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51. s 1 5 tan 2 1 1 2 tan 2 1 2 5 } p 4 } 2 tan 2 1 2 s 2 5 (tan 2 1 1 2 tan 2 1 2) 1 (tan 2 1 2 2 tan 2 1 3) 5 } p 4 } 2 tan 2 1 3 s 3 5 (tan 2 1 1 2 tan 2 1 2) 1 (tan 2 1 2 2 tan 2 1 3) 1 (tan 2 1 3 2 tan 2 1 4) 5 } p 4 } 2 tan 2 1 4 s n 5 } p 4 } 2 tan 2 1 ( n 1 1) S 5 lim n s n 5 } p 4 } 2 lim n tan 2 1 n 5 } p 4 } 2 } p 2 }52} p 4 } 52. } 1 2 1 x } 5 n 5 0 x n Differentiate: } (1 2 1 x ) 2 } 5 n 5 0 nx n 2 1 Multiply by x: } (1 2 x x ) 2 } 5 n 5 0 nx n Differentiate: } d d x } } (1 2 x x ) 2 }5 5 } (1 ( 2 1 2 x ) x 1 ) 3 2 x } 5 } (1 x 2 1 x 1 ) 3 } } (1 x 2 1 x 1 ) 3 } 5 n 5 0 n 2 x n 2 1 Multiply by x: } x (1 ( x 2 1 x 1 ) 3 ) } 5 n 5 0 n 2 x n Let x 5 } 1 2 } : 5 n 5 0 n 2 1 } 1 2 } 2 n 6 5 n 5 0 } 2 n 2 n } The sum is 6. Section 9.5 Testing Convergence at Endpoints (pp. 496–508) Exploration 1 The p -Series Test 1. We first note that the Integral Test applies to any series of the form } n 1 p } where p is positive. This is because the function f ( x ) 5 x 2 p is continuous and positive for all x . 0, and f 9 ( x ) 52 p ? x p 2 1 is negative for all x . 0. If p . 1: E 1 } x 1 p } dx 5 lim k E k 1 } x 1 p } dx 5 lim k 1 } 2 x 2 p p 1 1 1 1 } 42 = lim k 1 } 1 2 1 p } ? 1 } k p 1 2 1 } 2 1 22 5 0 1 } p 2 1 1 } (since p 2 1 . 0) 5 } p 2 1 1 },‘ . The series converges by the Intergral Test. 2. If 0 , p , 1: E 1 } x 1 p } dx 5 lim k E k 1 } x 1 p } dx 5 lim k 1 } 2 x 2 p p 1 1 1 1 } 5 lim k 1 } 1 2 1 p } ? ( k 1 2 p 2 1) 2 5‘ (since 1 2 p . 0). The series diverges by the Integral Test. If p # 0, the series diverges by the n th Term Test. This completes the proof for p , 1. 3. If p 5 1: E 1 } x 1 p } dx 5 lim k E k 1 } 1 x } dx 5 lim k 1 ln x 5 lim k ln k . The series diverges by the Integral Test. Exploration 2 The Maclaurin Series of a Strange Function 1. Since f ( n ) (0) 5 0 for all n , the Maclaurin Series for f has all zero coefficients! The series is simply n 5 0 0 ? x n 5 0. 2. The series converges (to 0) for all values of x . 3. Since f ( x ) 5 0 only at x 5 0, the only place that this series actually converges to its f -value is at x 5 0. k 1 k 1 k 1 } 1 2 } 1 } 3 2 } 2 } 1 } 1 2 } 2 3 (1 2 x ) 2 (1) 2 ( x )(2)(1 2 x )( 2 1) }}}} (1 2 x ) 4 386 Section 9.5

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Quick Review 9.5 1. Converges, since it is of the form E 1 } x 1 p } dx with p . 1 2. Diverges, limit comparison test with integral of } 1 x } 3. Diverges, comparison test with integral of } 1 x } 4. Converges, comparison test with integral of } x 2 2 } 5. Diverges, limit comparison test with integral of } ˇ 1 x w } 6. Yes, for N 5 0 7. Yes, for N 5 2 ˇ 2 w 8. No, neither positive nor decreasing for x . ˇ 3 w 9. No, oscillates 10. No, not positive for x \$ 1 Section 9.5 Exercises 1. Diverges by the Integral Test, since E 1 } x 1 5 1 } dx diverges. 2. Diverges because n 5 1 } ˇ 3 n w } 5 3 n 5 1 } n 1 1/2 } , which diverges by the p -series Test. 3. Diverges by the Direct Comparison Test, since } ln n n } . } 1 n } for n \$ 2 and n 5 2 } 1 n } diverges.
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