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# Exam 3 - Version 071 – Exam 3 – Gompf –(58370 1 This...

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Unformatted text preview: Version 071 – Exam 3 – Gompf – (58370) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Determine if the sequence { a n } converges when a n = 1 n ln parenleftbigg 3 2 n + 2 parenrightbigg , and if it does, find its limit. 1. the sequence diverges 2. limit = ln 3 2 3. limit = ln 3 4 4. limit = − ln 2 5. limit = 0 correct Explanation: After division by n we see that 3 2 n + 2 = 3 n 2 + 2 n , so by properties of logs, a n = 1 n ln 3 n − 1 n ln parenleftbigg 2 + 2 n parenrightbigg . But by known limits (or use L’Hospital), 1 n ln 3 n , 1 n ln parenleftbigg 2 + 2 n parenrightbigg −→ as n → ∞ . Consequently, the sequence { a n } converges and has limit = 0 . 002 10.0 points Determine if the sequence { a n } converges when a n = n 6 n ( n − 5) 6 n , and if it does, find its limit 1. limit = e − 5 6 2. sequence diverges 3. limit = e 5 6 4. limit = e − 30 5. limit = e 30 correct 6. limit = 1 Explanation: By the Laws of Exponents, a n = parenleftbigg n − 5 n parenrightbigg − 6 n = parenleftbigg 1 − 5 n parenrightbigg − 6 n = bracketleftBigparenleftBig 1 − 5 n parenrightBig n bracketrightBig − 6 . But parenleftBig 1 + x n parenrightBig n −→ e x as n → ∞ . Consequently, { a n } converges and has limit = ( e − 5 ) − 6 = e 30 . 003 10.0 points If the n th partial sum S n of an infinite series ∞ summationdisplay n = 1 a n is given by S n = 3 − n 5 n , find a n for n > 1. 1. a n = 3 parenleftbigg 6 n − 5 5 n parenrightbigg 2. a n = 3 parenleftbigg 4 n − 5 5 n parenrightbigg Version 071 – Exam 3 – Gompf – (58370) 2 3. a n = 4 n − 5 5 n correct 4. a n = 3 parenleftbigg n − 5 5 n − 1 parenrightbigg 5. a n = 6 n − 5 5 n 6. a n = n − 5 5 n − 1 Explanation: By definition, the n th partial sum of ∞ summationdisplay n = 1 a n is given by S n = a 1 + a 2 + ··· + a n . In particular, a n = braceleftbigg S n − S n − 1 , n > 1, S n , n = 1. Thus a n = S n − S n − 1 = n − 1 5 n − 1 − n 5 n = 5 ( n − 1) 5 n − n 5 n when n > 1. Consequently, a n = 4 n − 5 5 n for n > 1. 004 10.0 points Let f be a continuous, positive, decreasing function on [5 , ∞ ). Compare the values of the integral A = integraldisplay 19 5 f ( z ) dz and the series B = 19 summationdisplay n = 6 f ( n ) . 1. A > B correct 2. A = B 3. A < B Explanation: In the figure 5 6 7 8 9 . . . a 6 a 7 a 8 a 9 the bold line is the graph of f on [5 , ∞ ) and the areas of the rectangles the terms in the series ∞ summationdisplay n = 6 a n , a n = f ( n ) . Clearly from this figure we see that a 6 = f (6) < integraldisplay 6 5 f ( z ) dz, a 7 = f (7) < integraldisplay 7 6 f ( z ) dz , while a 8 = f (8) < integraldisplay 8 7 f ( z ) dz, a 9 = f (9) < integraldisplay 9 8 f ( z ) dz , and so on. Consequently, A > B . Version 071 – Exam 3 – Gompf – (58370) 3 keywords: Szyszko 005 10.0 points Which, if any, of the following statements are...
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Exam 3 - Version 071 – Exam 3 – Gompf –(58370 1 This...

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