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Pre-Calc Homework Solutions 367

# Pre-Calc Homework Solutions 367 - Section 9.1 22 Since tan...

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15. Converges; since } p e } < 0.865 , 1, n 5 1 1 } p e } 2 n 5 5 } p p 2 e } 16. Converges; n 5 0 } 6 5 n 1 n 1 } 5 n 5 0 } 1 6 } 1 } 5 6 } 2 n 5 5 1 17. Since n 5 0 2 n x n 5 n 5 0 (2 x ) n , the series converges when ) 2 x ) , 1 and the interval of convergence is 1 2 } 1 2 } , } 1 2 } 2 . Since the sum of the series is } 1 2 1 2 x } , the series represents the function f ( x ) 5 } 1 2 1 2 x } , 2 } 1 2 } , x , } 1 2 } . 18. Since n 5 0 ( 2 1) n ( x 1 1) n 5 n 5 0 [ 2 ( x 1 1)] n , the series converges when ) 2 ( x 1 1) ) , 1 and the interval of convergence is ( 2 2, 0). Since the sum of the series is } 1 2 [ 2 1 ( x 1 1)] } 5 } x 1 1 2 } , the series represents the function f ( x ) 5 } x 1 1 2 } , 2 2 , x , 0. 19. Since n 5 0 1 2 } 1 2 } 2 n ( x 2 3) n 5 n 5 0 1 } 3 2 2 x } 2 n , the series converges when ) } 3 2 2 x } ) , 1 and the interval of convergence is (1, 5). Since the sum of the series is 5 } x 2 2 1 } . the series represents the function f ( x ) 5 } x 2 2 1 } , 1 , x , 5. 20. For n 5 0 3 1 } x 2 2 1 } 2 n , the series converges when ) } x 2 2 1 } ) , 1 and the interval of convergence is ( 2 1, 3).
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