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L EXAM 3-solutions - Version 055 L EXAM 3 meth(54960 This...

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Version 055 – L EXAM 3 – meth – (54960) 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 = ln(3 n 4 ) ln(5 n 3 ) , and if it converges, find the limit. 1. converges with limit = 4 3 correct 2. converges with limit = 0 3. converges with limit = ln 3 ln 5 4. diverges 5. converges with limit = 3 5 Explanation: By properties of logs, ln(3 n 4 ) = ln 3 + 4 ln n , ln(5 n 3 ) = ln 5 + 3 ln n . Thus a n = ln 3 + 4 ln n ln 5 + 3 ln n = 4 + ln 3 ln n 3 + ln 5 ln n . On the other hand, lim n →∞ ln 3 ln n = lim n →∞ ln 5 ln n = 0 . Properties of limits thus ensure that the given sequence converges with limit = 4 3 . 002 10.0 points Determine if the sequence { a n } converges when a n = 5 + parenleftbigg - 10 7 π parenrightbigg n , and if it does, find its limit. 1. limit = 5 + 1 π 2. limit = 5 π 3. limit = 5 - 1 π 4. the sequence diverges 5. limit = 5 correct Explanation: It is known that x n -→ 0 as n whenever - 1 < x < 1. Now 10 < 7 π , so parenleftbigg - 10 7 π parenrightbigg n -→ 0 as n → ∞ . Consequently, the given sequence converges and has limit = 5 . 003 10.0 points Determine if the sequence { a n } converges, and if it does, find its limit when a n = 4 n + ( - 1) n 6 n + 5 . 1. converges with limit = 2 3 correct 2. sequence does not converge
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Version 055 – L EXAM 3 – meth – (54960) 2 3. converges with limit = 4 11 4. converges with limit = 1 2 5. converges with limit = 5 6 Explanation: After division by n we see that a n = 4 + ( - 1) n n 6 + 5 n . But ( - 1) n n , 5 n -→ 0 as n → ∞ , so a n 2 3 as n → ∞ . Conse- quently, the sequence converges and has limit = 2 3 . 004 10.0 points If the n th partial sum of an infinite series is S n = 3 n 2 - 1 5 n 2 + 1 , what is the sum of the series? 1. sum = 3 5 correct 2. sum = - 1 3. sum = - 1 5 4. series diverges 5. sum = 3 Explanation: By definition sum = lim n →∞ S n = lim n → ∞ parenleftBig 3 n 2 - 1 5 n 2 + 1 parenrightBig . Thus sum = 3 5 . 005 10.0 points Find the values of x for which the series summationdisplay n =1 x n 6 n converges, and then find the sum of the series for those values of x . 1. converges: - 6 < x < 6 , sum = 6 6 - x 2. converges: - 6 x < 6 , sum = 6 6 - x 3. converges: - 6 x 6 , sum = 6 6 - x 4. converges: - 6 < x 6 , sum = x 6 - x 5. converges: - 6 < x < 6 , sum = x 6 - x correct 6. converges: - 6 x 6 , sum = x 6 - x Explanation: The given series is a geometric series summationdisplay n =1 a r n 1 in which a = x 6 , r = x 6 . But such a geometric series (i) converges when | r | < 1, with sum a 1 - r , and (ii) diverges when | r | ≥ 1. Consequently, the given series converges when - 6 < x < 6 ,
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Version 055 – L EXAM 3 – meth – (54960) 3 and then has sum sum = x 6 - x . 006 10.0 points To apply the ratio test to the infinite series summationdisplay n a n , the value of λ = lim n → ∞ a n +1 a n has to be determined.
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