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Unformatted text preview: gompf Review exam 3 Gompf (58370) 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 3 n + 6 7 n + 5 parenrightbigg , and if it does, find its limit. 1. limit = 0 2. limit = 3 7 3. limit = 6 5 4. limit = 3 7 5. sequence diverges correct Explanation: After division, 3 n + 6 7 n + 5 = 3 + 6 n 7 + 5 n . Now 6 n , 5 n 0 as n , so lim n 3 n + 6 7 n + 5 = 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 . 002 10.0 points Determine whether the series 4 + 1 + 1 4 + 1 16 + is convergent or divergent, and if convergent, find its sum. 1. convergent with sum = 1 3 2. convergent with sum = 16 3 correct 3. convergent with sum = 3 4. convergent with sum = 3 16 5. divergent Explanation: The series 4 + 1 + 1 4 + 1 16 + = summationdisplay n =1 a r n 1 is an infinite geometric series in which a = 4 and r = 1 4 . 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 = 16 3 . 003 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 8 2. diverges correct 3. converges with sum = 1 2 gompf Review exam 3 Gompf (58370) 2 4. converges with sum = 1 4 5. converges with sum = 1 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 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 3 n + 1 5 n parenrightbigg 2 n . 1. = 3 5 2. = 1 9 3. = 9 25 correct 4. = 1 25 5. = 1 5 Explanation: After division, 3 n + 1 5 n = 3 + 1 /n 5 , so  a n  1 /n = parenleftBig 3 + 1 /n 5 parenrightBig 2 . On the other hand, lim n 3 + 1 /n 5 = 3 5 . Consequently, = 9 25 . keywords: /* If you use any of these, fix the comment symbols. 005 10.0 points Determine whether the series summationdisplay n =1 ( 1) n 1 cos parenleftBig 1 2 n parenrightBig is absolutely convergent, conditionally con vergent or divergent....
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 Spring '08
 Cepparo

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