Business Calc Homework w answers_Part_80

Business Calc Homework w answers_Part_80 - 396 Chapter 9...

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39. f (3) 5 } 1 x } ) x 5 3 5 } 1 3 } f 9 (3) 52 x 2 2 ) x 5 3 52} 1 9 } f 0 (3) 5 2 x 2 3 ) x 5 3 5 } 2 2 7 } , so } f 0 2 ( ! 3) } 5 } 2 1 7 } f - (3) 6 x 2 4 ) x 5 3 2 2 7 } , so } f - 3 ( ! 3) }52} 8 1 1 } } f ( n n ) ( ! 3) } 5 } ( 2 3 n 1 +1 ) n } } 1 x } 5 } 1 3 } 2 } 1 9 } ( x 2 3) 1 } 2 1 7 } ( x 2 3) 2 2 } 8 1 1 } ( x 2 3) 3 1 1 ( 2 1) n } ( x 3 2 n + 3 1 ) n } 40. f ( p ) 5 sin x ) x 5 p 5 0 f 9 ( p ) 5 cos x ) x 5 p 1 f 0 ( p ) sin x ) x 5 p 5 0, so } f 0 2 ( p ! ) } 5 0 f - ( p ) cos x ) x 5 p 5 1, so } f - 3 ( ! p ) } 5 } 1 6 } f ( k ) ( p ) 5 5 sin x ( x 2 p ) 1 } 3 1 ! } ( x 2 p ) 3 2 } 5 1 ! } ( x 2 p ) 5 1 } 7 1 ! } ( x 2 p ) 7 2 1 ( 2 1) n 1 1 } (2 n 1 1 1)! } ( x 2 p ) 2 n 1 1 1 41. Diverges, because it is 2 5 times the harmonic series: n 5 1 } 2 n 5 }52 5 n 5 1 } 1 n }52‘ 42. Converges conditionally. If u n 5 } ˇ 1 n w } , then { u n } is a decreasing sequence of positive terms with lim n u n 5 0, so n 5 1 } ( 2 ˇ 1 n w ) n } converges by the Alternating Series Test. The convergence is conditional because n 5 1 } ˇ 1 n w } is a divergent p -series 1 p 5 } 1 2 } 2 . 43. Converges absolutely by the Direct Comparison Test, since 0 # } ln n 3 n } , } n 1 2 } for n $ 1 and n 5 1 } n 1 2 } converges as a p -series with p 5 2. 44. Converges absolutely by the Ratio Test, since lim n ) } a a n 1 n 1 } ) 5 lim n } ( n n 1 1 1 2 )! } ? } n n 1 ! 1 } 5 lim n } ( n n 1 1 1 2 ) 2 } 5 0. 45. Converges conditionally: If u n 5 } ln ( n 1 1 1) } , then { u n } is a decreasing sequence of positive terms with lim n u n 5 0, so n 5 1 } ln ( ( 2 n 1 1) n 1) } converges by the Alternating Series Test. The convergence is conditional because } ln ( n 1 1 1) } . } 1 n } for n $ 1 and n 5 1 } 1 n } diverges, so n 5 1 } ln ( n 1 1 1) } diverges by the Direct Comparison Test. 46. Converges absolutely by the Integral Test, because E 2 } x (ln 1 x ) 2 } dx 5 lim b 3 2} ln 1 x } 4 5 } ln 1 2 } . 47. Converges absolutely the the Ratio Test, because lim n ) } a a n 1 n 1 } ) 5 lim n } ( ) 2 n 1 3 ) n 1 1 ) 1 ! } ? } ) 2 n 3 ! ) n } 5 lim n } n 1 3 1 } 5 0. 48. Converges absolutely by the Direct Comparison Test, since } 2 n n 3 n n } # 1 } 1 2 } 2 n for n $ 12 and n 5 1 1 } 1 2 } 2 n is a convergent geometric series. Alternately, we may use the Ratio Test or the n th-Root Test (see Exercise 57 and 58 in Section 9.5). 49. Diverges by the n th-Term Test, since lim n } ( 2 2 n 1 2 ) 1 n ( n n 2 2 1 1 1) } does not exist. 50. Converges absolutely by the Direct Comparison Test, since } ˇ n w ( n w 1 w 1 1 w )( w n w 1 w 2 w ) w },} n 1 3/2 } and n 5 1 } n 1 3/2 } converges as a p -series with p 5 } 3 2 } . 51. Converges absolutely by the Limit Comparison Test. Let a n 5 } n ˇ n w 1 2 w 2 w 1 w } and b n 5 } n 1 2 } . Then lim n } a b n n } 5 lim n } n ˇ n w n 2 2 w 2 w 1 w }5 1 and n 5 2 } n 1 2 } converges as a p -series ( p 5 2). Therefore n 5 2 a n converges. 52. Diverges by the n th-Term Test, since lim n 1 } n 1 n 1 } 2 n 5 lim n 1 1 1 } 1 n } 2 2 n 5 } 1 e } ± 0. b 2 if k is even if k 5 2 n 1 1, n even if k 5 2 n 1 1, n odd 0, 2 1, 1, 396 Chapter 9 Review
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53. This is a telescoping series.
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This document was uploaded on 10/31/2011 for the course MAC 2311 at University of Florida.

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Business Calc Homework w answers_Part_80 - 396 Chapter 9...

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