page 1 of 11Math 1B: Final ExamFriday, 14 August 2009Instructor: Theo Johnson-Freydhttp://math.berkeley.edu/~theojf/09Summer1B/Name: ANSWERSProblem Number 1 2 3 4 5 6 7 TotalScoreMaximum 20 20 10 15 10 10 15 100Please do not begin this test until 2:10 p.m. You may work on the exam until 4 p.m.Please do not leave during the last 15 minutes of the exam time.You must always justify your answers: show your work, show it neatly, and when in doubt,use words (and pictures!) to explain your reasoning. Please box your final answers.Calculators are not allowed. Please sign the following honor code:I, the student whose name and signature appear on this midterm, have completed the examby myself, without any help during the exam from other people, or from sources other thanmy allowed one-page hand-written cheat sheet. Moreover, I have not provided any aid toother students in the class during the exam. I understand that cheating prevents me fromlearning and hurts other students by creating an atmosphere of distrust. I consider myselfto be an honorable person, and I have not cheated on this exam in any way. I promise totake an active part in seeing to it that others also do not cheat.Signature:Name: ANSWERS Math 1B Final Exam: 14 August 2009 page 2 of 111. (20 pts total – 2 pts each) For each of the following statements, determine if the conclusionALWAYS follows from the assumptions, if the conclusion is SOMETIMES true given theassumptions, or if the conclusion is NEVER true given the assumptions. You do not need toshow any work or justify your answers for these question: only your answer will be graded.(a) (2 pts) If limn→∞an< 1, thenP∞1anconverges.ALWAYS SOMETIMESNEVER(b) (2 pts) If the seriesP∞1anandP∞1bnboth converge, thenP∞1(an+ bn) converges.ALWAYS SOMETIMES NEVER(c) (2 pts) If the sequences {an} and {bn} both diverge, then {anbn} diverges.ALWAYS SOMETIMES NEVER(d) (2 pts) IfPN1an≥ −10 for every N and if an≤ 0 for every n, thenP∞1anconverges.ALWAYS SOMETIMES NEVER(e) (2 pts) IfP∞1cn2nconverges absolutely, thenP∞1cn(−2)nconverges conditionally.ALWAYS SOMETIMES NEVERName: ANSWERS Math 1B Final Exam: 14 August 2009 page 3 of 11(f) (2 pts) IfP∞1cn(−2)nconverges conditionally, thenP∞1cnconverges absolutely.ALWAYS SOMETIMES NEVER(g) (2 pts) If {bn} is a decreasing positive sequence, thenP∞1(−1)nbnconverges.ALWAYS SOMETIMESNEVER(h) (2 pts) If p is a real number, then the Ratio Test can be used to determine whetherP∞11/npconverges.ALWAYS SOMETIMES NEVER(i) (2 pts) If 0 ≤ an≤ bnandP∞1anconverges, thenP∞1bnconverges.ALWAYS SOMETIMES NEVER(j) (2 pts) If limn→∞bnexists, thenP∞1(bn− bn+1) converges to b1.ALWAYS SOMETIMES NEVERName: ANSWERS Math 1B Final Exam: 14 August 2009 page 4 of 112. (20 pts total – 5 pts each) Determine whether each of the following series is ABSOLUTELYCONVERGENT, CONDITIONALLY CONVERGENT, or DIVERGENT. You must specifywhich test(s) you use for each series, and why the series satisfies the conditions of the test.(a) (5 pts)∞Xn=1(−1)n(n + 1)n!We use the ratio test. We havelimn→∞an+1an= limn→∞−(n + 2)/(n + 1)!(n + 1)/n!= limn→∞n + 2(n + 1)2= 0 < 1Therefore the series converges absolutely .(b) (5 pts)∞Xn=0(−1)nn2n + 1The series diverges by the divergence test: limn→∞n2/(n + 1) = ∞, and so the limitof the alternating sequence is not 0 (indeed, it does not exist).Name: ANSWERS Math 1B Final Exam: 14 August 2009 page 5 of 11(c) (5 pts)∞Xn=2(−1)nln nThe series satisfies the conditions of the Alternating Series Test, since ln n is an increasingpositive function that tends to +∞, and so 1/ ln n is descreasing, positive, and tends to 0.But the absolute sequenceP∞21/ ln n diverges by, for example, comparison withP1/n:ln n < n for every n, and so 1/ ln n > 1/n. Thus the series converges conditionally .(d) (5 pts)∞Xn=1(−1)nn3+√nThe seriesP∞n=11n3+√nconverges by Comparison or Limit Comparison withP1n3. Thus,the above series converges absolutely .Name: ANSWERS Math 1B Final Exam: 14 August 2009 page 6 of 113. (10 pts) Find the interval of convergence of the following power series:∞Xn=2(x − 4)nn ln n 2nWe begin with the ratio test to determine the radius of convergence. We let cn= 1/(n ln n 2n).ThenROC = limn→∞cncn+1if this limit exists= limn→∞(n + 1) ln(n + 1) 2n+1n ln n 2n= 2limn→∞n + 1nlimn→∞ln(n + 1)ln ncancel and split the limits= 2limn→∞11limn→∞1/(n + 1)1/nL’Hospital= 2 limn→∞nn + 1= 2 evaluate and L’HospitalThe interval of convergence is centered at 4, and so is either (2, 6), (2, 6], [2, 6), or [2, 6],depending on the convergence at the endpoints.We now test these endpoints. At x = 2, the series is∞Xn=2(2 − 4)nn ln n 2n=∞Xn=2(−1)nn ln nThis series satisfies the conditions of the alternating series test, as n ln n is the product ofincreasing positive functions and both tend to +∞. So the power series converges at x = 2.At x = 6, the series is∞Xn=2(6 − 4)nn ln n 2n=∞Xn=21n ln nWe use the integral test, since 1/(x ln x) is a positive decreasing function. We haveZ∞21x ln xdx =Z∞ln 21udu = ln u|∞ln 2= ∞where we substituted u = ln x. Since the integral diverges, the series does as well.Thus, the interval of convergence of the power series is x ∈ [2, 6) .Name: ANSWERS Math 1B Final Exam: 14 August 2009 page 7 of 114. (15 pts) Evaluate the following definite integral as a series.Z103p1 + x6dxYou must both:• Write your final answer in Σ notation, but you may leave your answer in terms of thebinomial coefficientskn.• Write out the first four terms of the series.We expand3√1 + x6with the binomial theorem:3p1 + x6=∞Xn=01/3n(x6)n=∞Xn=01/3nx6nZ103p1 + x6dx =Z10∞Xn=01/3nx6ndx=∞Xn=01/3nZ10x6ndx =∞Xn=01/3n16n + 1In the last line, we used the integralR10xkdx = xk+1/(k + 1)10= 1/(k + 1) for k 6= −1.We recall that1/3n=(13)(13− 1) . . . (13− n + 1)n!so that the numerator and denominator each has n terms in the product. Then the first fourterms of the above sum are:∞Xn=01/3n16n + 1=1/3010 + 1+1/3116 + 1+1/32112 + 1+1/33118 + 1+ . . .= 1 +1317+(13)(−23)2113+(13)(−23)(−53)6119+ . . .= 1 +121−1117+51539+ . . .Name: ANSWERS Math 1B Final Exam: 14 August 2009 page 8 of 115. (a) (5 pts) Find power series representations
View Full Document
Unlocking...