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  +  +  + n2 n n1 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 pseries 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. 415 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. 417 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. 419 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 = ...
View
Full Document
 Fall '06
 GROSS
 Calculus

Click to edit the document details