SCORE. Exam 2. MA 114 Exam 2 Fall 2017

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Exam Name: Sectio ad/or TA: Do ot remove this aswer page you will retur the whole exam. You will be allowed two hours to complete this test. No books or otes may be used. You may use a graphig calculator durig the exam, but NO calculator with a Computer Algebra System (CAS) or a QWERTY keyboard is permitted. Absolutely o cell phoe use durig the exam is allowed. The exam cosists of 0 multiple choice questios ad 5 free respose questios. Record your aswers to the multiple choice questios o this page by fillig i the circle correspodig to the correct aswer. Show all work to receive full credit o the free respose problems. The wise studet will show work for the multiple choice problems as well. Multiple Choice Questios A B C D E A B C D E 3 A B C D E 4 A B C D E 5 A B C D E 6 A B C D E 7 A B C D E 8 A B C D E 9 A B C D E 0 A B C D E SCORE Multiple Total Choice 3 4 5 Score 50 9 8 5 6 00

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Multiple Choice Questios. A series a is coverget if ad oly if a A. the limit lim + a is greater tha. B. its sequece of terms {a } coverges to 0. C. its sequece of partial sums {s } coverges to some real umber. D. its sequece of terms {a } is alteratig. E. its sequece of partial sums {s } is bouded.. The power series expaded aroud a = 0 for the fuctio f (x) = 3 + 4x is A. B. C. 3 D. E. ( ) 3 x. 3 x. 3 4 3 4 x. ( 4) x. ( ) x. Page 3 of

3. Cosider the followig statemets. I. The series ( ) = is absolutely coverget. II. By the alteratig series test the series III. The series ( ) = A. Oly I is true. B. Oly II is true. C. Oly III is true. D. Oly I ad III are true. E. Oly II ad III are true. is coditioally coverget. ( ) 3 is coverget. = 4. What is the value of A. B. C. 4 D. 6 = 3? E. The series diverges. Page 4 of

5. Determie whether the series = 4 + is coverget or diverget by expressig S as a telescopig sum. If coverget, fid its sum. A. diverges B. C. 4 D. E. 6. Give the iterval of covergece for the series (x 4) =. A. (3, 5) B. [3, 5) C. [3, 5] D. [3, 5) E. Noe of the above. Page 5 of

7. The sequece {a } is give recursively by a + = a + a ad a =. The first five terms of this sequece are: A., 3, 5, 7, 9 B., 3,, 60, 360 C., 3, 3 4, 4 5, 5 6 D., 3, 5,, 4 E., 3, 5, 5, 75 = si 8. Cosider the series + 3. Applyig the compariso test with the series = leads to the followig coclusio. A. The test is icoclusive. B. The series coverges absolutely. C. The series coverges coditioally. D. The series diverges. E. The test caot be applied to a = si + 3 ad b =. Page 6 of

9. The series = A. diverges by the Ratio Test. B. diverges by the Itegral Test. C. coverges by the Limit Compariso Test with the series D. diverges by the Limit Compariso Test with the series E. coverges by the Root Test. =. =. 0. Which oe of the followig series coverges for all real umbers x? A. B. C. D. E. x = x = = x e x! =!x e = Page 7 of

Free Respose Questios. Determie if the sequece is coverget or diverget. If coverget give its limit. (a) (3 poits) a = 3 + 5 + Solutio: The sequece coverges 3 + 5 lim + = lim 3 + 5 + = 5. (b) (3 poits) a = e /(+) Solutio: The sequece coverges. lim e/(+) = e. (c) (3 poits) a = arcta() Solutio: The sequece coverges. arcta() lim = π/ lim = 0. Page 8 of

. (a) (4 poits) State the ratio test. Be sure your coclusio describes all three cases. Solutio: Give a series a let = lim a + a = L.. If L < the series is absolutely coverget (ad therefore coverget).. If L > the series is diverget. 3. If L = the test is icoclusive. (b) (4 poits) Use the ratio test to determie if the series your reasoig. Solutio: L = lim a + a = lim Sice L < the series coverges. + ( + )!! = lim! = coverges. Explai + = 0 < Page 9 of

3. (5 poits) Use the itegral test to determie whether the series coverges or diverges. Show your work. e 3 = Solutio: Use the Itegral Test with f (x) = x e x3. Let u = x 3 ad du = 3x dx M x e x3 dx = lim M 3 M x e x3 dx = lim e u du M = lim M M 3 e u = 3e < Sice the itegral coverges, the series coverges. Page 0 of

4. (a) (4 poits) Determie the power series expasio about a = 0 for the fuctio f (x) = + x 4. Solutio: Sice x = x, replacig x by x 4, we have + x 4 = ( x 4 ) = ( ) x 4. (b) (4 poits) Determie the power series expasio about a = 0 for x 0 + t 4 dt. Solutio: We fid the power series expasio for this by itegratig the power series from (a) term-by-term x 0 + t 4 dt = x 0 ( ) t 4 dt = ( ) x4+ 4 + = x x5 5 + x9 9 x3 3 + (c) (4 poits) Use the first three ozero terms to estimate Solutio: / 0 / 0 + t 4 dt. + t 4 dt + 5 5 9 9 = 659 3040 = 0.50603986 (d) (4 poits) Usig the Alteratig Series Estimatio Theorem, estimate the error i usig the sum i part (c). Solutio: Sice this is a alteratig series, the error is bouded by the ext term i the series which is 3 3 = 06496 = 0.00000939004038465. Page of

5. A fuctio f is defied by f (x) = x x3 3 + x5 5 x7 7 + x9 x+ + + ( ) 9 + + = ( ) x+ + for all x i the iterval of covergece for the power series. (a) (4 poits) Fid the radius of covergece for the power series. Show your work. Solutio: Use the Ratio Test. lim a + a = lim + x +3 + 3 x + = x <. To coverge we must have x <, so the radius of covergece is. (b) (4 poits) Fid the iterval of covergece for the power series. work. Show your Solutio: The radius of covergece is, so we eed to check the edpoits: x = ad x =. At x =, we have ( ) 3+ + = ( ) + + which coverges by the Alteratig Series Test. At x = we have ( ) + which coverges by the Alteratig Series Test. Thus, the iterval of covergece is x or [, ]. (c) (4 poits) Fid the power series represetatio for f (x) ad state its radius of covergece. Solutio: f (x) = ( ) x = x + x 4 x 6 + + ( ) x +. The radius of covergece does ot chage ad remais at. Page of