EX_3[1] - Garza, Ernesto – Exam 3 – Due: May 1 2007,...

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Garza, Ernesto – Exam 3 – Due: May 1 2007, 11:00 pm – Inst: JEGilbert 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 n + 1 ¾ , B. 5 n 2 3 n + 4 ¾ . 1. B only 2. A only 3. neither oF them correct 4. both oF them Explanation: A. AFter simplifcation, 4 n 2 n + 4 = 2 n 1 + 4 2 n . Consequently, 4 n 2 n + 4 -→ ∞ as n → ∞ , so the sequence does not converge. B. AFter simplifcation, 5 n 2 3 n + 4 = 5 n 3 + 4 n . Consequently, 5 n 2 3 n + 4 -→ ∞ 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 = 3 n 2 - 5 4 n 2 + 1 , what is the sum oF the series? 1. sum = 3 4 correct 2. sum = 15 16 3. sum = 1 4. sum = 13 16 5. sum = 7 8 Explanation: By defnition sum = lim n →∞ S n = lim n →∞ 3 n 2 - 5 4 n 2 + 1 · . Thus sum = 3 4 . 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) X n | a n | is convergent , (B) lim n →∞ a n = 0 . 1. A only 2. B only correct 3. neither A nor B
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2 4. both A and B Explanation: (A) 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. (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 . keywords: 004 (part 1 of 1) 10 points Determine the convergence or divergence of the series ( A ) 1 + 1 8 + 1 27 + 1 64 + 1 125 + ..., and ( B ) X k = 1 k 3 e - k 4 . 1. both series convergent correct 2. both series divergent 3. A divergent, B convergent 4. A convergent, B divergent Explanation: ( A ) The given series has the form 1 + 1 8 + 1 27 + 1 64 + 1 125 + ... = X n =1 1 n 3 . This is a p -series with p = 3 > 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 3 e - x 4 . Note ±rst that f is continuous and positive on [1 , ); in addition, since f 0 ( x ) = e - x 4 (3 x 2 - 4 x 6 ) < 0 for x > 1, f is decreasing on [1 , ). Thus we can use the Integral Test. Now, by substitu- tion, Z t 1 x 3 e - x 4 dx = - 1 4 e - x 4 t 1 , and so Z 1 x 3 e - x 4 dx = 1 4 e . Since the integral converges, the series con- verges. This could also be established using the Ratio Test. keywords:
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EX_3[1] - Garza, Ernesto – Exam 3 – Due: May 1 2007,...

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