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9 Pages

### HW11-solutions[1]

Course: M 408, Spring 2012
School: University of Texas
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Word Count: 1543

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(kah2852) huang HW11 radin (54915) This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 assuming that the pattern of the rst few terms continues. 4 3 n 1. an = 3 4 n 2. an = 5 4 n 3. an = 5 4 n1 4. an = 3 4 n1 5. an = 4 3 n1 6. an = 10.0 points Find a formula for the general term an of the...

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(kah2852) huang HW11 radin (54915) This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 assuming that the pattern of the rst few terms continues. 4 3 n 1. an = 3 4 n 2. an = 5 4 n 3. an = 5 4 n1 4. an = 3 4 n1 5. an = 4 3 n1 6. an = 10.0 points Find a formula for the general term an of the sequence {a n } n=1 = 1 2, 6, 10, 14, . . . , assuming that the pattern of the rst few terms continues. 1. an = 5n 3 2. an = 4n 2 correct 3. an = 3n 1 correct Explanation: By inspection, consecutive terms an1 and an in the sequence 4. an = n + 4 5. an = n + 3 Explanation: By inspection, consecutive terms an1 and an in the sequence {a n } = n=1 64 4 16 , , ... 1, , 39 27 {a n } = n=1 have the property that 2, 6, 10, 14, . . . have the property that an = ran1 = an an1 = d = 4 . an = an1 + d = an2 + 2d = . . . an = ran1 = r 2 an2 = . . . = a1 + (n 1)d = 2 + 4(n 1) . Consequently, = r n1 a1 = an = keywords: 4 3 n1 4 3 n1 since a1 = 1. 10.0 points Find a formula for the general term an of the sequence 64 4 16 , , ... 1, , 39 27 Consequently, a n = 4n 2 . {a n } = n=1 4 an1 . 3 Thus Thus 002 , keywords: sequence, common ratio 003 10.0 points a1 . huang (kah2852) HW11 radin (54915) Determine if the sequence {an } converges when 1 4 an = , ln n 5n + 1 and if it does, nd its limit. 1. limit = ln 5 3. limit = 5 4 Explanation: After division by n4 we see that 2 3 4 4. limit = ln 5 3. limit = ln an 5. limit = 0 correct Explanation: After division by n we see that 4 = 5n + 1 4 n 5+ 1 n Now as n ; in particular, the denominator converges and has limit 7 = 0. Thus by properties of limits , 1 1 14 ln ln 5 + nnn n {an } diverges . But by known limits (or use LHospital), 14 ln , nn 1 1 ln 5 + n n 5 5 +4 n n. = 4 2 7+ 2 + 4 n n 6n 5 5 2 4 , 4 , 2 , 4 0 nn n n so by properties of logs, an = 5 2 4. limit = 0 5. limit = 2. the sequence diverges 2 0 as n . Consequently, the sequence {an } converges and has since the sequence {6n} diverges. 005 10.0 points Determine if the sequence {an }n converges when 5n an = 6n 5 and if it does, nd its limit when limit = 0 . 1. converges with limit = 004 10.0 points Determine if the sequence {an } converges, and if it does, nd its limit when an = 1. limit = 6 n5 5 n3 + 5 . 7 n4 + 2 n2 + 4 6 7 2. the sequence diverges correct 5 correct 6 2. converges with limit = 5 3. converges with limit = 6 5 4. the sequence diverges 5. converges with limit = 0 Explanation: huang (kah2852) HW11 radin (54915) After division by n we see that an = 5 5 6 n as n . Thus {an } converges and has 5 6 limit = 007 as n . Thus {an }n converges and has limit = 006 5 . 6 10 . 7 10.0 points Determine if the sequence {an } converges, and if it does, nd its limit when an = 10.0 points Determine whether the sequence {an } converges or diverges when 3 6n + (1)n . 4n + 3 1. converges with limit = 3 correct 2 2. converges with limit = 5 4 and if it does, nd its limit 3. converges with limit = 1. limit = 10 correct 7 6 7 4. converges with limit = 2. limit = 10 21 7 4 5. sequence does not converge an = 14 n2 2n2 + 3 , 7n + 2 n+1 3. limit = 0 Explanation: After division by n we see that 4. the sequence diverges 5. limit = an = 5 7 Explanation: After bringing the two terms to a common denominator we see that an 14 n3 + 14 n2 (7n + 2) 2n2 + 3 = (7n + 2) (n + 1) (1)n n. 3 4+ n (1)n , n 3 0 n 6+ But 3 as n , so an 2 as n . Consequently, the sequence converges and has 10n2 21n 6 . = 7n 2 + 9n + 2 Thus an = But 21 6 2 n n. 9 2 7+ + 2 nn 10 9 2 21 6 , 2, , 2 0 nn nn limit = 008 3 . 2 10.0 points Determine if the sequence {an } converges when n2 n , an = (n 8)2 n and if it does, nd its limit huang (kah2852) HW11 radin (54915) 16 1. limit = e Explanation: After simplication correct 5n =4 4n1 2. limit = e4 3. sequence diverges 5 4 n . But the sequence {r n } 16 4. limit = e (a) converges with limit 0 when |r | < 1, (b) and diverges for all |r | 1. 5. limit = 1 Consequently, the given sequence {an } 6. limit = e4 diverges . Explanation: By the Laws of Exponents, an = 4 n8 n 2 n = 8 1 n = 1 8 n 2 n 010 Determine whether the sequence {an } converges or diverges when n 2 . But xn 1+ ex n as n . Consequently, {an } converges and has an = (1)n 009 e8 2 = e16 1. limit = 8 5 10.0 points an = 5n 4n1 , and if it converges, nd the limit. 1. converges with limit = 5 3. limit = 5. limit = 0 Explanation: After division, 8 3+ 3n+8 n. = 5 4n+5 4+ n 3. converges with limit = 1 5. diverges correct 3 4 4. sequence diverges correct 2. converges with limit = 4 4. converges with limit = 0 , 3 4 . Determine whether the sequence {an } converges or diverges when 3n+ 8 4n+ 5 and if it does, nd its limit. 2. limit = limit = 10.0 points Now 85 , 0 as n , so nn lim n 3 3n+8 = = 0. 4n+5 4 huang (kah2852) HW11 radin (54915) Thus as n , the values of an oscillate be3 tween values ever closer to 4 . Consequently, the sequence diverges . 5 Determine if the sequence {an } converges when (2n + 1)! an = , 3n2 (2n 1)! and if it converges, nd the limit. 011 10.0 points Determine if the sequence {an } converges, when an = 9 cos 4n + 5 12 n + 6 2. does not converge 5 6 3 4 2 3 5. converges with limit = 2. limit = 3 3. converges with limit = 4. converges with limit = 5 6 3. limit = 9 cos 3 2 Explanation: By m! denition, is the product 4. sequence does not converge 9 correct 2 9 3 6. limit = 2 (2n 1)! = 1 2 3 . . . (2n 1) , while Explanation: After division, 5 4 + 4n +5 n. = 6 12 n + 6 12 + n 56 , 0 as n , so nn 4n +5 lim =. n 12 n + 6 3 Consequently, since cos x is continuous as a function of x, the sequence {an } converges and has limit = 9 cos 012 m! = 1 2 3 . . . m of the rst m positive integers. When m = 2n 1, therefore, 5. limit = But 4 correct 3 , and if it does, nd its limit. 1. limit = cos 1. converges with limit = 9 . = 3 2 10.0 points (2n + 1)! = 1 2 3 . . . (2n 1)2n(2n + 1) . when m = 2n + 1. But then, (2n + 1)! 2n(2n + 1) 4 = 2 (2n 1)! 2 3n 3n 3 as n . Consequently, the given sequence converges with limit = 013 4 . 3 10.0 points Determine whether the sequence {an } converges or diverges when an = ln(3n4) , ln(4n3) huang (kah2852) HW11 radin (54915) 6 and if it converges, nd the limit. 3. converges with limit = 0 correct 1. converges with limit = 0 3 2. converges with limit = 4 3. converges with limit = ln 3 ln 4 4. converges with limit = 4 correct 3 4. diverges 5. converges with limit = 4 Explanation: Since 0 cos2 n 1, we see that 4 5 an . 5 + 2n 5 + 2n 5. diverges Explanation: By properties of logs, But lim 4 ln(3n ) = ln 3 + 4 ln n , ln(4n3 ) = ln 4 + 3 ln n . Thus n 5 4 = 0 = lim , n 5 + 2n 5 + 2n so the Squeeze Theorem applies and ensures the sequence {an } converges with limit = 0 . an ln 3 4+ ln 3 + 4 ln n ln n . = = ln 4 ln 4 + 3 ln n 3+ ln n On the other hand, lim n ln 3 ln 4 = lim = 0. n ln n ln n Properties of limits thus ensure that the given sequence 4 converges with limit = . 3 014 015 10.0 points Determine if the limit lim n ( n + 7 n 5) n exists, and nd its value when it does. 1. limit doesnt exist 2. limit = 1 3. limit = 6 correct 10.0 points 4. limit = 2 Determine whether the sequence {an } converges or diverges when an = 4 + cos2 n . 5 + 2n 1. converges with limit = 4 5 2. converges with limit = 2 3 5. limit = 12 Explanation: By rationalization, n+7 (n + 7) (n 5) n5 = n+7+ n5 = 12 . n+7+ n5 huang (kah2852) HW11 radin (54915) On the other hand, n = n+7+ n5 for all n. But n 1+ 1+ 7 = lim n n 7 + n 1 5 . n 1 7 1+ + n lim as n . By the Pinching theorem, therefore, the sequence converges and limit = 0 . 5 = 1, n it thus follows by properties of limits that n 20 0 n 1 Since lim 7 5 1 n exists and has value 2. Consequently, again by properties of limits, the limit lim n ( n + 7 n 5) 017 10.0 points Determine if the sequence {6, 0, 6, 0, 0, 6, 0, 0, 0, 6, . . . } converges and, if it does, nd its limit. 1. limit = 0 n 2. sequence diverges correct exists and limit = 6 . 016 10.0 points Determine if the sequence 22552255 , , , , , , , , ... 13245768 converges and, if it does, nd its limit. 3. limit = 6 4. limit = 5 5. limit = 12 Explanation: Since a sequence {an }n converges and has limit L precisely when all the values of an are as close as we please to L for it all suciently large n, the 1. limit = 2 sequence diverges 2. limit = 0 correct because the values of an oscillate from 0 to 6 and back as n . 3. sequence diverges 4. limit = 018 7 2 10.0 points Determine whether the sequence an converges or diverges when 5. limit = 5 Explanation: Each term an in the sequence satises the inequalities 0 < an < 20 n an = tan1 4n 4n + 3 and if it converges, nd its limit. 1. diverges , huang (kah2852) HW11 radin (54915) 2. converges with limit = 3. converges with limit = 4. converges with limit = 5. converges with limit = 6 3 2 correct 4 Explanation: Since 4 9 14 (5n 1) (5n)n = 9 14 5n 1 4 4 5n 5n 5n 5n 5n for all n 1, we see that the inequalities Explanation: Now 0 an 4n = lim n n 4n + 3 4 lim 4+ 3 n =1 On the other hand, lim f n 4n 4n + 3 =f 4n lim n 4n + 3 limit = tan1 (1) = 019 . 4 10.0 points Determine whether the sequence {an } converges or diverges when 4 9 14 (5n 1) , (5n)n and if it converges, nd its limit. 1. converges with limit = 1 6 2. converges with limit = 6 3. converges with limit = 5 4. converges with limit = 0 correct lim n limit = 0 . keywords: 020 10.0 points Determine if the sequence {an }n converges and, if it does, nd its limit when an = n f n+1 n and f (x) = 2x2 + 4x + 2. 1. limit = 9 2. limit doesnt exist 3. limit = 10 4. limit = 8 correct 5. limit = 7 1 5 4 = 0. 5n The Squeeze Theorem thus ensures that the given sequence converges and has 5. sequence diverges 6. converges with limit = 4 5n hold for all n 1. But whenever f is a continous function. Consequently, the sequence an converges and has an = 8 Explanation: f (1) huang (kah2852) HW11 radin (54915) By the Newtonian quotient, f (x) = lim h0 f (x + h) f (x) . h 1 Setting h = n , we thus see that lim n f 1 n x+ n For the given limit therefore, lim an = f (1) n where f (x) = 4x + 4. Consequently, the sequence {an }n converges and limit = 8 . 021 10.0 points Determine if the sequence {an } converges, and if it does, nd its limit when 6n an = n 1 dx . 3x+4 1. limit = ln 6 2. the sequence diverges 3. limit = 1 ln 6 correct 3 4. limit = 1 ln 2 3 5. limit = 1 1 ln 3 6 Explanation: After integration, an = 1 ln(3 x + 4) 3 = Now as n . But ln x is continuous as a function of x. We thus see that {an } converges as n and has limit = = f ( x) . f ( x) 6n n 18 n + 4 1 ln 3 3n+ 4 4 18 + 18 n + 4 n 6 = 4 3n+ 4 3+ n . 9 1 ln 6 . 3
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