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Unformatted text preview: Rehman (aar638) Makeup Exam 3 sachse (56620) 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 Determine if the sequence { a n } converges, and if it does, find its limit when a n = parenleftBig n + 1 n + 5 parenrightBig n . 1. limit = 1 2. limit = e 1 5 3. limit = e 6 4. limit = e 4 correct 5. does not converge Explanation: By the Laws of Exponents, a n = ( n + 1) n ( n + 5) n = parenleftBig 1 + 1 n parenrightBig n parenleftBig 1 + 5 n parenrightBig n . But lim n parenleftBig 1 + x n parenrightBig n = e x . Consequently, by Properties of Limits the given limit exists and limit = e 4 . 002 10.0 points Determine whether the sequence { a n } con verges or diverges when a n = ( 1) n parenleftbigg 3 n 7 n + 8 parenrightbigg , and if it does, find its limit. 1. sequence diverges correct 2. limit = 3 7 3. limit = 0 4. limit = 3 8 5. limit = 3 8 6. limit = 3 7 Explanation: After division, 3 n 7 n + 8 = 3 7 + 8 n . Now 8 n 0 as n , so lim n 3 n 7 n + 8 = 3 7 negationslash = 0 . Thus as n , the values of a n oscillate be tween values ever closer to 3 7 . Consequently, the sequence diverges . 003 10.0 points If the n th partial sum of n =1 a n is given by S n = 4 n + 1 n + 2 , what is a n when n 2? 1. a n = 9 ( n + 2)( n + 3) 2. a n = 9 ( n + 2)( n + 1) 3. a n = 7 n ( n + 2) 4. a n = 7 ( n + 2)( n + 3) Rehman (aar638) Makeup Exam 3 sachse (56620) 2 5. a n = 7 ( n + 2)( n + 1) correct 6. a n = 9 n ( n + 2) Explanation: By definition S n = n summationdisplay k 1 a n = a 1 + a 2 + . . . + a n . Thus, for n 2, a n = S n S n 1 = 4 n + 1 n + 2 4( n 1) + 1 ( n 1) + 2 . Consequently, a n = 7 ( n + 2)( n + 1) . 004 10.0 points Determine whether the infinite series summationdisplay n =1 ( n + 1) 2 n ( n + 2) converges or diverges, and if converges, find its sum. 1. converges with sum = 1 2. converges with sum = 1 8 3. converges with sum = 1 2 4. diverges correct 5. converges with sum = 1 4 Explanation: By the Divergence Test, an infinite series n a n diverges when lim n a n negationslash = 0 . Now, for the given series, a n = ( n + 1) 2 n ( n + 2) = n 2 + 2 n + 1 n 2 + 2 n . But then, lim n a n = 1 negationslash = 0 . Consequently, the Divergence Test says that the given series diverges . keywords: infinite series, Divergence Test, ra tional function 005 10.0 points Find the sum of the series summationdisplay n = 0 (2 x 5) n 3 n for those values of x for which it converges. 1. sum = 3 8 2 x correct 2. sum = 3 2 2 x 3. sum = 8 2 x 3 4. sum = 2 + 2 x 3 5. sum = 3 8 + 2 x 6. sum = 3 2 + 2 x Explanation: When the geometric series n =0 ar n con verges it has sum = a 1 r . Rehman (aar638) Makeup Exam 3 sachse (56620) 3 Now for the given series, a = 1 , r = 2 x...
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 Fall '09
 GOGOLEV
 Calculus

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