exam3'06 - Bryson, Spencer Exam 3 Due: May 1 2007, 11:00 pm...

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Bryson, Spencer – Exam 3 – Due: May 1 2007, 11:00 pm – Inst: Mar Gonzalez 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. 001 (part 1 oF 1) 10 points Which oF the Following sequences converge? A. 4 n 2 5 n 2 + 3 ¾ , B. 4 n 2 n + 1 ¾ . 1. both oF them 2. neither oF them 3. A only correct 4. B only Explanation: A. AFter simplifcation, 4 n 2 5 n 2 + 3 = 4 5 + 3 n 2 . Consequently, 4 n 2 5 n 2 + 3 -→ 4 5 as n → ∞ , so the sequence converges. B. AFter simplifcation, 4 n 2 n + 3 = 2 n 1 + 3 2 n . Consequently, 4 n 2 n + 3 -→ ∞ as n → ∞ , so the sequence does not converge. keywords: 002 (part 1 oF 1) 10 points IF the n th partial sum oF an infnite series is S n = 5 n 2 - 2 2 n 2 + 3 , what is the sum oF the series? 1. sum = 19 8 2. sum = 5 2 correct 3. sum = 21 8 4. sum = 9 4 5. sum = 17 8 Explanation: By defnition sum = lim n →∞ S n = lim n →∞ 5 n 2 - 2 2 n 2 + 3 · . Thus sum = 5 2 . keywords: partial sum, defnition oF series 003 (part 1 oF 1) 10 points IF n a n converges, which, iF any, oF the Following statements are true: (A) lim n →∞ a n = 0 , (B) X n | a n | is convergent . 1. A only correct 2. B only 3. neither A nor B
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2 4. both A and B Explanation: (A) 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 . (B) FALSE: set a n = ( - 1) n n . Then X n | a n | = X n 1 n , so by the p -series test with p = 1, the series X | a n | diverges. On the other hand, X n a n = X n ( - 1) n n converges by the Alternating Series Test. keywords: 004 (part 1 of 1) 10 points Determine the convergence or divergence of the series ( A ) 1 + 1 4 + 1 9 + 1 16 + 1 25 + ..., and ( B ) X k = 1 k 2 e - k 3 . 1. both series convergent correct 2. A convergent, B divergent 3. both series divergent 4. A divergent, B convergent Explanation: ( A ) The given series has the form 1 + 1 4 + 1 9 + 1 16 + 1 25 + ... = X n =1 1 n 2 . This is a p -series with p = 2 > 1, so the series converges. ( B ) The given series has the form X k = 1 f ( k ) with f de±ned by f ( x ) = x 2 e - x 3 . Note ±rst that f is continuous and positive on [1 , ); in addition, since f 0 ( x ) = e - x 3 (2 x 1 - 3 x 4 ) < 0 for x > 1, f is decreasing on [1 , ). Thus we can use the Integral Test. Now, by substitu- tion, Z t 1 x 2 e - x 3 dx = - 1 3 e - x 3 t 1 , and so Z 1 x 2 e - x 3 dx = 1 3 e . Since the integral converges, the series con- verges. This could also be established using the Ratio Test. keywords:
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exam3'06 - Bryson, Spencer Exam 3 Due: May 1 2007, 11:00 pm...

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