fa07hw9 - 3. n=5 1 (2 + )2n 1 1 1 + + + 10 12 (2 + ) (2 + )...

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Unformatted text preview: 3. n=5 1 (2 + )2n 1 1 1 + + + 10 12 (2 + ) (2 + ) (2 + )14 1 1 1 = + + 1+ 10 2 (2 + ) (2 + ) (2 + )4 1 1 1 . = = 1 (2 + )10 (2 + )8 (2 + )2 - 1 1- (2 + )2 = 9. n=1 3 + 2n diverges to because 2n+2 3 + 2n 2n+2 3 +1 1 2n = lim = > 0. n 4 4 n lim 11. Since 1 1 = n(n + 2) 2 sn = 1 1 - , therefore n n+2 1 1 1 1 + + ++ 13 24 35 n(n + 2) 1 1 1 1 1 1 1 1 1 = - + - + - + - + 2 1 3 2 4 3 5 4 6 1 1 1 1 1 1 - + - + - + n-2 n n-1 n+1 n n+2 1 1 1 1 - 1+ - . = 2 2 n+1 n+2 3 Thus lim sn = , and 4 n=1 3 1 = . n(n + 2) 4 19. sn = -1 + 1 - 1 + + (-1)n = Thus lim sn does not exist, and -1 if n is odd . 0 if n is even n diverges. (-1) 3. n2 + 1 diverges to infinity by comparison with n3 + 1 1 n2 + 1 1 , since 3 > . n n +1 n 5. Since sin x x for x 0, we have sin 1 1 1 = sin 2 2 , 2 n n n 1 . n2 so sin 1 converges by comparison with n2 11. 1 + n 4/3 diverges to infinity by comparison with the divergent p-series 2 + n 5/3 1 + n 4/3 lim n 2 + n 5/3 1 n 1/3 n 1/3 + n 5/3 = lim = 1. 2 + n 5/3 1 n 1/3 , since 19. n! diverges to infinity by the ratio test, since n 2 en = lim 1 n2 n 2 en (n + 1)! = lim = . n! e n+1 (n + 1)2 en+1 23. (2n)! converges by the ratio test, since (n!)3 (2n + 2)! (n!)3 (2n + 2)(2n + 1) = lim = lim = 0 < 1. 3 (2n)! ((n + 1)!) (n + 1)3 25. 2n converges by the ratio test since 3n - n 3 = lim 2n+1 3n - n 3 3n+1 - (n + 1)3 2n n3 1- n 2 2 2 3n - n 3 3 = lim = lim = < 1. 3 3 3 3 3 (n + 1) (n + 1) 3n - 1- 3 3n+1 15. x = t 3 - 4t, y = t 2 , (-2 t 2). 2 Area = -2 t 2 (3t 2 - 4) dt 2 =2 0 (3t 4 - 4t 2 ) dt 2 3t 5 4t 3 =2 - 5 3 y = 0 256 sq. units. 15 x=t 3 -4t y=t 2 A x Fig. 4-15 17. x = sin4 t, y = cos4 t, 0 t /2 . 2 Area = 0 (cos4 t)(4 sin3 t cos t) dt cos5 t (1 - cos2 t) sin t dt 1 1 - 6 8 Let u = cos t du = - sin t dt 1 = sq. units. 6 /2 =4 0 1 =4 0 (u 5 - u 7 ) du = 6 y x=sin4 t y=cos4 t 0t/2 A x Fig. 4-17 19. x = (2 + sin t) cos t, y = (2 + sin t) sin t, (0 t 2 ). This is just the polar curve r = 2 + sin . 2 Area = - 0 2 (2 + sin t) sin t d (2 + sin t) cos t dt dt =- 0 2 (2 sin t + sin2 t)(cos2 t - 2 sin t - sin2 t) dt = 0 4 sin2 t + 4 sin3 t + sin4 t - 2 sin t cos2 t - sin2 t cos2 t dt 2 = 0 2(1 - cos 2t) + 2 1 - cos 2t (- cos 2t) dt 2 + 0 sin t 4 - 6 cos2 t dt = 4 + 9 +0 = sq. units. 2 2 y x=(2+sin t) cos t y=(2+sin t) sin t 0t2 A x Fig. 4-19 3. 4- (-1)n n 7 13 = 5, , , . . . is bounded, positive, and converges to 4. 2 3 5. n2 - 1 1 = n- n n diverges to infinity. 3 8 15 = 0, , , , . . . is bounded below, positive, increasing, and 2 3 4 17. lim(-1)n n3 n = 0. +1 25. lim( n 2 + n - n 2 - 1) n 2 + n - (n 2 - 1) = lim n2 + n + n2 - 1 n+1 = lim 1 1 n 1+ + 1- 2 n n 1+ = lim 1+ 1 + n 1 n 1- 1 n2 1 . 2 = ...
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This homework help was uploaded on 02/07/2008 for the course MATH 1120 taught by Professor Gross during the Fall '06 term at Cornell.

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