solution_pdf2 - suleimenov (bs26835) – HW 02 – rusin...

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Unformatted text preview: suleimenov (bs26835) – HW 02 – rusin – (55565) 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 if the sequence { a n } converges when a n = 1 n ln parenleftbigg 5 2 n + 4 parenrightbigg , and if it does, find its limit. 1. limit = ln 5 6 2. limit = ln 5 2 3. limit = − ln 2 4. limit = 0 correct 5. the sequence diverges Explanation: After division by n we see that 5 2 n + 4 = 5 n 2 + 4 n , so by properties of logs, a n = 1 n ln 5 n − 1 n ln parenleftbigg 2 + 4 n parenrightbigg . But by known limits (or use L’Hospital), 1 n ln 5 n , 1 n ln parenleftbigg 2 + 4 n parenrightbigg −→ as n → ∞ . Consequently, the sequence { a n } converges and has limit = 0 . 002 10.0 points Determine if { a n } converges when a n = 5 √ n 3 − 2 √ n 2 , and if it does, find its limit. 1. limit = + 2 3 2. limit = 2 3. limit = 5 4. sequence diverges correct 5. limit = 3 Explanation: After simplification, a n = 5 n √ n − 2 n = n ( 5 √ n − 2 ) , so a n → ∞ as n → ∞ . But then the sequence is unbounded, and hence diverges . 003 10.0 points Determine if the sequence { a n } converges, and if it does, find its limit when a n = parenleftBig n + 1 n − 1 parenrightBig − 3 n . 1. limit = 1 2. does not converge 3. limit = e 3 4. limit = e 6 5. limit = e − 6 correct 6. limit = e − 3 Explanation: By the Laws of Exponents, a n = parenleftBig n − 1 n + 1 parenrightBig 3 n = parenleftBig 1 − 1 n parenrightBig 3 n parenleftBig 1 + 1 n parenrightBig 3 n . suleimenov (bs26835) – HW 02 – rusin – (55565) 2 But lim n →∞ parenleftBig 1 + x n parenrightBig n = e x . Consequently, by Properties of Limits the given limit exists and limit = e − 6 . 004 10.0 points Determine whether the sequence { a n } con- verges or diverges when a n = ( − 1) n − 1 n n 2 + 5 , and if it converges, find the limit. 1. converges with limit = − 1 5 2. sequence diverges 3. converges with limit = − 5 4. converges with limit = 0 correct 5. converges with limit = 1 5 6. converges with limit = 5 Explanation: After division, a n = ( − 1) n − 1 n n 2 + 5 = ( − 1) n − 1 n + 5 n . Consequently, ≤ | a n | = 1 n + 5 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: 005 10.0 points Determine if the sequence { a n } converges when a n = (2 n − 1)! (2 n + 1)! , and if it converges, find the limit. 1. converges with limit = 0 correct 2. converges with limit = 1 3. converges with limit = 1 4 4. converges with limit = 4 5. does not converge Explanation: By definition, m ! is the product m ! = 1 . 2 . 3 . . . . .m of the first m positive integers. When m = 2 n − 1, therefore, (2 n − 1)! = 1 . 2 . 3 . . . . (2 n − 1) , while (2 n + 1)! = 1 . 2 . 3...
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This note was uploaded on 03/20/2011 for the course M 408 C taught by Professor Treisman during the Spring '07 term at University of Texas at Austin.

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solution_pdf2 - suleimenov (bs26835) – HW 02 – rusin...

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