3 8 3! 13 x 48 f 0x 9 cos x, c f f0 1 1 4 f0 0.75 Rn x P3 x 1 2 f0 sin 61. ln 1.75 1 f0 2 sin 95 f f0 x 2 2 2 n 1 n 12 1nx 2 2 n lim n→ un 1 un lim n→ n 1!x 2 n! x 2 n n 1 which implies that the series converges only at the center x 2. 1 Center: 2 Since the series converges when x 1 and when x the interval of convergence is 1 ≤ x ≤ 3. n 2 0 3, R eview Exercises for Chapter 9 71. y n 1 n 1 n n n 1 n 0 n 1 n 0 4 0 n 4n 1 x 2x 33 n 1n 4n 0 2 2n n 1! 1 2n 2 2n 4n 1 n 1 ! n! n 1 1 1 n 1 1 n 0 1 1 1 1 4n n! 2 n 1 1n 4n n 1 4n n! n x 2n 2 x 2n 2 2 1 n 0 n x2n 2 4n n! 2 n 1 4n n! 2 1 4n n! 2 x2n 1!2 2n n 2 1! x 2n 2 1 1 2n 1 2 4n 12 n 1!2 1 2 4n 1 2 1 1 x2n 2 2n 1 n 1!2 1 2 x2n 1!2 2n n 2 2 2n 1 x2n n 1!2 1 1 1 2 x 2n 2 2 2 0 a 23 x3 1 n 2n 4n 1 n 1 n 1 2n 0 n 2 1 4 0 n 3 1 n 1 73. 2n n 1 x2y 1 4n 0 n 2 n 2n x2n 4n n! 2 1 y xy 4n 0 y x2y x2n n! n 1 2 75. g x r 3 x . Power series n 0 n 2 x 3 77. 1 n 1 n ... n 0 2x 3 n 1 2x 3 1 3 3 2x 1 3 n 2x n 93 1 2 n 09 n 1 1 n x 3 3 3 <x< 2 2 , sin x fx Derivative: 1 83 x 27 42 x 9 fx 0 2x n 3 13 n 2x 03 3 n n 2x n 3n 497 cos x 79. fx sin x fx cos x, . . . f sin x n n eln 3 3 4 n n! 0 2 2 81. 3x xx 2 x 2 x 3 4 2 x 2 2! ex ln 3 and since ex n 83. fx f 2 2n ... xn , we have 3x 0 n! n 0 1 nn 12 x n! 0 x ln 3 n! n 1 x ln 3 1 x2 2 x3 6 ... , x4 x 1 x 2 1 x fx fx 3 4 f n 0 n 1x n! 1n n 0 n! x 1 n! n x n 0 1 n, 2<x<0 3 4 x2 ln 3 2! n 2 x3 ln 3 3! 3 x 4 ln 3 4! 4 . . .. 498 Chapter 9 85. 1 x k 1 x 15 Infinite Series 1 1 x2 kk kx 1 x 5 1 x 5 1 1 x 5 n 1 x 5 22 x 25 ln x 1 n 5 4 ln n n 1 n 6 xn ... ex 89. xn , 0 n! n 1 e1 <x< 12 n! 2 n 0 n n 22n 2n ! 0.7859 n f0 1 fx 2 e2 x f0 fx 4e 2 x f0 fx f0 8 4x 2 2! 8x 3 3! 1 2x n sin t t 0 arctan x arctan x x 2n x3 3 x arctan x x lim 2n 0 0 1 2x 2 2x 43 x 3 x 1 1 nx 2n 1 2n x5 5 1! x x9 2 5 ln t x13 7 ... 2 x17 9 2 ... 0 1 arctan x By L’Hôpital’s Rule, lim x→0 x lim x→0 x2 1 1 2x lim x→0 2x 1 x2 1 t 0 x9 9 ... 0. 1 x2 2! x ... 43 x 3 2x 2 1 ntn n 0 1 1 t 1 ntn n1 dt n 0 1 nn 1 t 1 x7 7 x2 2! t ln t 0 1! 43 x 3 1 t 2x n n! 0 2x2 2x 1 x 1 n x 1 ln 1 1 1 2n x5 2 3 ex P 1 n t 2n 2n 1 ! 0 n n x→0 2x 99. n t 2n sin t dt t 101. (c) e x 1 nt 2n 1 2n 1 ! 0 n (b) Px sin t x 1 4 8e 2 x xn 2x ,e 0 n! ex 2 Px 1.6487 93. The series for Exercise 41 converges very slowly because the terms approach 0 at a slow rate. <x< e2x 95. (a) f x 97. 1 2 n n! 0 0.2231 32n 0 14 . . . 5n 5nn! 9 ... n 1 4nn n 0 9 5 x3 0<x≤2 , x2n , 2n ! n 1 2 3 14 n 54 n 1 1 n cos 1 ... ... 63 x 125 n 1 cos x n 45 3! 4 9x3 533! 1 15 2 x 1 1 1 91. n n 1 4x2 52 2! 1 1 n 4 5 x2 2! 15 2 x3 1k 3! 2! 1 87. kk n n 0 dt n 0 1t 1 1 nt n 1 n 12 x 0 n 0 1 nxn 1 n 12 P roblem Solving for Chapter 9 Problem Solving for Chapter 9 1 3 1. (a) 1 1 9 2 4 1 27 n 12 03 3 ... n 13 1 23 1 12 (b) 0, , , 1, etc. 33 (c) lim Cn 1 n→ n 0 12 33 3. If there are n rows, then an n 1 nn 1 1 2 0 . For one circle, 1 3 1 and r1 a1 3 2 1 3 2 3 6 1 . 23 r1 1 2 For three circles, 3 and 1 a2 r2 2 3r2 2r2 1 . 23 2 2r2 r2 3 r2 1 For six circles, a3 6 and 1 r3 1 23 2 3 r3 4 4r3 . 2r3 23 rn2 an Total Area An lim An 2 n→ 5. (a) an x n 1 4 1 2n 23 Continuing this pattern, rn 1 2n nn 22 3 . 2 nn 1 1 2 1 2 2n 1 2x4 3x5 8 1 2x 3x2 x3 1 x3 x6 ... 1 x3 x6 ... 1 1 R 1 r3 3 r3 1 2x 3x2 2x 2x ... x4 x7 3x2 1 1 x3 1 because each series in the second line has R —CONTINUED— ... 1. 3 x2 x5 x8 ... 499 500 Chapter 9 Infinite Series 5. —CONTINUED— a0 a1x ... a0 1 xp ... a0 a1 x ... ap 1x p 1 a0 a1 x ... ap 1x p 1 an x n (b) R ap 1x p 1 a0 x p ... xp a1 x 1 a1x p ... ... ap 1x ... p 1 ... xp 1 ... xp 1 1 1 xp 1 1 Assume all an > 0. ex 7. (a) 1 n xe x n e ex xn C n n 0 1 n 1 n 0 2 2 n! 1 2 n 1 n 1 n n 1, 2 n! 0 1, 2e 1. Letting x 2 n! 0 1 n! 5.4366. . 1 . 2 2 n! 2 sin x dx, a2 x 9. Let a1 n 1 1 Thus, 0 Letting x 1 xn . n! n ex n 0, you have C e (b) Differentiating, xe x n Letting x xn n! 0 ... xn 1 0 n! xe x xe x dx x2 2! x sin x dx, a3 x 3 2 sin x dx, etc. x Then, 0 sin x dx x Since lim an a2 0 and an n→ 11. (a) a1 a1 1 a3 a4 . . .. < an, this series converges. 3.0 a2 1.73205 a3 2.17533 a4 2.27493 a5 2.29672 a6 2.30146 1 lim an 13 2 n→ [See part (b) for proof.] (b) Use mathematical induction to show the sequence is increasing. Clearly, a2 Now assume an > an an a > an an a> an 1 1. a1 a a> Then a 1 an a 1 a > an. Use mathematical induction to show that the sequence is bounded above by a. Clearly, a1 —CONTINUED— a < a. a a1. P roblem Solving for Chapter 9 501 11. —CONTINUED— Now assume an < a. Then a > an and a 1 > 1 implies 1 > an 1 aa a2 a > an a2 > an a> a an a an 1. Hence, the sequence converges to some number L. To find L, assume an L L⇒ L2 1 2 1± L 1 2 1 Hence, L 1 13. (a) n 1 4a a 4a 1 2n n 1 21 L⇒ an 1 L: . a L a 0 . 1 1 L2 1 22 1 23 1 1 24 1 1 25 ... 1 (b) 1n 2n an 1 an 2 n 2 1n 1 1 1n 1n 1 21 1 1 This sequence is 8, 2, 8, 2, . . . which diverges. S1 1 20 S1 1 1 8 9 8 S3 9 8 1 4 11 8 S4 11 8 1 32 45 32 S5 45 32 1 16 47 32 40 240 1 (c) 15. S6 130 70 S7 240 130 70 S8 440 240 130 810 440 240 1490 810 1 2n 1 1 2n 1n 2 because 2 and n 1 2 2n 1n 1 2 n 1 2, 1n 1n 1n → 1 < 1 converges 2 2, 1, 2, . . . and 2 n 1 2 →1 2 → 1. 1490 S10 1 2n 810 S9 n 17. (a) n (b) Let f x fx 440 440 2740 1 diverges (harmonic) 1n sin x. By the Mean Value Theorem, fy fcx y cos c x y≤x y, where c is between x and y. Thus, 1 2n 0 ≤ sin ≤ 1 2n sin 1 2n 1 1 2n 1 2n 2n Because n 1 1 1 2n 1 2n 1 sin converges, the Comparison Theorem tells us that n 1 1 2n sin 1 2n 1 converges....
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## This note was uploaded on 01/11/2013 for the course MATH 111 taught by Professor Man during the Spring '13 term at University of Washington-Tacoma Campus.

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