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University of Texas M 408D Rusin HW 02

# University of Texas M 408D Rusin HW 02 - byrne(cmb3744 HW...

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byrne (cmb3744) – 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 10.0points Determine if the sequence { a n } converges when a n = 1 n ln parenleftbigg 5 6 n + 6 parenrightbigg , and if it does, find its limit. 1. limit = ln 5 12 2. limit = ln 5 6 3. limit = 0 correct 4. the sequence diverges 5. limit = ln 6 Explanation: After division by n we see that 5 6 n + 6 = 5 n 6 + 6 n , so by properties of logs, a n = 1 n ln 5 n 1 n ln parenleftbigg 6 + 6 n parenrightbigg . But by known limits (or use L’Hospital), 1 n ln 5 n , 1 n ln parenleftbigg 6 + 6 n parenrightbigg −→ 0 as n → ∞ . Consequently, the sequence { a n } converges and has limit = 0 . 002 10.0points Determine if { a n } converges when a n = 2 n 3 6 n 2 , and if it does, find its limit. 1. limit = 2 2. limit = 4 3. sequence diverges correct 4. limit = 7 3 5. limit = 6 Explanation: After simplification, a n = 2 n n 6 n = n ( 2 n 6 ) , so a n → ∞ as n → ∞ . But then the sequence is unbounded, and hence diverges . 003 10.0points Determine if the sequence { a n } converges, and if it does, find its limit when a n = parenleftBig n 1 n + 1 parenrightBig 4 n . 1. limit = e 8 correct 2. limit = e 4 3. limit = e 8 4. limit = e 4 5. limit = 1 6. does not converge Explanation: By the Laws of Exponents, a n = parenleftBig n + 1 n 1 parenrightBig 4 n = parenleftBig 1 + 1 n parenrightBig 4 n parenleftBig 1 1 n parenrightBig 4 n .

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byrne (cmb3744) – 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 8 . 004 10.0points Determine whether the sequence { a n } con- verges or diverges when a n = ( 1) n 1 n n 2 + 4 , and if it converges, find the limit. 1. converges with limit = 0 correct 2. converges with limit = 1 4 3. converges with limit = 4 4. sequence diverges 5. converges with limit = 4 6. converges with limit = 1 4 Explanation: After division, a n = ( 1) n 1 n n 2 + 4 = ( 1) n 1 n + 4 n . Consequently, 0 ≤ | a n | = 1 n + 4 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 10.0points 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 = 4 2. converges with limit = 0 3. does not converge correct 4. converges with limit = 1 5. converges with limit = 1 4 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 . . . . . (2 n 1)2 n (2 n + 1) . when m = 2 n + 1. But then, (2 n + 1)! (2 n 1)! = 2 n (2 n + 1) . Consequently, the given sequence does not converge .
byrne (cmb3744) – HW 02 – rusin – (55565) 3 006 10.0points Determine whether the sequence { a n } con- verges or diverges when a n = 1 + cos 2 n 5 + 2 n .

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University of Texas M 408D Rusin HW 02 - byrne(cmb3744 HW...

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