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Cal 2 Test 3

# Cal 2 Test 3 - Version 041 L EXAM 3 Henry(54974 This...

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Version 041 – L EXAM 3 – Henry – (54974) 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(4 n 5 ) ln(3 n 2 ) , and if it converges, find the limit. 1. converges with limit = ln 4 ln 3 2. diverges 3. converges with limit = 4 3 4. converges with limit = 5 2 correct 5. converges with limit = 0 Explanation: By properties of logs, ln(4 n 5 ) = ln 4 + 5 ln n , ln(3 n 2 ) = ln 3 + 2 ln n . Thus a n = ln 4 + 5 ln n ln 3 + 2 ln n = 5 + ln 4 ln n 2 + ln 3 ln n . On the other hand, lim n →∞ ln 4 ln n = lim n →∞ ln 3 ln n = 0 . Properties of limits thus ensure that the given sequence converges with limit = 5 2 . 002 10.0 points Determine if the sequence { a n } converges when a n = 5 + parenleftbigg - 8 7 π parenrightbigg n , and if it does, find its limit. 1. limit = 5 - 1 π 2. limit = 5 correct 3. the sequence diverges 4. limit = 5 π 5. limit = 5 + 1 π Explanation: It is known that x n -→ 0 as n whenever - 1 < x < 1. Now 8 < 7 π , so parenleftbigg - 8 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 = 7 n + ( - 1) n 4 n + 2 . 1. sequence does not converge 2. converges with limit = 7 6

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Version 041 – L EXAM 3 – Henry – (54974) 2 3. converges with limit = 3 2 4. converges with limit = 7 4 correct 5. converges with limit = 2 Explanation: After division by n we see that a n = 7 + ( - 1) n n 4 + 2 n . But ( - 1) n n , 2 n -→ 0 as n → ∞ , so a n 7 4 as n → ∞ . Conse- quently, the sequence converges and has limit = 7 4 . 004 10.0 points If the n th partial sum of an infinite series is S n = 3 n 2 - 2 n 2 + 3 , what is the sum of the series? 1. sum = - 2 3 2. series diverges 3. sum = 1 4. sum = 3 correct 5. sum = - 2 Explanation: By definition sum = lim n →∞ S n = lim n → ∞ parenleftBig 3 n 2 - 2 n 2 + 3 parenrightBig . Thus sum = 3 . 005 10.0 points Find the values of x for which the series summationdisplay n =1 x n 4 n converges, and then find the sum of the series for those values of x . 1. converges: - 4 < x < 4 , sum = x 4 - x correct 2. converges: - 4 < x < 4 , sum = 4 4 - x 3. converges: - 4 x 4 , sum = 4 4 - x 4. converges: - 4 x < 4 , sum = 4 4 - x 5. converges: - 4 x 4 , sum = x 4 - x 6. converges: - 4 < x 4 , sum = x 4 - x Explanation: The given series is a geometric series summationdisplay n =1 a r n 1 in which a = x 4 , r = x 4 . 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 - 4 < x < 4 ,
Version 041 – L EXAM 3 – Henry – (54974) 3 and then has sum sum = x 4 - 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. Compute λ for the series summationdisplay n =1 3 n n !

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Cal 2 Test 3 - Version 041 L EXAM 3 Henry(54974 This...

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