Andrew HW 12 - rabinovich(ahr454 HW12 um(55915 This print-out should have 20 questions Multiple-choice questions may continue on the next column or page

# Andrew HW 12 - rabinovich(ahr454 HW12 um(55915 This...

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rabinovich (ahr454) – HW12 – um – (55915) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Which one of the following series is conver- gent? 1. n = 1 ( - 1) n - 1 1 + n correct 2. n = 1 ( - 1) 2 n 5 4 + n 3. n = 1 4 5 + n 4. n = 1 ( - 1) 3 5 4 + n 5. n = 1 ( - 1) n - 1 1 + n 5 + n Explanation: Since n = 1 ( - 1) 3 5 4 + n = - n = 1 5 4 + n , use of the Limit Comparison and p -series Tests with p = 1 2 shows that this series is divergent. Similarly, since n = 1 ( - 1) 2 n 5 4 + n = n = 1 5 4 + n , the same argument shows that this series as well as n = 1 4 5 + n is divergent. On the other hand, by the Divergence Test, the series n = 1 ( - 1) n - 1 1 + n 5 + n is divergent because lim n → ∞ ( - 1) n - 1 1 + n 5 + n = 0 . This leaves only the series n = 1 ( - 1) n - 1 1 + n . To see that this series is convergent, set b n = 1 1 + n . Then (i) b n +1 b n , (ii) lim n → ∞ b n = 0 . Consequently, by the Alternating Series Test, the series n = 1 ( - 1) n - 1 1 + n is convergent. 002 (part 1 of 3) 10.0 points Decide whether the series n = 1 ( - 1) n - 1 3 n - 1 converges or diverges. 1. diverges 2. converges correct Explanation: The given series has the form n = 1 ( - 1) n - 1 f ( n ) with f defined by f ( x ) = 1 3 x - 1 .
rabinovich (ahr454) – HW12 – um – (55915) 2 Now f ( n ) = 1 3 n - 1 > 1 3 n + 2 = f ( n + 1) ; on the other hand, lim n → ∞ 1 3 n - 1 = 0 . Consequently, by the Alternating Series Test, the given series converges . 003 (part 2 of 3) 10.0 points Decide whether the series n = 1 ( - 1) n - 1 3 n 2 + n converges or diverges. 1. converges 2. diverges correct Explanation: The given series has the form n = 1 ( - 1) n - 1 f ( n ) with f defined by f ( x ) = 3 x 2 + x . Now lim x → ∞ 3 x 2 + x = 3 . Consequently, by the Divergence Test, the given series diverges . 004 (part 3 of 3) 10.0 points Decide whether the series n = 1 ( - 1) n - 1 sin 1 n converges or diverges. 1. converges correct 2. diverges Explanation: The given series has the form n = 1 ( - 1) n - 1 f ( n ) with f defined by f ( x ) = sin 1 x . Now by the Chain Rule f ( x ) = - 1 x 2 cos 1 x , so f ( x ) < 0 for large, positive x . Thus f ( n ) = sin 1 n > sin 1 n + 1 = f ( n + 1) , while lim x → ∞ f ( x ) = sin(0) = 0 . Consequently, by the Alternating Series Test, the given series converges . 005 10.0 points Determine whether the series n = 0 n n 2 + 5 cos(2 n π ) converges or diverges.
rabinovich (ahr454) – HW12 – um – (55915) 3 1. series converges 2. series diverges correct Explanation: Since cos(2 n π ) = 1, the given series can be rewritten as the series n = 0 n n 2 + 5 = n = 0 f ( n ) of positive, non-alternating terms where f ( x ) = x x 2 + 5 . Now lim x → ∞ xf ( x ) = lim x → ∞ x 2 x 2 + 5 = 1 . Thus by the Limit Comparison Test, the given series converges if and only if the series n = 1 1 n converges. But, by applying the p -series Test with p = 1, we see that this last series di- verges. Consequently, the given series diverges .

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• Fall '11
• Gramlich
• Accounting, Mathematical Series, lim

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