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Unformatted text preview: lawrence (cdl678) – Homework 4 – ODELL – (56280) 1 This printout should have 31 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Which, if any, of the following statements are true? A. If lim n →∞ a n = 0, then summationdisplay a n converges. B. If summationdisplay a n is divergent, then summationdisplay  a n  is divergent. C. The Ratio Test can be used to determine whether summationdisplay 1 /n 3 converges. 1. B and C only 2. A and B only 3. C only 4. all of them 5. A only 6. none of them 7. A and C only 8. B only correct Explanation: A. False: when a n = 1 /n , then lim n →∞ a n = 0, but ∞ summationdisplay n = 1 a n = ∞ summationdisplay n = 1 1 n diverges by the Integral Test. B. True: if summationdisplay  a n  were convergent, then summationdisplay a n would be absolutely convergent, hence convergent. C. False: when a n = 1 /n 3 , then vextendsingle vextendsingle vextendsingle vextendsingle a n +1 a n vextendsingle vextendsingle vextendsingle vextendsingle = n 3 ( n + 1) 3 −→ 1 as n → , ∞ , so the Ratio Test is inconclu sive. 002 10.0 points Which, if any, of the following statements are true? A. The Ratio Test can be used to deter mine whether the series summationdisplay n = ∞ 1 /n ! converges or diverges. B. The Root Test can be used to determine whether the series ∞ summationdisplay k = 1 k 3 + k 2 converges or diverges. 1. both of them 2. A only correct 3. B only 4. neither of them Explanation: A. True: when a n = 1 /n !, then vextendsingle vextendsingle vextendsingle vextendsingle a n +1 a n vextendsingle vextendsingle vextendsingle vextendsingle = 1 n + 1 −→ as n → ∞ , so ∑ a n is convergent by Ratio Test. B. False: when a k = k 3 + k 2 , then (  a k  ) 1 /k = k 1 /k / (1 + k 2 ) 1 /k −→ 1 lawrence (cdl678) – Homework 4 – ODELL – (56280) 2 so the Root Test is inconclusive. 003 10.0 points If ∑ n a n converges, which, if any, of the following statements are true: (A) summationdisplay n  a n  is convergent , (B) lim n →∞ a n = 0 . 1. B only correct 2. neither A nor B 3. A only 4. both A and B Explanation: (A) FALSE: set a n = ( − 1) n n . Then summationdisplay n  a n  = summationdisplay n 1 n , so by the pseries test with p = 1, the series summationdisplay  a n  diverges. On the other hand, summationdisplay n a n = summationdisplay n ( − 1) n n converges by the Alternating Series Test. (B) TRUE: to say that ∑ n a n converges is to say that the limit lim n →∞ S n of its partial sums S n = a 1 + a 2 + . . . + a n converges. But then lim n →∞ a n = lim n →∞ ( S n − S n − 1 ) = 0 ....
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 Spring '10
 odell
 Mathematical Series, Lawrence

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