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Course: M 408D 408D, Spring 2012
School: University of Texas
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Word Count: 2165

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gilbert HW03 (55035) This print-out should have 18 questions. Multiple-choice questions may continue on beforethenextanswering.column or page find all choices 1 (ii) while if 0 &lt; ck bk, Xk ck converges, then the Comparison Test is inconclusive be- X 001 10.0 points If ak , bk , and ck satisfy the inequalities 0 &lt; ak ck bk , for all k, what can we say about the series (A) converges ,...

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gilbert HW03 (55035) This print-out should have 18 questions. Multiple-choice questions may continue on beforethenextanswering.column or page find all choices 1 (ii) while if 0 < ck bk, Xk ck converges, then the Comparison Test is inconclusive be- X 001 10.0 points If ak , bk , and ck satisfy the inequalities 0 < ak ck bk , for all k, what can we say about the series (A) converges , (B) need not converge . X (A) : cause bk could converge, but it could diverge - we cant say precisely without further restrictions on bk. Consequently, what we can say is X k =1 ak, (B) : k =1 b k 002 10.0 points Determine whether the series if we know that the series n =1 3 n(n + 1)(n + 5) p converges or diverges. (C) : X ck k =1 is convergent but know nothing else about ak and bk? 1. (A) diverges , (B) converges 1. series is convergent 2. series is divergent correct Explanation: Note first that 2. (A) need not converge , (B) converges lim 3. (A) converges , correct 2 X (B) need not converge 4. (A) converges , (B) converges 5. (A) converges , (B) diverges n 3 n n(n + 1)(n + 5) Thus by the limit comparison test, the given series X 2 n = 3 n(n + 1)( n + 5) p converges if and only if the series X n 6. (A) diverges , (B) diverges Explanation: Lets try applying the Comparison Test: (i) if 0 < ak ck, Xk c k = 1 > 0. 2 =1 n converges. But by the p-series test with p = 1 (or use the comparison test applied to the harmonic series), this last series diverges. Consequently, the given converges, series is divergent then the Comparison Test applies and says that X ak converges; 003 10.0 points . HW03 gilbert (55035) X Determine whether the series k (k + 2)3 1. A and B only 1. series is divergent 2. A, B, and C 2. series is convergent correct Explanation: We use the Limit Comparison Test with k = (k + 2)3 k b , 4 converge(s)? converges or diverges. ak 3 n = 15 k k=1 n (C) X 2 k 1 =. k 3 For k lim ak = lim = 1 > 0. k bk k k + 2 Thus the series 3. B and C only correct 4. C only 5. B only Explanation: (A) Because of the way the nTH term is defined as a quotient of polynomials in the series, use of the integral test is suggested. Set f (x ) = X k k =1 (k + 2)3 Then f is continuous, positive and decreasing on [1, ); thus X X1 3 1 converges if and only if the improper integral 2 1 3 2n 2 n = 1 3n + 6 k converges. But this last series is a geometric series with 3x2 + 6 10.0 points n =1 (B) 2n 3n 2 + 6 X n =1 5 n 2 x dx . 3x2 + 6 1 2 Now after substitution (set u = x ), we see that Which of the following series 2 In = series is convergent . (A) dx converges, which requires us to evaluate the integral n X x < 1, hence convergent. Consequently, the given series is 004 3x + 6 k=1 |r| = . 2 k converges if and only if the series 2x In = 3 1 2 ln(3x + 6) n 1 =3 But 1 2 ln(3n + 6) ln(3 + 6) . HW03 gilbert (55035) Consequently, of the three infinite series, 2 ln(3n + 6) as n , so the infinite series only B and C converge 2n 3n 2 + 6 X n =1 005 diverges. n 5 X=1 2 n 2 n 2 = =3 2 1 We apply the Limit Comparison Test with n 43 = tan 1 n , 3 5+n bn 1 =. 3 n For is a geometric series, but the summmation starts at n = 15 instead of at n = 1, as it usually does. We can still show, however, that this series converges. Indeed, X Explanation: an X 43 1. series is convergent correct 5 n =15 converges or diverges. 2. series is divergent 2 because the initial term in the series 2 is 5 . (C) The series 3 5+ n n=1 5 tan1 n X is a geometric series whose common 2 ratio is r = 5 . Since 0 < r < 1, the series converges; in fact, we know that 5 10.0 points nX=1 . Determine whether the series (B) The series 3 n nlim n 3tan 1 5+ n 3 n = nlim tan 1 n= 2. Thus the given series X tan1 n 3 5+n n=1 n =15 = 4 315 +4 16 3 + 43 17 is convergent if and only if the series This shows that the series is a geometric series 3 whose common ratio is r = 4 and whose 3 15 initial term is ( 4 ) . Since 0 < r < 1, the geometric series converges, and we know that 43 15 + 4 3 16 + 43 17 + . . . = 43 15 3 14 =4 3 4 +. . . . 15 X n=1 1 3 n is convergent. But by the p-series test, this last series converges because p = 3 > 1. Con-sequently, the given series is convergent . . 006 10.0 points HW03 gilbert (55035) 4 Find all values of r for which the infinite series X (7n n =1 converges. 1. r > 7 007 6 4n 8 + 4) r Which one of the following series is convergent? correct 8 1. 2. r > 2 2 2. 2 7 8 3. n =1 5+ n X 1 2n (1) 2 5 + n X 8 r 8r6 4 (7n + 4) n correct X 4. Explanation: After simplification we see that, 6 4n 4 so X ( 1) 6 + n n =1 n =1 3. r < 5. r < n 1 2 1 4. r > 10.0 points (7n = n 6 8r6 4)r 8 n+ 4 7+ n 7 X 5. 8 r , 4n X (7n 8 r> 0 + 4)r will converge if and only if the infinite series X 1 n=1 n 8r6 converges. But by the Integral test we know that the series X 1 n=1 n converges if and only if p > 1. Consequently, the given series will converge if and only if 8r 6 > 1, I.E., when r> 8 n =1 . n n1 6 (1) 5 + + n Explanation: Since X n =1 X 35 (1) 2 + n = n =1 5 2 + n , use of the Limit Comparison and p1 series Tests with p = 2 shows that this series is divergent. Similarly, since X 2n n =1 (1) 5 2 + n = X n =1 5 2 + n , the same argument shows that this series as well as X p 7 5 + n 4 6 n =1 3 (1) 2 as n . Thus by the Limit Comparison test, the infinite series n =1 2 5+ n n=1 is divergent. On the other hand, by the Divergence Test, the series n X n =1 (1) n1 6 5 + + n HW03 gilbert (55035) is divergent because 5 Now 3 6 + n n 1 5 + n 6 = 0 . nlim (1) f (n) = n n =1 6 + n + 6 = f (n + 1) 3 n + 5 = 0 . nlim f (n) = n Consequently, by the Alternating Series Test, the given series . To see that this series is convergent, set 1 b n lim n 1 X ( 1) > for all n, while This leaves only the series +5 3 = 6+ n n converges . . 009 Then 10.0 points Determine whether the series (i) bn+1 bn, lim bn = 0 . (ii) 2 2 2 2 2 3 4 + 5 6 + 7 ... n Consequently, by the Alternating Series Test, the series is conditionally convergent, absolutely con-vergent or divergent. n 1 X ( 1) 6+ n =1 1. series is absolutely convergent n 2. is convergent. series is conditionally convergent cor- rect 008 10.0 points Determine whether the series X n =0 n 3 + 5 cos n 3. series is divergent Explanation: In summation notation, 2 2+2 34 with converges or diverges. 5 2+2 67 ... = ( 1)n1f (n) , X n =1 f (x ) = 1. series diverges 2 x +2 . Now the improper integral 2. series converges correct Explanation: n Since cos n = (1) , the given series can be rewritten as the alternating series X (1) 3 n n +5 = n=0 with f (x) = X n=0 3 x +5 1 f (x) dx = x+2 1 dx is divergent, so by the Integral Test, the given series is not absolutely convergent. the On other hand, (1)nf (n) 2 f (n) = . 2 while n+2 2 > n+1+2 lim n 2 n+ 2 = f (n + 1) , = 0. HW03 gilbert (55035) Consequently, by the Alternating Series Test, the given series 6 diverges, and so by the Comparison Test, the series e1/n X is conditionally convergent . keywords: alternating series, Alternating se-ries test, conditionally convergent, absolutely convergent, divergent 010 too diverges; in other words, the given series is not absolutely convergent. To check for conditional convergence, con-sider the series X 10.0 points where e1/x 1/n X n =1 (1) n+1 e n (1) f (n) n=1 Determine whether the series 5n n=1 f (x) = 5x . Then f (x) > 0 on (0, ). On the other hand, 5n is absolutely convergent, conditionally con-vergent or divergent. f (x) = 1 3 5x e1/x e1/x 5x 2 1+ x = e1/x . 3 5x 2. absolutely convergent Thus f (x) < 0 on (0, ), so f (n) > f (n + 1) for all n. Finally, since 1/x = 1, lim e 3. divergent we see that f (n) 0 as n . By the 1. conditionally convergent correct x Alternating Series Test, therefore, the series Explanation: Since X n =1 1/n n+1 (1) e 5n X 1X = 5 n =1 (1) 1/n n e n=1 n , we have to decide if the series X n 1/n n =1 (1) e First we check for absolute convergence. 1/n Now, since e 1 for all n 1, 5n 1 5n > 0 . But by the p-series test with p = 1, the series X 1 n=1 5n is convergent. Consequently, the given series is conditionally convergent . n is absolutely convergent, conditionally con-vergent, or divergent. e1/n (1)nf (n) keywords: 011 10.0 points Determine whether the series X n =2 (1) n 4 n ln n converges conditionally, converges absolutely, or diverges. 1. series converges absolutely HW03 gilbert (55035) 2. series converges conditionally correct 7 Consequently, by the Alternating series Test, the given series is 3. series diverges conditionally convergent Explanation: The given series can be rewritten as 012 10.0 points X n =2 (1) n n ln 4 = 4 nX=2 (1)nf (n) , n P To apply the root test to an infinite series n an the value of = lim (an) 1/n where f (x) = 1 n = (x ln x) 1 > 0 x ln x has to be determined. Compute the value of for the series for all x > 1. On the other hand, by the Chain and Product Rules, X 1 for all x > 1. Thus f (x) is positive and decreasing for all x > 1. But the improper integral 2 f ( x) d x = 2 1 x ln x dx does not converge. By the Integral Test, therefore, the given series is not absolutely convergent. It could still converge conditionally, however. To use the Alternating Series Test to show that the series n (1) f (n) X 1. = is convergent, we have to check that 7 6 correct 36 2. = 7 35 3. = 6 4. = 7 5. = 67 Explanation: After division, 5n + 6 1 + 6 , 5n =5 n n=2 so (i) f (n) > f (n + 1) for n 2, 1/n (an) = lim f (x) = 0 . x Since f is decreasing for all x 2, however, we see that n2 = f (n) > f (n + 1) . x 1/n 5 1 + 56 n 1 = 0. x ln x 7 6. But 6 1/n 1 + 5 n1/n = 1 nlim 5 as n . Consequently, On the other hand, lim 5 n n + 6 67 n . n =1 (ln x + 1) < 0 f (x ) = (x l n x )2 (ii) . = 7 6 . HW03 gilbert (55035) 8 Consequently, for the given series, 013 10.0 points = 1. To apply the ratio test to the infinite series Xn an, the value = lim n+1 an n has 014 10.0 points a Decide whether the series Computetobedetermined,fortheseries X n =1 1 3n + 7 X 1 n sin n =1 . converges or diverges. 1. converges 1 10 1. = 2. diverges correct 1 2. = Explanation: The given series has the form 3 1 3. = 7 4. = 0 X But then an = an , 2 1 n+ 2 n n2 n . n =1 n 1 n sin 1 n +1 e lim . n lim sin x = 1 , x x0 1 n sin n+1 1 n+1 while lim n also. Thus = nlim 1 n 1 1/n |an| nlim n + n2 = 2> 1 , n = nlim 1 + n = 1, 2 n 2 2 =e>2 . Consequently, the Root Test ensures that the given series 1 diverges . =1 sin n 015 n n + 1 3(n3+n+1)7+ 7 = nlim 2 since n so n+ n , in which case n+1 lim n 2 Explanation: By algebra, 1 sin sin 1 n+1 n+1 1= 1 sin n 1 n + 2 n |an|1/n = 5. = 1 correct But 2 21n n +n 2 n 3n + 7 1 3n + 3 + 7 10.0 points The terms of a series are specified recur-sively by the equations . a 1 = 4, a n+1 9n + 1 = 11n + 7 an . HW03 gilbert (55035) Determine whether the series X 9 Consequently, the Ratio Test ensures that the given series an n=1 converges . converges or diverges. 1. converges correct 017 Determine which, if any, of the following series diverge. n (5n) 2. neither converges nor diverges 3. diverges Explanation: From the recursive formula we see that a 9n + 1 =9 (A) X n =1 (B) an n 11n + 7 11 Consequently, by the Ratio Test the series converges. n 5 X (C) = lim n+1 n! . lim 10.0 points (n + 6) n =1 n X n 6n5+n 4 n 45 n n =1 1. C only 016 10.0 points 2. A only Decide whether the series 3. none of them X n=1 (2n)! 2n (n!) 5 4. all of them converges or diverges. 5. B and C 1. converges correct 6. A and C correct 2. diverges 7. B only Explanation: The given series has the form 8. A and B X n =1 an , an = (2 n )! (n !) 2 5n . But then a n+1 a = n! n! (2n + 2)! 5 (n + 1)! (n + 1)! (2n)! 2 4n + 6n + 2 (2n + 1)(2n + 2) = = , 2 5(n + 2n + 1) (n + 1)(n + 1) 5 lim n a n a = 4. 5 for each of the given series. (A) The ratio test is the better one to use because a n+1 n in which case Explanation: To check for divergence we shall use either the Ratio test or the Root test which means computing one or other of 1/n a n+1 nlim a n , nlim |an| an+1 n =5 n! (n + 1) n+1 (n + 1)! nn . HW03 gilbert (55035) 10 Now n! 1 (n + 1)! = n+1 3. B and C , 4. A only while (n + 1) n n Thus n+1 = (n + 1) a n+1 a = 5 n +n 1 n n n n + n 1 5. C only . 5e > 1 as n , so series (A) diverges. 6. none of them 7. all of them 8. A and C correct Explanation: We compute one of (B) The root test is the better one to apply because 1/n 5 |an| = 0 n + 6 a lim n as n , so series (C) diverges. Consequently, of the given infinite series, (A) The ratio test is the better one to use: a n+1 = 3 n + 18 n + 3 3 <1 an 8nn+1+38 as n , so series (A) converges. (B) The ratio test is the better one to apply: a n+1 a only A and C n an for each of the given series. as n , so series (B) converges. (C) Again the root test is the better one to th apply because of the n powers. For then 5 5n 25 |an|1/n = >1 4 6n + 4 24 1/n lim (an) , n+1 n = ( 47 nn + 1 n + 7) 4)( + 1 + 7) ( diverge. 4)( n 7 as n , 4 so series (B) diverges. 018 10.0 points Decide which of the following series con-verge. 8 n n (A) X n + 3 83 n =1 (B) n X n + n 74 47 n =1 (C) X n =1 1. A and B 2. B only 4 nn3 ++ 87 n (C) The root test is the better one to apply: 4n + 7 (an)1/n = , 3 0 n +8 as n , so series (C) converges. Consequently, of the given infinite series, only A and C converge .
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University of Texas - M 408D - 408D
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University of Texas - M 408D - 408D
HW06a gilbert (55035)This print-out should have 10 questions.Multiple-choice questions may continueon the next column or page find allchoices before answering.110.0 pointsA rectangular box is constructed in 3space with one corner at the origin and
University of Texas - M 408D - 408D
HW06 gilbert (55035)This print-out should have 17 questions.Multiple-choice questions may continueon the next column or page find allchoices before answering.13. y =4. y =00110.0 pointsFind a Cartesian equation for thecurve given in parametric
University of Texas - M 408D - 408D
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University of Texas - M 408D - 408D
HW08 gilbert (55035)This print-out should have 21 questions.Multiple-choice questions may continueon the next column or page find allchoices before answering.001100210.0 pointsIdentify the quadric surface10.0 pointsClassify the quadric surface
University of Texas - M 408D - 408D
HW09 gilbert (55035)This print-out should have 18 questions.Multiple-choice questions may continueon the next column or page find allchoices before answering.00110.0 pointsFind an equation for the plane passingthrough the origin that is parallel t
University of Texas - M 408D - 408D
HW10 gilbert (55035)This print-out should have 16 questions.Multiple-choice questions may continueon the next column or page find allchoices before answering.00110.0 pointsIn the contour map below identify thepoints P, Q, and R as local minima, lo
University of Texas - M 408D - 408D
DiscGQ02 gilbert (55035)This print-out should have 4 questions.Multiple-choice questions may continueon the next column or page find allchoices before answering.00110.0 points1and the series11B=F (N) .N=61. A &lt; BDetermine whether the series
University of Texas - M 408D - 408D
EXAM 02 gilbert (55035)This print-out should have 14 questions.Multiple-choice questions may continueon the next column or page find allchoices before answering.0011I.E., y = x. Thus by the constraint equation,g (x, x) = 2x + x 6 = 0 ,I.E., x = 2
University of Texas - M 408D - 408D
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University of Texas - M 408D - 408D
DiscGQ07 gilbert (55035)1This print-out should have 4 questions.Multiple-choice questions may continue onthe next column or page find all choicesbefore answering.z001 10.0 pointsWhich one of the following equations hasgraphxBut this circle has
University of Texas - M 408D - 408D
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University of Texas - M 408D - 408D
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University of Texas - M 408D - 408D
EXAM 01 gilbert (55035)This print-out should have 14 questions.Multiple-choice questions may continueon the next column or page find allchoices before answering.00141. I =9= 29 correct3. IDetermine whether the series= 894. integral doesnt con
University of Texas - M 408D - 408D
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University of Texas - M 408D - 408D
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Summary of Key Learning Objectives forClass 3: Seligram CaseThe Seligram case provides an example of how traditional cost systems that meet financial accountingneeds for inventory valuation, fail in dynamic competitive environments characterized by rap
UC Davis - MGT ` - 170
Summary of Key Learning Objectives forClass 2: Bridgeton CaseThe Bridgeton Case introduces cost accounting by providing a simple example of the design andfunctioning of traditional product cost systems. The case setting and managerial decisions covered
UC Davis - MGT ` - 170
MGT 170: Class 1Summary of Key Learning Objectives forP&amp;G Polska CaseThe P&amp;G Polska case introduces the topic of management accounting, including both costmanagement and management control (i.e., performance measurement and evaluation). The caseillus
UC Davis - MGT ` - 170
5-10 Year Plan DevelopmentJanF ebM arA prM ayJ unJ ulA ugS epO ct2 pageB u r k e L e t t erCE O J &amp; JDis c u s sw it hE xec .C om mStr ateg yReviewExec . C om mStolz erDeb atean dA dju s tCod m anBoar dFunc . D epts .P r ior Yr
UC Davis - MGT ` - 170
JanFebMarAprMayJunJulAugSepO ctCEO J&amp;JExec. CommStolzerDebateandAdjustCodmanBoardFunc. Depts.Prior Yr2nd YearFcstDebateandAdjustMissionState.5/10 Plan-by area-bus. seg-strategies-mktg plan-other deptsInitialPrepApproved
NUCES - FINANCIAL - 101
CHAPTER 10ACQUISITION AND DISPOSITION OFPROPERTY, PLANT, AND EQUIPMENTIFRS questions are available at the end of this chapter.TRUE-FALSEConceptualAnswerFTFTFTFFFTTTTFFTTFFTNo.Description1.2.3.4.5.6.7.8.9.10.11.12.13
McMaster - CELL BIO - 2b03
Cancer biology: tumorigenesis can stem from errors in signaling pathways and cell cycle regulationmechanisms Cancer is result of uncontrolled cell divisiono Leading cause of premature death in Canadao In young animal, cell proliferation exceeds cell
McMaster - CELL BIO - 2b03
McMaster - CELL BIO - 2b03
Final Exam Practice QuestionsThese questions can be used as a guide to help you focus your studying to cover key points that willlikely be on the final exam. Two disclaimers: 1) This guide should not be used as a substitute for thelecture notes, althou
Utah Valley University - HIST - 2710
AmistadAmistad We are all born into something, a certain belief or ideology. Others create their ownideologies, either out of conviction or for the purpose of convenience. Our lives revolve aroundthese ideas, these ideologies, for it is through them th
Utah Valley University - HIST - 2710
Brazil TourismSubject: Tourism in Brazil Statement: A brief overview of Brazil. Introduction: With its 8,512thousands km2, Brazil is the fifth largest country in the world and it is today America Latin'smost important country, if considering the produc
Utah Valley University - HIST - 2710
Brazil, Japan, UkExecutive Summary A cross analysis between three countries around the world were arrangedto determine which of those countries present the best investment opportunities. An evaluationamong the countries was structured through the compa
Utah Valley University - HIST - 2710
BRAZIL For years Brazil has been a featured country in South America. Not only is Brazil avery pretty country, it is filled with an abundance of natural resources. For three centuries Brazilwas ruled by Portugal. The country finally gained its independe
Utah Valley University - HIST - 2710
Brazilians and North AmericansBrazilians have a strikingly distinct culture which favors physical contact to show love andaffection compared to the North American culture. The personal space that Brazilians have isnot as great as the North Americans ha
Utah Valley University - HIST - 2710
The world's most luxurious hotel, Burj Al-Arab or the Arabian Tower, is the second modernwonder of the world. Ever since its completion in November 1997, the hotel has become an iconfor the technologically expanding city of Dubai. The jewel of the Middl
Utah Valley University - HIST - 2710
Child Prostitution EssayThe word &quot;prostitution&quot; dates back to the year 1553. Webster's dictionary defines it as the act orpractice of indulging in promiscuous sexual relations especially for money or the state of beingprostituted. Prostitution probably
Utah Valley University - HIST - 2710
China 1Abstract the purpose of this paper is to present China as a nation that has gone through severaleconomic phases until it reaches a point that has made several consider this country as apromising future power in the region. While Japan is falling