Math 132 Exam 3 Fall 2016• 15 multiple choice questions worth 5 points each.• 2 hand graded questions worth 12 and 13 points each.• Exam covers sections 11.1-11.6:Sequences, Series, Integral, Comparison, Alternating, Absolute (not Root or Ratio)• No calculators!• For the multiple choice questions, mark your answer on the answer card.• Show all your work for the written problems. Your ability to make your solution clearwill be part of the grade.Useful FormulasPni=1i =n(n+1)2Pni=1i2=n(n+1)(2n+1)6Pni=1i3=n(n+1)22sin2θ + cos2θ = 11 + tan2θ = sec2θ 1 + cot2θ = csc2θsin(A ± B) = sin A cos B ± sin B cos A cos(A ± B) = cos A cos B ∓ sin A sin Btan(A ± B) =tan A±tan B1∓tan A tan Bsin A sin B =12[cos(A − B) − cos(A + B)]cos A cos B =12[cos(A − B) + cos(A + B)] sin A cos B =12[sin(A + B) + cos(A − B)]sin2x =12(1 − cos 2x) cos2x =12(1 + cos 2x)sin(2θ) = 2 sin θ cos θ cos(2θ) = cos2θ − sin2θRcsc x dx = −ln |csc x + cot x| + CRsec x dx = ln |sec x + tan x| + CMath 132 Exam 3 Page 2 of 201. Determine whether the sequence defined byan= ln(2n3+ 2) − ln(5n3+ 2n2+ 4)converges or diverges. If it converges, find the limit.A. 0B. ln25C. −ln25D.25E. 2F. −5G. DivergesSolution:limn→∞an= limn→∞ln(2n3+ 2) − ln(5n3+ 2n2+ 4) = limn→∞ln2n3+ 25n3+ 2n2+ 4= lnlimn→∞2n3+ 25n3+ 2n2+ 4= ln252. Find ALL possible values of x for which the series∞Xn=09 + xn5nconverges.A. It is not possible to find such x because the series diverges.B. x > 0C. |x| < 1D. −3 < x < 3E. −3 < x < 5F. −5 < x < 5Math 132 Exam 3 Page 3 of 20G. −∞ < x < ∞Solution:∞Xn=09 + xn5n= ∞Xn=095n!+ ∞Xn=0xn5n!="∞Xn=0915n#+"∞Xn=0x5n#These are both geometric series. The first has r = 1/5 so it converges. The secondhas r = x/5 so it converges whenx5< 1, or −5 < x < 5.Math 132 Exam 3 Page 4 of 203. Determine whether the sequence defined byan= n2cos2n2+π2converges or diverges. If it converges, find the limit.A. −2B. −1C. 0D. 1E. 2F. πG. DivergesSolution:limn→∞an= limn→∞n2cos2n2+π2= limn→∞cos2n2+π21/n2Apply L’Hopital’s Rule= limn→∞−sin2n2+π2· (−4/n3)−2/n3= limn→∞−2 sin2n2+π2= −2 sin (π/2) = −24. Which of the following sequences converge?an=(2n + 1)!(n + 4)!, bn=πnn100, cn=ln(n10)√n, dn=n4(n + 1)!A. {dn} onlyB. {an}, {bn} onlyC. {cn}, {dn} onlyD. {an}, {dn} onlyMath 132 Exam 3 Page 5 of 20E. {an}, {bn}, {dn} onlyF. {an}, {cn}, {dn} onlyG. All of themSolution: For an, write out the factorial and see what cancels.an=(2n + 1)!(n + 4)!=(2n + 1)(2n)(2n − 1) ···(n + 5)(n + 4)(n + 3) ···1(n + 4)(n + 3)(n + 2) ···1= (2n + 1)(2n)(2n − 1) ···(n + 5) → ∞ (Diverges)For bn=πnn100, you can apply L’Hopital’s Rule 100 times to getlimn→∞bn= limn→∞(ln π)nπn100!= ∞For cn, apply L’Hopital:limn→∞cn= limn→∞ln(n10)√n= limn→∞10 ln nn1/2Apply L’Hopital= limn→∞10/n(1/2)n−1/2= limn→∞20n1/2= 0For dn, consider what happens when n > 8:dn=n4(n + 1)!=n4(n + 1)(n)(n − 1)(n − 2)(n − 3) ···1≤n4(n − 2)(n − 2)(n − 2)(n − 2)(n − 3) ···1=n4(n − 2)4(n − 3)!=nn − 241(n − 3)!→ 1 · 0 = 0Math 132 Exam 3 Page 6 of 205. Assume the terms of a sequence {an} are given by the following formulaan=13n3+223n3+323n3+ ··· +n23n3Find the limit of the sequence or conclude that it diverges.A. 0B. 1C.12D.23E.16F.19G. DivergesSolution: We use the formula on the formula sheet:limn→∞an= limn→∞13n3+223n3+323n3+ ··· +n23n3= limn→∞13n31 + 22+ 32+ ··· + n2= limn→∞13n3 nXi=1i2!= limn→∞13n3n(n + 1)(2n + 1)6= limn→∞n(n + 1)(2n + 1)18n3=196. Determine the value of the series∞Xn=26n(n + 3)or conclude that it diverges.A. 0B.132Math 132 Exam 3 Page 7 of 20C.53D.136E.133F.102G. DivergesSolution: This is telescoping so you need to use partial fractions to write out theseries.6n(n + 3)=2n−2n + 3∞Xn=26n(n + 3)=∞Xn=22n−2n + 3=22−25+23−26+24−27+25−28+26−29+ ···SoSN= 1 +23+24−2N + 2−2N + 3−2N + 4and limn→∞SN= 1 +23+24=136.Math 132 Exam 3 Page 8 of 207. Determine the value of the series∞Xn=02n−2+ 3n+14nor conclude that it diverges.A. 4B.92C.132D.374E.252F.978G. DivergesSolution: This is two geometric series:∞Xn=02n−2+ 3n+14n= ∞Xn=02n−24n!+ ∞Xn=03n+14n!= ∞Xn=01412n!+ ∞Xn=03 ·34n!=1/41 − 1/2+31 − 3/4=12+ 12 =2528. Assume∞Xn=1anis an infinite series with partial sums given by SN= 4 +2N. What is a5?A.225B.25C.110D.310E. -92Math 132 Exam 3 Page 9 of 20F. -52G. -110H. -310Solution: Note thata5= S5− S4=4 +25−4 +24=25−12=110Math 132 Exam 3 Page 10 of 209. Which of the following series converge?I.∞Xn=1(−1)n1 +1nII.∞Xn=21ln(n4)III.∞Xn=1n√n2+ 1A. None of themB. I onlyC. II onlyD. III onlyE. I and IIF. I and IIIG. II and IIIH. All of themSolution: For Series I note that limn→∞|an| = 1 6= 0 so the series diverges by the testfor divergence.Similarly, for series III, limn→∞an= 1 6= 0 so diverges by the test for divergence.For series II, note that1ln n4=14 ln 4≥14nSinceP14ndiverges, series II must divergetoo.10. The series∞Xn=11n4=π490. Find the value of the series∞Xn=22n4.A.8π445B.8π445− 1C. 16π490− 1D. −2E.8π490Math 132 Exam 3 Page 11 of 20F.8π490− 16G. DivergesSolution: Must take into account that our series starts at n = 2 instead of at n = 1.∞Xn=22n4=∞Xn=224n4= 16∞Xn=21n4= 16 −1 +∞Xn=11n4!= 16−1 +π490Math 132 Exam 3 Page 12 of 2011. Which of the following series converge?I.∞Xn=12n+ n44n+ n2II.∞Xn=14n5n+ nIII.∞Xn=1n!3nA. None of themB. I onlyC. II onlyD. III onlyE. I and IIF. I and IIIG. II and IIIH. All of themSolution: For I you can do a limit comparison with bn=2n4n. Note that limn42n= 0lim2n+n44n+n22n/4n= lim2n+n42n4n+n24n= lim1 +n42n1 +n24n= 1This series I converges.For II4n5n+ n≤4n5n=45nAnd so the series converges by comparison to a geometric series with r = 4/5.For III, limn!3n= ∞ 6= 0 so the series diverges.12. Which of the following series converge?I.∞Xn=1ln nen+ 2II.∞Xn=2sin2(n)n + n3/2III.∞Xn=1(−1)nn2/3A. None of themMath 132 Exam 3 Page 13 of 20B. I onlyC. II onlyD. III onlyE. I and IIF. I and IIIG. II and IIIH. All of themSolution: For I,ln nen+ 2≤nenand the
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