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Unformatted text preview: Miller, Kierste Exam 3 Due: May 1 2007, 11:00 pm Inst: Gary Berg 1 This printout should have 18 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Which of the following sequences converge? A. 2 n 2 n + 1 , B. 3 n 2 5 n + 1 . 1. both of them 2. neither of them 3. B only 4. A only correct Explanation: A. After simplification, 2 n 2 n + 1 = 1 1 + 2 n . Consequently, 2 n 2 n + 1 as n , so the sequence converges. B. After simplification, 3 n 2 5 n + 1 = 3 n 5 + 1 n . Consequently, 3 n 2 5 n + 1 as n , so the sequence does not converge. keywords: 002 (part 1 of 1) 10 points If the n th partial sum of an infinite series is S n = 3 n 2 5 n 2 + 2 , what is the sum of the series? 1. sum = 3 correct 2. sum = 5 2 3. sum = 9 4 4. sum = 13 4 5. sum = 11 4 Explanation: By definition sum = lim n S n = lim n 3 n 2 5 n 2 + 2 . Thus sum = 3 . keywords: partial sum, definition of series 003 (part 1 of 1) 10 points If lim n a n = 0, which, if any, of the following statements are true: (A) X n a n is convergent , (B) lim n ( a n ) 2 = 0 . 1. both A and B 2. A only 3. B only correct 4. neither A nor B Explanation: Miller, Kierste Exam 3 Due: May 1 2007, 11:00 pm Inst: Gary Berg 2 (A) FALSE: when a n = 1 /n , then lim n a n = 0, but X n = 1 a n = X n = 1 1 n diverges by the Integral Test. (B) TRUE: by Properties of Limits, lim n ( a n ) 2 = lim n a n 2 = 0 . 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 divergent 2. both series convergent correct 3. A convergent, B 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 pseries with p = 2 > 1, so the series converges. ( B ) The given series has the form X k = 1 f ( k ) with f defined by f ( x ) = x 2 e x 3 . Note first that f is continuous and positive on [1 , ); in addition, since f ( x ) = e x 3 (2 x 1 3 x 4 ) < 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: 005 (part 1 of 1) 10 points Determine which, if any, of the following series diverge. ( A ) X n = 1 (5 n ) n n !...
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This note was uploaded on 05/08/2008 for the course M 408 L taught by Professor Cepparo during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Cepparo

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