ReviewCalc3 - nav277 Review plus (exam 3) Odell (58340) 1...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: nav277 Review plus (exam 3) Odell (58340) 1 This print-out should have 23 questions. Multiple-choice 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 5 n + 6 6 n + 2 parenrightbigg , and if it does, find its limit. 1. limit = 5 6 2. limit = 5 6 3. limit = 0 4. limit = 3 5. sequence diverges correct Explanation: After division, 5 n + 6 6 n + 2 = 5 + 6 n 6 + 2 n . Now 6 n , 2 n 0 as n , so lim n 5 n + 6 6 n + 2 = 5 6 negationslash = 0 . Thus as n , the values of a n oscillate be- tween values ever closer to 5 6 . Consequently, the sequence diverges . 002 10.0 points Determine whether the series 1 + 3 + 9 + 27 + is convergent or divergent, and if convergent, find its sum. 1. convergent with sum = 9 2 2. divergent correct 3. convergent with sum = 1 2 4. convergent with sum = 2 5. convergent with sum = 2 9 Explanation: The series 1 + 3 + 9 + 27 + = summationdisplay n =1 a r n 1 is an infinite geometric series in which a = 1 and r = 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 divergent . 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. diverges correct 2. converges with sum = 1 3. converges with sum = 1 2 nav277 Review plus (exam 3) Odell (58340) 2 4. converges with sum = 1 8 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 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 6 n + 3 2 n parenrightbigg 2 n . 1. = 3 2. = 3 2 3. = 9 correct 4. = 1 4 5. = 9 4 Explanation: After division, 6 n + 3 2 n = 6 + 3 /n 2 , so | a n | 1 /n = parenleftBig 6 + 3 /n 2 parenrightBig 2 . On the other hand, lim n 6 + 3 /n 2 = 3 . Consequently, = 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....
View Full Document

This note was uploaded on 04/15/2011 for the course MATH 408 L taught by Professor Zheng during the Fall '10 term at University of Texas at Austin.

Page1 / 12

ReviewCalc3 - nav277 Review plus (exam 3) Odell (58340) 1...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online