11.7 Strategies for Testing Series-solutions

11.7 Strategies for Testing Series-solutions - im(gi768...

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im (gi768) – 11.7 Strategies for Testing Series – sadun – (53088) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Since there’s no preclass due on Tuesday night, I’ll give you until midnight to finish the postclass. 001 10.0points Determine whether the series summationdisplay m =1 ( - 1) m 1 m 5 m + 8 is absolutely convergent, conditionally con- vergent or divergent. 1. divergent 2. conditionally convergent correct 3. absolutely convergent Explanation: Set a m = m 5 m + 8 , b m = 1 m . Then given series is of the form summationdisplay m =1 ( - 1) m 1 a m , while lim m → ∞ a m b m = lim m → ∞ m 5 m + 8 = 1 5 . But by the p -series test with p = 1 / 2, the series summationdisplay m b m diverges, so the series summationdisplay m =1 ( - 1) m 1 m 5 m + 8 is not absolutely convergent. On the other hand, if f ( x ) = x 5 x + 8 , then f is continuous and positive on (0 , ); furthermore, lim x → ∞ f ( x ) = lim x → ∞ x 5 x + 8 = 0 and f ( x ) is ultimately decreasing because f ( x ) = 8 - 5 x 2 x (5 x + 8) 2 < 0 for all x > 8 / 5. Consequently, by the Alter- nating Series test, the series summationdisplay m =1 ( - 1) m 1 f ( m ) is conditionally convergent, which in turn means that the given series is conditionally convergent . keywords: absolutely convergent, condition- ally convergent, divergent, p -series test, Al- ternating Series test, 002 10.0points Determine whether the series summationdisplay m =5 m 2 - 25 m 2 + 5 m is convergent or divergent. 1. the series is divergent correct 2. the series is convergent Explanation: Set a m = m 2 - 25 m 2 + 5 m . Then lim n → ∞ a m = lim m → ∞ m 2 - 25 m 2 + 5 m = lim m → ∞ 1 - 25 m 2 1 + 5 m = 1 .
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im (gi768) – 11.7 Strategies for Testing Series – sadun – (53088) 2 By the Divergence Test, therefore, the series is divergent .
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