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int cal exam 3

int cal exam 3 - Version 060 EXAM 3 Radin(57410 This...

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Version 060 – EXAM 3 – Radin – (57410) 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 whether the sequence { a n } con- verges or diverges when a n = ( 1) n 1 n n 2 + 2 , and if it converges, find the limit. 1. converges with limit = 0 correct 2. converges with limit = 1 2 3. converges with limit = 1 2 4. converges with limit = 2 5. converges with limit = 2 6. sequence diverges Explanation: After division, a n = ( 1) n 1 n n 2 + 2 = ( 1) n 1 n + 2 n . Consequently, 0 ≤ | a n | = 1 n + 2 n 1 n . But 1 /n 0 as n → ∞ , so by the Squeeze theorem, lim n → ∞ | a n | = 0 . But −| a n | ≤ a n ≤ | a n | , so by the Squeeze theorem again the given sequence { a n } converges and has limit = 0 . keywords: 002 10.0 points Which of the following sequences converge? A . braceleftbigg 5 e n 4 + e n bracerightbigg B . braceleftbigg e n + 5 3 n + 2 bracerightbigg 1. A only correct 2. neither of them 3. B only 4. both A and B Explanation: A . After simplification, 5 e n 4 + e n = 5 4 e n + 1 . Thus 5 e n 4 + e n −→ 5 as n → ∞ since e x 0 as x → ∞ . B . To test for convergence of the sequence braceleftbigg e n + 5 3 n + 2 bracerightbigg we use L’Hospital’s Rule with f ( x ) = e x + 5 , g ( x ) = 3 x + 2 since f ( x ) −→ ∞ , g ( x ) −→ ∞ as x → ∞ . Thus lim n → ∞ e n 3 n + 2 = lim x → ∞ f ( x ) g ( x ) .

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Version 060 – EXAM 3 – Radin – (57410) 2 But f ( x ) g ( x ) = e x 3 −→ ∞ as n → ∞ . Consequently, only A converges . 003 10.0 points Determine if the sequence { a n } converges, and if it does, find its limit when a n = parenleftbigg 1 3 4 n parenrightbigg 5 n . 1. limit = e 15 4 correct 2. limit = e 3 4 3. limit = e 15 4 4. limit = 1 5. the sequence diverges 6. limit = e 3 4 7. limit = e 12 5 Explanation: We know that lim n → ∞ parenleftBig 1 x n parenrightBig n = e x . Now a n = bracketleftbiggparenleftbigg 1 3 4 n parenrightbigg n bracketrightbigg 5 . By the properties of limits, therefore, lim n → ∞ a n = bracketleftbigg lim n → ∞ parenleftbigg 1 3 4 n parenrightbigg n bracketrightbigg 5 , so the sequence converges and has limit = parenleftBig e 3 4 parenrightBig 5 = e 15 4 . 004 10.0 points Determine whether the infinite series summationdisplay n =1 ln parenleftbigg 4 n 5 n + 4 parenrightbigg is convergent or divergent, and if convergent, find its sum. 1. converges with sum = ln parenleftbigg 4 5 parenrightbigg 2. converges with sum = ln parenleftbigg 4 9 parenrightbigg 3. converges with sum = ln parenleftbigg 5 4 parenrightbigg 4. diverges correct 5. converges with sum = ln parenleftbigg 9 4 parenrightbigg Explanation: An infinite series summationdisplay a n diverges when lim n →∞ a n negationslash = 0 . For the given series a n = ln parenleftbigg 4 n 5 n + 4 parenrightbigg .
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