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Unformatted text preview: fh2825 Practice Exam 3 Radin (58415) 1 This printout should have 23 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Determine whether the sequence { a n } con verges or diverges when a n = ( 1) n parenleftbigg 4 n + 1 5 n + 8 parenrightbigg , and if it does, find its limit. 1. limit = 0 2. sequence diverges correct 3. limit = 1 8 4. limit = 4 5 5. limit = 4 5 Explanation: After division, 4 n + 1 5 n + 8 = 4 + 1 n 5 + 8 n . Now 1 n , 8 n 0 as n , so lim n 4 n + 1 5 n + 8 = 4 5 negationslash = 0 . Thus as n , the values of a n oscillate be tween values ever closer to 4 5 . Consequently, the sequence diverges . 002 10.0 points Determine whether the series 3 + 1 + 1 3 + 1 9 + is convergent or divergent, and if convergent, find its sum. 1. convergent with sum = 2 9 2. convergent with sum = 2 3. divergent 4. convergent with sum = 9 2 correct 5. convergent with sum = 1 2 Explanation: The series 3 + 1 + 1 3 + 1 9 + = summationdisplay n =1 a r n 1 is an infinite geometric series in which a = 3 and r = 1 3 . But such a series is (i) convergent with sum a 1 r when  r  < 1, (ii) divergent when  r  1 . Thus the given series is convergent with sum = 9 2 . 003 10.0 points Determine whether the infinite series summationdisplay n =1 2( n + 1) 2 n ( n + 2) converges or diverges, and if converges, find its sum. 1. converges with sum = 1 2. converges with sum = 2 3. converges with sum = 1 4 fh2825 Practice Exam 3 Radin (58415) 2 4. converges with sum = 1 2 5. diverges correct 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 = 2( n + 1) 2 n ( n + 2) = 2 n 2 + 4 n + 2 n 2 + 2 n . But then, lim n a n = 2 negationslash = 0 . Consequently, the Divergence Test says that the given series diverges . keywords: infinite series, Divergence Test, ra tional function 004 10.0 points To apply the root test to an infinite series n a n the value of = lim n  a n  1 /n has to be determined. Compute the value of for the series summationdisplay n = 1 parenleftbigg 2 n + 3 6 n parenrightbigg 2 n . 1. = 9 4 2. = 1 4 3. = 1 9 correct 4. = 1 2 5. = 1 3 Explanation: After division, 2 n + 3 6 n = 2 + 3 /n 6 , so  a n  1 /n = parenleftBig 2 + 3 /n 6 parenrightBig 2 . On the other hand, lim n 2 + 3 /n 6 = 1 3 . Consequently, = 1 9 . keywords: /* If you use any of these, fix the comment symbols. 005 10.0 points Determine whether the series summationdisplay k = 1 ( 1) k 1 cos parenleftBig 1 5 k parenrightBig is absolutely convergent, conditionally con vergent or divergent....
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 Spring '08
 RAdin

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