116 - MATH 153 Selected Solutions for 11.6 Exercise 10:...

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Selected Solutions for § 11.6 Exercise 10 : Determine whether the series is absolutely convergent, condi- tionally convergent or divergent. X n =1 ( - 1) n n n 3 + 2 Solution : First, we can say that the series does not converge absolutely by comparison with n n 3 / 2 = 1 n 1 / 2 : lim n →∞ n n 3 +2 1 n 1 / 2 = lim n →∞ n · n 1 / 2 n 3 + 2 · 1 /n 3 / 2 1 /n 3 / 2 = lim n →∞ 1 p 1 + 2 /n 3 = 1 1 + 0 = 1 . Thus, since 1 /n 1 / 2 diverges, so does n n 3 +2 . The series does pass the Alternating Series Test, though; let b n = n n 3 +2 . Then b n 0 as n → ∞ . It remains to show that b n decreases. We know that d dx x x 3 + 2 = x 3 + 2 - x ( 1 2 3 x 2 x 3 +2 ) x 3 + 2 . As the denominator is positive for all x > 0, if we can show the numerator is negative we’ll be done. But x 3 + 2 - x ± 1 2 3 x 2 x 3 + 2 ² < 0 x 3 + 2 - 3 2 x 3 < 0 2 - 1 2 x 3 < 0 3 4 < x. So the
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116 - MATH 153 Selected Solutions for 11.6 Exercise 10:...

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