chapter_10 - January 31, 2005 12:53 L24-ch10 Sheet number 1...

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January 31, 2005 12:53 L24-ch10 Sheet number 1 Page number 424 black 424 CHAPTER 10 Infnite Series EXERCISE SET 10.1 1. (a) 1 3 n 1 (b) ( 1) n 1 3 n 1 (c) 2 n 1 2 n (d) n 2 π 1 / ( n +1) 2. (a) ( r ) n 1 ;( r ) n (b) ( 1) n +1 r n 1) n r n +1 3. (a) 2 , 0 , 2 , 0 (b) 1 , 1 , 1 , 1 (c) 2(1+( 1) n ); 2 + 2 cos 4. (a) (2 n )! (b) (2 n 1)! 5. 1 / 3, 2 / 4, 3 / 5, 4 / 6, 5 / 7 ,... ; lim n + n n +2 = 1, converges 6. 1 / 3, 4 / 5, 9 / 7, 16 / 9, 25 / 11 ; lim n + n 2 2 n +1 =+ , diverges 7. 2 , 2 , 2 , 2 , 2 ; lim n + 2 = 2, converges 8. ln 1, ln 1 2 ,ln 1 3 1 4 1 5 ; lim n + ln(1 /n )= −∞ , diverges 9. ln 1 1 , ln 2 2 , ln 3 3 , ln 4 4 , ln 5 5 ; lim n + ln n n = lim n + 1 n =0 µ apply L’Hˆ opital’s Rule to ln x x , converges 10. sin π , 2 sin( π/ 2), 3 sin( 3), 4 sin( 4), 5 sin( 5) ; lim n + n sin( π/n ) = lim n + sin( π/n ) 1 /n = lim n + ( π/n 2 ) cos( π/n ) 1 /n 2 = π , converges 11. 0 , 2 , 0 , 2 , 0 ; diverges 12. 1, 1 / 4, 1 / 9, 1 / 16, 1 / 25 ; lim n + ( 1) n +1 n 2 = 0, converges 13. 1, 16 / 9, 54 / 28, 128 / 65, 250 / 126 ; diverges because odd-numbered terms approach 2, even-numbered terms approach 2. 14. 1 / 2, 2 / 4, 3 / 8, 4 / 16, 5 / 32 ; lim n + n 2 n = lim n + 1 2 n ln 2 = 0, converges 15. 6 / 2, 12 / 8, 20 / 18, 30 / 32, 42 / 50 ; lim n + 1 2 (1+1 /n )(1+2 /n )=1 / 2, converges 16. 4, π 2 / 4 2 , π 3 / 4 3 , π 4 / 4 4 , π 5 / 4 5 ; lim n + ( 4) n = 0, converges 17. cos(3), cos(3 / 2), cos(1), cos(3 / 4), cos(3 / 5) ; lim n + cos(3 /n ) = 1, converges
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January 31, 2005 12:53 L24-ch10 Sheet number 2 Page number 425 black Exercise Set 10.1 425 18. 0, 1, 0, 1, 0 ,... ; diverges 19. e 1 ,4 e 2 ,9 e 3 ,16 e 4 ,25 e 5 ; lim x + x 2 e x = lim x + x 2 e x = 0, so lim n + n 2 e n = 0, converges 20. 1, 10 2, 18 3, 28 4, 40 5 ; lim n + ( p n 2 +3 n n ) = lim n + 3 n n 2 n + n = lim n + 3 p 1+3 /n +1 = 3 2 , converges 21. 2, (5 / 3) 2 ,(6 / 4) 3 ,(7 / 5) 4 ,(8 / 6) 5 ; let y = · x x ¸ x , converges because lim x + ln y = lim x + ln x x 1 /x = lim x + 2 x 2 ( x + 1)( x +3) = 2, so lim n + · n n ¸ n = e 2 22. 1, 0, (1 / 3) 3 ,(2 / 4) 4 ,(3 / 5) 5 ; let y =(1 2 /x ) x , converges because lim x + ln y = lim x + ln(1 2 /x ) 1 /x = lim x + 2 1 2 /x = 2, lim n + (1 2 /n ) n = lim x + y = e 2 23. ½ 2 n 1 2 n ¾ + n =1 ; lim n + 2 n 1 2 n = 1, converges 24. ½ n 1 n 2 ¾ + n =1 ; lim n + n 1 n 2 = 0, converges 25. ½ ( 1) n 1 1 3 n ¾ + n =1 ; lim n + ( 1) n 1 3 n = 0, converges 26. { ( 1) n n } + n =1 ; diverges because odd-numbered terms tend toward −∞ , even-numbered terms tend toward + . 27. ½ 1 n 1 n ¾ + n =1 ; lim n + µ 1 n 1 n = 0, converges 28. © 3 / 2 n 1 ª + n =1 ; lim n + 3 / 2 n 1 = 0, converges 29. © n n +2 ª + n =1 ; converges because lim n + ( n n + 2) = lim n + ( n +1) ( n +2) n +1+ n = lim n + 1 n n =0 30. © ( 1) n +1 / 3 n +4 ª + n =1 ; lim n + ( 1) n +1 / 3 n +4 = 0, converges 31. a n = ± +1 k even 1 k odd oscillates; there is no limit point which attracts all of the a n .
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chapter_10 - January 31, 2005 12:53 L24-ch10 Sheet number 1...

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