Math 116 Practice for Exam 3

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1 Math 6 Practice for Exam Geerated October 0, 207 Name: SOLUTIONS Istructor: Sectio Number:. This exam has 7 questios. Note that the problems are ot of equal difficulty, so you may wat to skip over ad retur to a problem o which you are stuck. 2. Do ot separate the pages of the exam. If ay pages do become separated, write your ame o them ad poit them out to your istructor whe you had i the exam.. Please read the istructios for each idividual exercise carefully. Oe of the skills beig tested o this exam is your ability to iterpret questios, so istructors will ot aswer questios about exam problems durig the exam. 4. Show a appropriate amout of work (icludig appropriate explaatio) for each exercise so that the graders ca see ot oly the aswer but also how you obtaied it. Iclude uits i your aswers where appropriate. 5. You may use ay calculator except a TI-92 (or other calculator with a full alphaumeric keypad). However, you must show work for ay calculatio which we have leared how to do i this course. You are also allowed two sides of a 5 ote card. 6. If you use graphs or tables to obtai a aswer, be certai to iclude a explaatio ad sketch of the graph, ad to write out the etries of the table that you use. 7. You must use the methods leared i this course to solve all problems. Semester Exam Problem Name Poits Score Witer Fall Witer Fall Witer Fall Fall Total 84 Recommeded time (based o poits): 0 miutes

2 Math 6 / Fial (April 2, 200) page 4 2. [8 poits] For each of the followig series, write whether the series Coverges or Diverges o the space provided ext to the series. Support your aswer by statig the test(s) you used to prove covergece or divergece, ad show complete work ad justificatio. a. [6 poits] = Diverges Solutio: We ote that 2 > ad 2 + > for 2, so >. We kow diverges by the p-test, so by the compariso test, the series diverges. =2 b. [6 poits] =!(+)! (2)! Solutio: We use the ratio test to get Coverges After some cacellatio we get Thus it coverges. (2)!(+)!(+2)! (2(+))!(+)!!. (+2)(+) (2+)(2+2) = /4. c. [6 poits] =2 si() 2 Coverges Solutio: We will show that this is absolutely coverget, ad hece coverget. Additioally, sice si() we ca boud the sum of the absolute values by =2 2. The we apply the it compariso test with =2 2. Sice 2 2 = both series coverge or both diverge. Sice =2 2 coverges by p-test, they both coverge. Uiversity of Michiga Departmet of Mathematics Witer, 200 Math 6 Exam Problem 2 Solutio

3 Math 6 / Fial (December 7, 200) page 9 7. [4 poits] For each of the followig sequeces.compute a. 2.Decide if a coverges or diverges. Circle your aswer. Support your aswer by statig the test(s) or facts you used to prove covergece or divergece, ad show complete work ad justificatio. a. [4 poits] a = ( ) π a = a : Coverges Diverges Solutio: Sequece: a = r where r = =.8 < hece π a = 0. Series: a = r is a geometric series with r < the it coverges. b. [4 poits] a = a = a : Coverges Diverges Solutio: Sequece: a = = = 4. Series: Sice a does ot coverge to 0 the a diverges. Note to the graders: The criteria for the justificatio of the divergece of the series has bee give differet ames i some sectios (th term test ad some others). If you see these kid of justificatios, please ask the istructor before cosiderig ay deductios. c. [6 poits] a = 4 +5 a = a : Coverges Diverges Solutio: Sequece: a = Series: a = as goes to ifiity Limit Compariso Test: a a = Hece diverges. = 4 +5 = = 4 +5 = 2 =. 2 Uiversity of Michiga Departmet of Mathematics Fall, 200 Math 6 Exam Problem 7 Solutio

4 Math 6 / Fial (April 20) page 4 x t y t 2 t 2 t 2. [7 poits] For, cosider the followig sequeces a = ( ) +. b = + ( ). c = ( 6 5 ). s = k=. k 2 Circle your aswers. No justificatio is eeded.. Which sequeces are bouded? a b c s Noe. 2. Which sequeces are icreasig? a b c s Noe.. Which sequeces are coverget? a b c s Noe. Solutio:. a b s 2. c s. b s Uiversity of Michiga Departmet of Mathematics Witer, 20 Math 6 Exam Problem 2 Solutio

5 Math 6 / Fial (December 5, 20) page 7. [2 poits] Determie whether the followig series coverge or diverge (circle your aswer). Be sure to metio which tests you used to justify your aswer. If you use the compariso test or it compariso test, write a appropriate compariso fuctio. a. [ poits] ( ) +2 = Solutio: Sice +2 = 2 = 2, the the sequece a = ( ) +2 does ot coverge to zero (it oscillates closer to 2 ad 2. Sice the termsa does ot coverge to0, the the series = a diverges. b. [4 poits] = e 2 Solutio: Let f(x) = xe x2. The fuctio f(x) > 0 ad f (x) = e x2 ( 2x 2 ) < 0 for x. Hece by the Itegral test e 2 behaves as = xe x2 dx = b xe x2 dx = 2 e x2 b = 2e the series coverges. c. [5 poits] = cos( 2 ) 2 Solutio: This series has positive ad egative terms. Sice cos( 2 ) 2 2, the the series of the absolute values satisfies cos( 2 ) 2 2. = The series o the right coverges by the p series test with p = 2, hece the series of absolute values coverges. Sice the series of absolute values coverges, the cos( 2 ) = 2 coverges. = Uiversity of Michiga Departmet of Mathematics Fall, 20 Math 6 Exam Problem 7 Solutio

6 Math 6 / Fial (April 9, 202) page 6. [2 poits] a. [6 poits] State whether each of the followig series coverges or diverges. Idicate which test you use to decide. Show all of your work to receive full credit.. l =2 Solutio: The fuctio is decreasig ad positive for 2, the the l Itegral test says that behaves as l x lx dx. 2 Hece 2. x lx =2 = =2 b dx = 2 l diverges. cos 2 () x lx dx = 2 lb l2 u 2 du = 2 u lb l2 =. Solutio: Sice 0 cos2 (), ad 2 2 > ), the compariso test yields the covergece of 2 coverges by p-series test (p = = cos 2 (). b. [6 poits] Decide whether each of the followig series coverges absolutely, coverges coditioally or diverges. Circle your aswer. No justificatio required.. ( ) Coverges absolutely Coverges coditioally Diverges 2. ( 2) 5 Coverges absolutely Coverges coditioally Diverges Uiversity of Michiga Departmet of Mathematics Witer, 202 Math 6 Exam Problem Solutio

7 Math 6 / Fial (April 9, 202) page 7 Solutio: ( ) = behaves as = sice = = > 0. Sice diverges (p-series test p = ), the by it compariso test = ( ) diverges. ( ) The covergece of 2 + follows from alteratig series test sice for a = : a = 0. a is decreasig for large. r = 8 5 ( 2) 5 = d d ( ) 2 + = ( 2 ++8) 2 < 0 ( 8 5) is a geometric series with ratio <, hece it diverges. Uiversity of Michiga Departmet of Mathematics Witer, 202 Math 6 Exam Problem Solutio

8 Math 6 / Fial (December 4, 202) page 7 5. [9 poits] Cosider the followig power series: = 5 (x 4)+ a. [4 poits] Fid the radius of covergece of the power series. Show all your work. Solutio: so R = 5. (+)5 + x 4 +2 x 4 = x x 4 < x 4 < 5, 5 + = 5 x 4. b. [5 poits] For which values of x does the series coverge absolutely? For which values of x does it coverge coditioally? Solutio: Coverges absolutely iside radius: (, 9). Left edpoit: x =, = 5 ( 5)+ = = ( ) +, = coverges coditioally (alteratig harmoic series). Right edpoit: x = 9, = , diverges. So, coverges coditioally for x =, absolutely for < x < 9. = Uiversity of Michiga Departmet of Mathematics Fall, 202 Math 6 Exam Problem 5 Solutio

9 Math 6 / Fial (December 4, 202) page 8 6. [2 poits] I the followig problems, support all you aswers by statig the test(s) or facts you used to prove covergece or divergece. Show all your work. a. [4 poits] Circle your aswer: Coverges Diverges + = Solutio: Compare to: = so series coverges by LCT. 5/2, coverges by p-test, p = 5 2 >. + = + = > 0, 5/2 b. [4 poits] = 2+cos 2 () Solutio: 2+cos 2 () Circle your aswer: Coverges Diverges 0 (does ot exist, i fact), so series diverges. c. [4 poits] For which values of a does the series = a = a + a2 9 + a 27 + coverge? For the values of a where the series coverges, fid the sum of the series. Solutio: Geometric series, coverges whe r = a <, so coverges o the iterval < a <. Coverges to ( ) a = a ( ) = a a a. a Uiversity of Michiga Departmet of Mathematics Fall, 202 Math 6 Exam Problem 6 Solutio