Math 116 — Practice for Exam 2

1 Math 116 Practice for exam 2 Generated October 29, 2018 Name:SOLUTIONSI nstructor:Section Number:1. This exam has 9 questions. Note that the problems are not of equal difficulty, so you may want to skipover and return to a problem on which you are Do not separate the pages of the exam. If any pages do become separated, write your name on themand point them out to your instructor when you hand in the Please read the instructions for each individual exercise carefully. One of the skills being tested onthis exam is your ability to interpret questions, so instructors will not answer questions about examproblems during the Show an appropriate amount of work (including appropriate explanation) for each exercise so that thegraders can see not only the answer but also how you obtained it. Include units in your answers You may use any calculator except a TI-92 (or other calculator with a full alphanumeric keypad).However, you must show work for any calculation which we have learned how to do in this course.

2 Youare also allowed two sides of a 3 5 note If you use graphs or tables to obtain an answer, be certain to include an explanation and sketch of thegraph, and to write out the entries of the table that you You must use the methods learned in this course to solve all 20123312 Fall 20133211 Winter 20143410 Fall 201431010 Winter 20153110 Winter 20163212 Winter 2017339 Winter 20182612 Winter 20133914 Total100 Recommended time (based on points): 117 minutesMath 116 / Final (April 19, 2012)page 63. [12 points]a. [6 points] State whether each of the following series converges or diverges. Indicatewhich test you use to decide. Show all of your work to receive full n=21n lnnSolution:The function1n lnnis decreasing and positive forn 2, then theIntegral test says that n=21n lnnbehaves as 21x lnxdx. 21x lnxdx= limb b21x lnxdx= limb lnbln 2u 12du= limb 2 u lnbln 2= :Hence n=21n n=1cos2(n) n3 Solution:Since 0 cos2(n) n3 1n32, and n=01n32converges byp-series test (p=32>1), then comparison test yields the convergence of n=1cos2(n) [6 points] Decide whether each of the following series converges absolutely, con-verges conditionally or diverges.

3 Circle your answer. No justification n=0( 1)n n2+ 1n2+n+ 8 Converges absolutelyConverges conditionallyDiverges2. n=0( 2)3n5nConverges absolutely Converges conditionallyDivergesUniversity of Michigan Department of MathematicsWinter, 2012 Math 116 Exam 3 Problem 3 SolutionMath 116 / Final (April 19, 2012)page 7 Solution: n=0 ( 1)n n2+ 1n2+n+ 8 = n=0 n2+ 1n2+n+ 8behaves as n=11nsincelimn n2+1n2+n+81n= limn n n2+ 1n2+n+ 8= 1>0:Since n=11ndiverges (p-series testp= 1), then by limit comparison test n=0 ( 1)n n2+ 1n2+n+ 8 convergence of n=0( 1)n n2+ 1n2+n+ 8follows from alternating series test since foran= n2+1n2+n+8: limn an= 0. anis decreasingddn( n2+ 1n2+n+ 8)= 1 + 6n n3 1 +n2(n2+n+ 8)2<0fornlarge. n=0( 2)3n5n= n=0( 85)nis a geometric series with ratior= 85< 1, hence it of Michigan Department of MathematicsWinter, 2012 Math 116 Exam 3 Problem 3 SolutionMath 116 / Final (December 17, 2013)page 42.

4 [11 points] Determine the convergence or divergence of the following series. In parts (a) and(b), support your answers by stating and properly justifying any test(s), facts or computationsyou use to prove convergence or divergence. Circle your final answer. Show all your [3 points] n=19ne n+nCONVERGESDIVERGESS olution:limn 9ne n+n= limn 9nn= 96= 0:Since limn an6= 0 then n=19ne n+ [4 points] n=24n(lnn)2:CONVERGESDIVERGESS olution:The functionf(n) =4n(lnn)2is positive and decreasing forn >2, then byIntegral Test the convergence or divergence of n=24n(lnn)2can be determined with theconvergence or divergence of 24x(lnx)2dx 4x(lnx)2dx= 4u2duwhereu= lnx:= 4u+C= 4lnx+CHence 24x(lnx)2dx= limb 4lnx|b2= 4ln 24x(lnx)2dx= 4 ln 21u2duconverges byp-test withp= 2> of Michigan Department of MathematicsFall, 2013 Math 116 Exam 3 Problem 2 SolutionMath 116 / Final (December 17, 2013)page 5c. [4 points] Letrbe arealnumber.

5 For which values ofris the series n=1( 1)nn2nr+ 4absolutely convergent? Conditionally convergent? No justification is :Absolutely convergent if :r >3 Conditionally convergent if : 2< r 3 Justification (not required): Absolute convergence:The series n=1 ( 1)nn2nr+ 4 = n=1n2nr+ 4behaves like n=1n2nr= n=11nr 2. The lastseries is ap-series withp=r 2 which converges ifr 2>1. Hence the seriesconverges absolutely ifr >3. Conditionally convergence:The functionn2nr+ 4is positive and decreasing (for large values ofn) whenr > by the Alternating series test n=1( 1)nn2nr+ 4converges in this of Michigan Department of MathematicsFall, 2013 Math 116 Exam 3 Problem 2 SolutionMath 116 / Final (April 28, 2014)page 54. [10 points] Determine whether the following series converge or diverge. Show all of your workand justify your [5 points] n=18n+ 10n9nSolution:limn 8n+10n9n= therefore by thenthterm test the series [5 points] n=41n3+n2cos(n)Solution: n=41n3+n2cos(n) n=41n2(n 1) n=41n2.

6 The final series is a convergentpseries sincep= 2>1. Therefore the original series converges by of Michigan Department of MathematicsWinter, 2014 Math 116 Exam 3 Problem 4 SolutionMath 116 / Final (December 12, 2014)page 1110. [10 points] Determine whether the following series converge or diverge. Justify your [5 points] n=2( 1)nln(n)nSolution:Note that this series is alternating and that the absolute values of the terms|ln(n)||n|form a decreasing sequence that converges to 0 asnapproaches . Therefore,this series converges by the alternating series [5 points] n=1n n3+ 2 Solution:This can be done with a comparison or limit comparison test. For comparison:n n3+ 2>n3 n3=(13)1 nBy thep-test withp= 1/2<1, we have that n=11 ndiverges. By the comparison test,the series n=1n n3+ 2also of Michigan Department of MathematicsFall, 2014 Math 116 Exam 3 Problem 10 SolutionMath 116 / Final (April 23, 2015)page 21. [10 points] Show that the following series converges.

7 Also, determine whether the seriesconverges conditionally or converges absolutely. Circle the appropriateanswer show all your work and indicate any theorems you use to show convergenceand to determine the type of convergence. n=2( 1)nln(n)nCONVERGES CONDITIONALLYCONVERGES ABSOLUTELYS olution:The series we obtain when we take the absolute value of the terms in theseries above is n=2ln(n)n. Now consider the integral 2ln(x)xdx. By making a change of variables we seethat 2ln(x)xdx= limb b2ln(x)xdx= limb ln(b)ln(2)udu= limb (ln(b))22 (ln(2))22= + and so the integral above diverges. Thus, the integral test implies that n=2ln(n) (n+ 1)n+ 1 ln(n)nand limn ln(n)n= 0, we have that n=2( 1)nln(n)nconverges by thealternating series we have shown that the series n=2( 1)nln(n)nis conditionally of Michigan Department of MathematicsWinter, 2015 Math 116 Exam 3 Problem 1 SolutionMath 116 / Final (April 21, 2016)DO NOT WRITE YOUR NAME ON THIS PAGE page 32.

8 [12 points] In this problemyou must give full evidence supporting your answer,showing all your work and indicating any theorems about series you [7 points] Show that the following seriesconverges. Does it converge conditionally orabsolutely? Justify. n=1( 1)nn! + 2nSolution:We notice that ( 1)nn! + 2n =1n! + 2n 12nfor alln 1 ( )The series n=112nconverges because it is a geometric series with ratio 1/2 and|1/2|<1. Thus, using theinequality ( ), the series n=1 ( 1)nn! + 2n converges by the comparison test. Since this series converges, the series n=1( 1)nn! + 2nconverges absolutely. (This also shows that it converges).b. [5 points] Determine whether the following series converges or diverges: n=21nlnnSolution:The functionf(x) =1xlnxis decreasing and has limx f(x) = 0. Moreover, 21xlnxdx= limb b21xlnxdx= limb lnbln 21udu= limb (ln(lnb) ln(ln 2)) = where in the second equality we used the substitutionu= lnx.

9 So the integral 21xlnxdxdiverges. Therefore, the integral test applies and tells us that theseries in question di-verges as of Michigan Department of MathematicsWinter, 2016 Math 116 Exam 3 Problem 2 SolutionMath 116 / Final (April 24, 2017)do not write your name on this exampage 43. [9 points] In this problem you must give full evidence supportingyour answer, showing allyour work and indicating any theorems or tests about series you use. (Remark: Youcannotuse any results about convergence from the team homework without justification.)a. [4 points] Determine whether the series below converges or diverges, and circle youranswer clearly. Justify your answer as described above. Xn=1sin 1 n ConvergesDivergesb. [5 points] Determine if the following infinite series convergesabsolutely, convergesconditionally, or diverges, and circle your answer clearly. Justify your answer asdescribed above. Xn=2( 1)nnln(n)Converges AbsolutelyConverges ConditionallyDivergesUniversity of Michigan Department of MathematicsWinter, 2017 Math 116 Exam 3 Problem 3 SolutionMath 116 / exam 2 (March 19, 2018)page 66.

10 [12 points] Determine whether the following series converge or justify your answer. Show all work and indicate any convergence tests [6 points] n=1n2+ncos(n) n8 n+ 1 ConvergesDivergesJustification:Solution: Sincen2+ncos(n) n8 n+ 1 1n2;we use the limit comparison test comparing n2+ncos(n) n8 n+11n2= limn n4+n3cos(n) n8 n+ 1= 1:We can see this is true by using domination arguments: the numerator isdominated byn4, while n8 n+ 1 is dominated by n8 Since n=11n2converges byp-test (p= 2),our original series converges by the [6 points] n=0sin(n)enConvergesDivergesJustificatio n:Solution:The terms in this series are not positive, but it is also not an alternatingseries. We will consider the series n=0 sin(n)en = n=0|sin(n)|en:(1)Since|sin(n)| 1 we have n=0|sin(n)|en n=01en:The larger series n=01enconverges by the geometric series test since the common ratio1/eis less than 1. (Note that there are many other ways to show that this series con-verges.)