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Unformatted text preview: sl7433 – Review 3 – Radin – (58415) 1 This printout should have 20 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Find the n th term, a n , of an infinite series ∑ ∞ n =1 a n when the n th partial sum, S n , of the series is given by S n = 3 n n + 1 . 1. a n = 5 2 n 2. a n = 3 2 n 2 3. a n = 3 n ( n + 1) correct 4. a n = 5 2 n 2 5. a n = 3 2 n 6. a n = 5 n ( n + 1) Explanation: Since S n = a 1 + a 2 + ··· + a n , we see that a 1 = S 1 , a n = S n − S n − 1 ( n > 1) . But S n = 3 n n + 1 = 3 − 3 n + 1 . Thus a 1 = 3 2 , while a n = 3 n − 3 n + 1 , ( n > 1) . Consequently, a n = 3 n − 3 n + 1 = 3 n ( n + 1) for all n . 002 10.0 points Let h be a continuous, positive, decreasing function on [3 , ∞ ). Compare the values of the integral A = integraldisplay 18 3 h ( t ) dt and the series B = 17 summationdisplay n = 3 h ( n ) . 1. A = B 2. A > B 3. A < B correct Explanation: In the figure 3 4 5 6 7 . . . a 3 a 4 a 5 a 6 the bold line is the graph of h on [3 , ∞ ) and the areas of the rectangles the terms in the series ∞ summationdisplay n = 3 a n , a n = h ( n ) . Clearly from this figure we see that h (3) > integraldisplay 4 3 h ( t ) dt, h (4) > integraldisplay 5 4 h ( t ) dt , sl7433 – Review 3 – Radin – (58415) 2 while h (5) > integraldisplay 6 5 h ( t ) dt, h (6) > integraldisplay 7 6 h ( t ) dt , and so on. Consequently, A < B . keywords: 003 10.0 points To apply the root test to an infinite series ∑ k a k , the value of ρ = lim k →∞  a k  1 /k has to be determined. Compute the value of ρ for the series ∞ summationdisplay k =1 3 k k (ln k + 6) k . 1. ρ = 0 correct 2. ρ = 6 3. ρ = ∞ 4. ρ = 3 5. ρ = 18 Explanation: For the given series ( a k ) 1 /k = 3 1 /k parenleftbigg ln k + 6 k parenrightbigg = 3 1 /k parenleftbigg ln k k + 6 k parenrightbigg . But 3 1 /k −→ 1 , ln k k −→ as k → ∞ . Consequently, ρ = 0 . 004 10.0 points Determine whether the series ∞ summationdisplay n =0 parenleftbigg 2 5 parenrightbigg n/ 2 is convergent or divergent, and if convergent, find its sum. 1. convergent with sum = √ 5 − √ 2 √ 2 2. convergent with sum = √ 5 − √ 2 √ 5 3. convergent with sum = √ 2 √ 5 − √ 2 4. convergent with sum = √ 5 √ 5 − √ 2 cor rect 5. divergent Explanation: The infinite series ∞ summationdisplay n =0 parenleftbigg 2 5 parenrightbigg n/ 2 is an infinite geometric series ∑ ∞ n =0 ar n with a = 1 and r = √ 2 / √ 5. But ∑ ∞ n =0 ar n is (i) convergent with sum a 1 − r when  r  < 1, and (ii) divergent when  r  ≥ 1 . So the given series is convergent with sum = √ 5 √ 5 − √ 2 ....
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This note was uploaded on 03/29/2009 for the course M 408L taught by Professor Radin during the Spring '08 term at University of Texas.
 Spring '08
 RAdin
 Calculus

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