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Business Calc Homework w answers_Part_79

Business Calc Homework w answers_Part_79 - Section 9.5...

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(b) S 5 n 5 1 } 3 n 2 n 1 1 } ? } 3 n } 5 n 5 1 } 3 n 2 3 1 1 } . This series converges by the Direct Comparison Test, since } 3 n 2 3 1 1 } , } n 1 2 } and n 5 1 } n 1 2 } is convergent as a p -series with p 5 2. 50. (a) From the list of Maclaurin series in Section 9.2, ln (1 1 x ) 5 x 2 } x 2 2 } 1 } x 3 3 } 2 1 ( 2 1) n 1 1 } x n n } 1 . (b) 2 1 , x # 1 (c) To estimate ln } 3 2 } , we would let x 5 } 1 2 } The truncation error is less than the magnitude of the sixth nonzero term, or ) 2 } x 6 6 } ) 5 } 2 6 1 ? 6 } 5 } 3 1 84 } , 0.002605 Thus, a bound for the (absolute) truncation error is 0.002605. (d) n 5 1 } ( 2 1) 2 n n 1 1 x 2 n } 5 } 1 2 } n 5 1 } ( 2 1) n 1 n 1 ( x 2 ) n } 5 } 1 2 } ln (1 1 x 2 ) 51. lim k ) } a a k 1 k 1 } ) 5 lim k } l 2 n k 1 ( k 1 ) x 1 ) k 1 3 1 ) } ? } ln 2 ( k k ) x 1 ) k 2) } 5 2 ) x ) The series converges absolutely for ) x ) , } 1 2 } , or 2 } 1 2 } , x , } 1 2 } . Check x 5 2 } 1 2 } : k 5 0 } ln ( ( 2 k 1 1 ) k 2) } converges by the Alternating Series Test. Check x 5 } 1 2 } : k 5 0 } ln ( k 1 1 2) } diverges by the Direct Comparison Test, since } ln ( k 1 1 2) } . } 1 k } for k \$ 2 and k 5 2 } 1 k } diverges. The original series converges for 2 } 1 2 } # x , } 1 2 } . 52. (a) The series converges by the Direct Comparison Test, since } n p l 1 n n } , } n 1 p } for n \$ 3, and n 5 3 } n 1 p } converges as a p -series when p . 1. (b) For p 5 1, the series is n 5 2 } n l 1 n n } , which diverges by the Integral Test, since E 2 } x l 1 n x } dx 5 lim b 3 ln (ln x ) 4 5 ‘ . (c) For 0 # p , 1, we have } n p 1 ln n } . } n l 1 n n } , so n 5 2 } n p 1 ln n } diverges by the Direct Comparison Test with } n l 1 n n } from part (b). 53. ln (1 1 x ) 5 Σ n 5 1 ( 2 1) n 1 1 } x n n } , so at x 5 1, the series is Σ n 5 1 } ( 2 1 n ) n 1 1 } . This series converges by the Alternating Series Test. 54. arctan x 5 n 5 0 ( 2 1) n } 2 x n 2 n 1 1 1 1 } At x 5 2 1, the sequence is n 5 0 } ( 2 1 2 ) n n ( 2 1 1 1 ) 2 n 1 1 } 5 2 n 5 0 } 2 ( n 2 1 1) n 1 } , which converges by the Alternating Series Test. At x 5 1, the sequence is n 5 0 } 2 ( n 2 1 1) n 1 } , which converges by the Alternating Series Test. 55. (a) It fails to satisfy u n \$ u n 1 1 for all n \$ N . (b) The sum is 1 n 5 1 } 3 1 n } 2 2 1 n 5 1 } 2 1 n } 2 5 } 1 2 1/3 1/3 } 2 } 1 2 1/2 1/2 } 5 } 1 2 } 2 1 5 2 } 1 2 } . 56. Answers will vary. 57. (a) lim n ˇ n a w n w 5 lim n ! n } n 2 § 2 n } § 5 lim n } ( ˇ n 2 n w ) 2 } 5 } 1 2 } The series converges. (b) lim n ˇ n a w n w 5 lim n ! n 1 } § 2 n § n 2 § 1 } § 2 n § 5 lim n } 2 n n 2 1 } 5 } 1 2 } The series converges. (c) lim n , n odd ˇ n a w n w 5 lim n , n odd ! n } 2 n § n } § 5 lim n , n odd } ˇ n 2 n w } 5 } 1 2 } lim n , n even ˇ n a w n w 5 lim n , n even ! n } 2 1 § n } § 5 } 1 2 } Thus, lim n ˇ n a w n w 5 } 1 2 } , so the series converges. 58. (a) lim n ˇ n ) a w n ) w 5 lim n ! n } ) x § 2 4 § n 1 § ) n } § 5 } ) x 2 4 1 ) } The series converges absolutely if } ) x 2 4 1 ) } , 1, or 2 3 , x , 5.

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Business Calc Homework w answers_Part_79 - Section 9.5...

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