Practice Test Problems for Test IV, with Solutions

Size: px
Start display at page:

Download "Practice Test Problems for Test IV, with Solutions"

Transcription

1 Practice Test Problems for Test IV, with Solutios Dr. Holmes May, 2008 The exam will cover sectios 8.2 (revisited) to 8.8. The Taylor remaider formula from 8.9 will ot be o this test. The fact that sums, products, itegrals, atiderivatives of Taylor series are also Taylor series is i 8.9 ad you are expected to kow it for this test (though I do t promise to ask about it; there is plety to do ad the test will ivolve a lot of compressio). REVISED: o 8.9 o this test at all. This practice test cotais problems similar i spirit to those that will be o your test. I m ot worryig about how log it is, though (ad I will as much as I ca cotrol that o the real test). Be aware that actual problems o the test may cover more tha oe of these issues; otice how log the practice documet is! For example, the Ratio ad Root tests might appear oly i tests for radius of covergece of power series.

2 . 8.2 (geometric series) (a) Evaluate the series or explai why it diverges: =0 2 3 = this is a geometric series with first term a = 2, (ot 2! Notice that the first value of is 0, ot!) ad commo ratio 3 r = < (it coverges sice r < ). The sum is 2 = (b) (this might also be viewed as 8.7) Write a geometric series which 4 coverges to for suitable values of x. Explai what the rage 4 x 2 of values of x is for which it coverges. Hit: start by dividig the top ad bottom of the expressio by 4; the it will be clearer how to apply the formula! 4 = which is 4 x 2 x2 4 a r for this has first term, commo ratio x2 4. x2 for a =, r =, so a geometric series 4 So the series is + x2 4 + ( x2 4 )2 + ( x2 4 )3... = + x2 4 + x x or i=0 x 2 4. It coverges if the commo ratio is less tha, that is if x2 4 <, that is x 2 < 4, that is 2 < x < 2. 2

3 (a) Use the Itegral Test to determie whether = 2 coverges or diverges. Do ot use aythig else (yes, its a p-series, ad the Itegral Test is how we showed the results about p-series i the first place: do t say aythig about p-series, just do the Itegral Test. ) Be sure to state the facts about the fuctio x 2 which esure that the Itegral Test applies! The fuctio is positive, decreasig, ad approaches zero as x x 2 approaches ifiity, so the Itegral Test does apply. This meas that the give series coverges if ad oly if the improper itegral coverges. I ask you to actually evaluate it: this is lim b [2 x] b or lim b [2 b 2] = : sice the improper itegral diverges, the series also diverges. (b) How may terms of the series = must be added to esure that the partial sum is withi 0 6 of the exact value of the series? The upper boud o the error i N = 2 2 as a estimate is 3 N x 2 dx

4 . Evaluate this ad you will fid that it is lim b [ ( )] =. b N N < iff N > 0 6 : to get error less tha oe oe millioth, you N 0 6 eed to add more tha oe millio terms. This series does ot coverge very fast! 4

5 For each series, determie whether it coverges or diverges by compariso with a series whose covergece or divergece behavior is kow (you should quickly idicate the status of the kow series as geometric, p-series with p > or p : you must also state whether the kow series is coverget or diverget). State the iequalities you use for compariso (ot as iequalities betwee whole series whose covergece behavior you do t kow yet!). Of course you eed to give a fial aswer: does the give series coverge or diverge? You might wat to use the limit compariso test o oe or more of these. (a) (b) e < e = +e 2 e 2 e The series =0 e + e 2 =0 coverges because it is a geometric series with r = <. So the e origial series coverges by compariso because it is term by term less tha this coverget geometric series. =0 e should be very close to = for large so we wat to compare this with the diverget series (p ). Ufortuately, = <, so direct compariso doest work. Limit compariso 3 + does, though: lim ( 2 ) = lim 3 + = lim = 5

6 so our series coverges if ad oly if = coverges: we kow that the secod series diverges, so the first does as well by the limit compariso test. 6

7 Use the Ratio Test or the Root Test to settle the covergece behavior of each of these series. Show all work!!! O oe of them the Ratio ad Root Tests will be icoclusive: say so, ad explai whether the series coverges or diverges (usig some other fact about it). If you use special limits, please idicate explicitly which special limits you used. (a) (b) (c) = Ratio Test or Root Test works. I ll use the Root Test: L = lim = lim e = < so the series coverges e e by the Root Test. Use the Ratio Test. lim = e + ( + ) 2 = lim ( + 2) = The Ratio Test gives o iformatio sice L =. But this is easily see to diverge, sice lim the series diverges by the th term test. =! 3 + = 0, so Use the Ratio Test (the presece of a factorial meas we are ot likely to be able to use the Root Test). L = lim (+)! (+) 3! 3 ( + )! 3 = lim!( + ) = lim ( + ) 3 3 ( + ) = 3 7

8 Sice L diverges to ifiity, this series diverges by the Ratio Test. Also, this shows that the terms are strictly icreasig after a while, so it must diverge by the th term test as well. 8

9 5. 8.5b Give two series, both of which have covergece behavior which caot be decided by the Ratio Test or Root Test (L = ), oe of which coverges ad oe of which does ot coverge. (There are very familiar examples of this!) Show the work verifyig that the Ratio Test is icoclusive for each series ad idicate why oe coverges ad the other does ot. the obvious examples are ad = =. Both of 2 these have L = with either the ratio or root test; both are p-series, the first oe divergig ad the secod oe covergig. 9

10 Two series are give. Oe satisfies the coditios of the Alteratig Series Test ad oe does ot. Explai i each case. Particularly i the case of the oe which does satisfy the coditios, you eed to spell out the three thigs that are true (o the other oe you eed oly poit out what fails). That the three coditios hold is obvious (i the part where AST works): you do ot eed to give verificatios of them, just spell out clearly what they are. For the oe which does satisfy the coditios of the AST, determie whether it coverges absolutely ad explai why it does or does ot. You do ot eed to say aythig about whether the other oe coverges or ot. Further, how may terms of the series that does satisfy AST do you eed to add to approximate the exact value of the series withi.0? (a) (b) The coditios = ( ) (positive), (decreasig) ad lim = 0 (coverget to 0) o the terms which are alterately added ad subtracted are all obviously true, so this coverges by the AST. It coverges oly coditioally because = ( ) + = = is a diverget p-series (p = ). The error i the sum of the 2 first N terms of a alteratig series is bouded by the (N +)-th term: N+ <.0 is true iff N + > 00 that is N + > 0000 or N > You eed to add at least 0000 terms to get error less tha.0. This series has rotte covergece behavior.... = + si() ( ) 3 0

11 This oe does ot satisfy the coditios of the test, as the terms si() which are alterately added ad subtracted are ot all positive. No further work is requested by the directios, though this 3 ca be show to coverge absolutely.

12 Use the Ratio Test or Root Test to determie the radius of covergece ad iterval of covergece of each of the followig power series. Remember that you do eed to check separately the behavior of the edpoits. Remember whe usig the Ratio or Root Test that you are checkig for absolute covergece. (a) (b) = + x ( ) We use the Root Test to test for absolute covergece. L = lim x = lim x = x So the series will coverge absolutely if x < ad diverge if x >, by the Root Test. At x = ad x = we eed to check separately because the Root Test is icoclusive (L = ). coverges because it is a alteratig series. x = : = ( ) + x = : + = ( ) = = ( ) diverges because it is the result of multiplyig the diverget harmoic series by. So the radius of covergece is ad the iterval is < x. = Seeig a factorial, we thik the Ratio Test is more likely to help. x 2! L = lim x 2(+) (+)! x 2! x 2(+)! = lim x 2 ( + )! = lim x 2 + = 0 < Sice L = 0 < o matter what x is, this series coverges for all values of x. The radius of covergece is ad the iterval of ocvergece is (, ). 2

13 (c) = 2 x 2 Here we use the Root Test. L = lim 2 x 2 2x = lim ( ) = 2x 2 so L < whe 2x < that is x < that is < x <. For these values of x the series coverges absolutely. If L = 2x > the the series diverges. We oly eed to check x = ad x =. 2 2 If x = we get the series 2 = 2 which is a coverget p-series ad if x = 2 we get ( ) = 2 which coverges absolutely (drop the mius sigs ad we have the p-series agai). So the radius of covergece is 2 ad the iterval is 2 x 2. 3

14 8. 8.7b Determie the first five terms of the power series for l( + x) by writig dow a series for (this is a 8.2 problem!) the itegratig +x term by term. What is the radius ad iterval of covergece for this series (do t forget to check edpoits!) +x is would be the value of a geometric series with a =, r = x, that x + x 2 x 3 + x 4 x 5... Itegratig this term by term gives x x2 2 + x3 3 x = = ( ) + x Applyig the Ratio or Root Test shows that the radius of covergece is (Im ot repeatig this, it is very much like oe of the problems above). We eed to check for covergece at x = ad x = : at x = it coverges coditioally by AST, while at x = it diverges because it is the harmoic series times mius oe. 4

15 Compute the first five terms of the Maclauri series of (+x) 2. Show all work. f(x) = ( + x) 2 f (x) = ( 2)( + x) 3 f (x) = ( 2)( 3)( + x) 4 f (x) = ( 2)( 3)( 4)( + x) 5 f (x) = ( 2)( 3)( 4)( 5)( + x) 6 so f(0) = ( + 0) 2 = f (0) = ( 2)( + 0) 3 = 2 f (0) = ( 2)( 3)( + 0) 4 = 6 f (0) = ( 2)( 3)( 4)( + 0) 5 = 24 f (0) = ( 2)( 3)( 4)( 5)( + 0) 6 = 20 so the series is or equivaletly 2x + 6 2! x2 24 3! x ! x4... 2x + 3x 2 4x 3 + 5x 3... The obvious patter does cotiue to hold. Write dow the power series for ad multiply it by itself. Check (+x) that you get the same first few terms of a series for this way as (+x) 2 you got by computig the Taylor series. I m omittig this part: I didt put it o the test ad while it is easy to show i hadwritte form, it is very hard to typeset the solutio i LaTeX. Of course you get the same series as i the first part! 5

4x 2. (n+1) x 3 n+1. = lim. 4x 2 n+1 n3 n. n 4x 2 = lim = 3

4x 2. (n+1) x 3 n+1. = lim. 4x 2 n+1 n3 n. n 4x 2 = lim = 3 Exam Problems (x. Give the series (, fid the values of x for which this power series coverges. Also =0 state clearly what the radius of covergece is. We start by settig up the Ratio Test: x ( x x ( x x

More information

= lim. = lim. 3 dx = lim. [1 1 b 3 ]=1. 3. Determine if the following series converge or diverge. Justify your answers completely.

= lim. = lim. 3 dx = lim. [1 1 b 3 ]=1. 3. Determine if the following series converge or diverge. Justify your answers completely. MTH Lecture 2: Solutios to Practice Problems for Exam December 6, 999 (Vice Melfi) ***NOTE: I ve proofread these solutios several times, but you should still be wary for typographical (or worse) errors..

More information

Math 113 Exam 3 Practice

Math 113 Exam 3 Practice Math Exam Practice Exam will cover.-.9. This sheet has three sectios. The first sectio will remid you about techiques ad formulas that you should kow. The secod gives a umber of practice questios for you

More information

Carleton College, Winter 2017 Math 121, Practice Final Prof. Jones. Note: the exam will have a section of true-false questions, like the one below.

Carleton College, Winter 2017 Math 121, Practice Final Prof. Jones. Note: the exam will have a section of true-false questions, like the one below. Carleto College, Witer 207 Math 2, Practice Fial Prof. Joes Note: the exam will have a sectio of true-false questios, like the oe below.. True or False. Briefly explai your aswer. A icorrectly justified

More information

Testing for Convergence

Testing for Convergence 9.5 Testig for Covergece Remember: The Ratio Test: lim + If a is a series with positive terms ad the: The series coverges if L . The test is icoclusive if L =. a a = L This

More information

Math 113 Exam 4 Practice

Math 113 Exam 4 Practice Math Exam 4 Practice Exam 4 will cover.-.. This sheet has three sectios. The first sectio will remid you about techiques ad formulas that you should kow. The secod gives a umber of practice questios for

More information

Math 163 REVIEW EXAM 3: SOLUTIONS

Math 163 REVIEW EXAM 3: SOLUTIONS Math 63 REVIEW EXAM 3: SOLUTIONS These otes do ot iclude solutios to the Cocept Check o p8. They also do t cotai complete solutios to the True-False problems o those pages. Please go over these problems

More information

MIDTERM 3 CALCULUS 2. Monday, December 3, :15 PM to 6:45 PM. Name PRACTICE EXAM SOLUTIONS

MIDTERM 3 CALCULUS 2. Monday, December 3, :15 PM to 6:45 PM. Name PRACTICE EXAM SOLUTIONS MIDTERM 3 CALCULUS MATH 300 FALL 08 Moday, December 3, 08 5:5 PM to 6:45 PM Name PRACTICE EXAM S Please aswer all of the questios, ad show your work. You must explai your aswers to get credit. You will

More information

SOLUTIONS TO EXAM 3. Solution: Note that this defines two convergent geometric series with respective radii r 1 = 2/5 < 1 and r 2 = 1/5 < 1.

SOLUTIONS TO EXAM 3. Solution: Note that this defines two convergent geometric series with respective radii r 1 = 2/5 < 1 and r 2 = 1/5 < 1. SOLUTIONS TO EXAM 3 Problem Fid the sum of the followig series 2 + ( ) 5 5 2 5 3 25 2 2 This series diverges Solutio: Note that this defies two coverget geometric series with respective radii r 2/5 < ad

More information

6.3 Testing Series With Positive Terms

6.3 Testing Series With Positive Terms 6.3. TESTING SERIES WITH POSITIVE TERMS 307 6.3 Testig Series With Positive Terms 6.3. Review of what is kow up to ow I theory, testig a series a i for covergece amouts to fidig the i= sequece of partial

More information

Series Review. a i converges if lim. i=1. a i. lim S n = lim i=1. 2 k(k + 2) converges. k=1. k=1

Series Review. a i converges if lim. i=1. a i. lim S n = lim i=1. 2 k(k + 2) converges. k=1. k=1 Defiitio: We say that the series S = Series Review i= a i is the sum of the first terms. i= a i coverges if lim S exists ad is fiite, where The above is the defiitio of covergece for series. order to see

More information

SCORE. Exam 2. MA 114 Exam 2 Fall 2016

SCORE. Exam 2. MA 114 Exam 2 Fall 2016 MA 4 Exam Fall 06 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

More information

Chapter 10: Power Series

Chapter 10: Power Series Chapter : Power Series 57 Chapter Overview: Power Series The reaso series are part of a Calculus course is that there are fuctios which caot be itegrated. All power series, though, ca be itegrated because

More information

Math 132, Fall 2009 Exam 2: Solutions

Math 132, Fall 2009 Exam 2: Solutions Math 3, Fall 009 Exam : Solutios () a) ( poits) Determie for which positive real umbers p, is the followig improper itegral coverget, ad for which it is diverget. Evaluate the itegral for each value of

More information

MTH 142 Exam 3 Spr 2011 Practice Problem Solutions 1

MTH 142 Exam 3 Spr 2011 Practice Problem Solutions 1 MTH 42 Exam 3 Spr 20 Practice Problem Solutios No calculators will be permitted at the exam. 3. A pig-pog ball is lauched straight up, rises to a height of 5 feet, the falls back to the lauch poit ad bouces

More information

SCORE. Exam 2. MA 114 Exam 2 Fall 2017

SCORE. Exam 2. MA 114 Exam 2 Fall 2017 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

More information

The Interval of Convergence for a Power Series Examples

The Interval of Convergence for a Power Series Examples The Iterval of Covergece for a Power Series Examples To review the process: How to Test a Power Series for Covergece. Fid the iterval where the series coverges absolutely. We have to use the Ratio or Root

More information

SCORE. Exam 2. MA 114 Exam 2 Fall 2016

SCORE. Exam 2. MA 114 Exam 2 Fall 2016 Exam 2 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

More information

SUMMARY OF SEQUENCES AND SERIES

SUMMARY OF SEQUENCES AND SERIES SUMMARY OF SEQUENCES AND SERIES Importat Defiitios, Results ad Theorems for Sequeces ad Series Defiitio. A sequece {a } has a limit L ad we write lim a = L if for every ɛ > 0, there is a correspodig iteger

More information

CHAPTER 10 INFINITE SEQUENCES AND SERIES

CHAPTER 10 INFINITE SEQUENCES AND SERIES CHAPTER 10 INFINITE SEQUENCES AND SERIES 10.1 Sequeces 10.2 Ifiite Series 10.3 The Itegral Tests 10.4 Compariso Tests 10.5 The Ratio ad Root Tests 10.6 Alteratig Series: Absolute ad Coditioal Covergece

More information

INFINITE SEQUENCES AND SERIES

INFINITE SEQUENCES AND SERIES INFINITE SEQUENCES AND SERIES INFINITE SEQUENCES AND SERIES I geeral, it is difficult to fid the exact sum of a series. We were able to accomplish this for geometric series ad the series /[(+)]. This is

More information

PLEASE MARK YOUR ANSWERS WITH AN X, not a circle! 1. (a) (b) (c) (d) (e) 3. (a) (b) (c) (d) (e) 5. (a) (b) (c) (d) (e) 7. (a) (b) (c) (d) (e)

PLEASE MARK YOUR ANSWERS WITH AN X, not a circle! 1. (a) (b) (c) (d) (e) 3. (a) (b) (c) (d) (e) 5. (a) (b) (c) (d) (e) 7. (a) (b) (c) (d) (e) Math 0560, Exam 3 November 6, 07 The Hoor Code is i effect for this examiatio. All work is to be your ow. No calculators. The exam lasts for hour ad 5 mi. Be sure that your ame is o every page i case pages

More information

Please do NOT write in this box. Multiple Choice. Total

Please do NOT write in this box. Multiple Choice. Total Istructor: Math 0560, Worksheet Alteratig Series Jauary, 3000 For realistic exam practice solve these problems without lookig at your book ad without usig a calculator. Multiple choice questios should

More information

Z ß cos x + si x R du We start with the substitutio u = si(x), so du = cos(x). The itegral becomes but +u we should chage the limits to go with the ew

Z ß cos x + si x R du We start with the substitutio u = si(x), so du = cos(x). The itegral becomes but +u we should chage the limits to go with the ew Problem ( poits) Evaluate the itegrals Z p x 9 x We ca draw a right triagle labeled this way x p x 9 From this we ca read off x = sec, so = sec ta, ad p x 9 = R ta. Puttig those pieces ito the itegralrwe

More information

Math 152 Exam 3, Fall 2005

Math 152 Exam 3, Fall 2005 c IIT Dept. Applied Mathematics, December, 005 PRINT Last ame: KEY First ame: Sigature: Studet ID: Math 5 Exam 3, Fall 005 Istructios. For the multiple choice problems, there is o partial credit. For the

More information

Math 113 Exam 3 Practice

Math 113 Exam 3 Practice Math Exam Practice Exam 4 will cover.-., 0. ad 0.. Note that eve though. was tested i exam, questios from that sectios may also be o this exam. For practice problems o., refer to the last review. This

More information

In this section, we show how to use the integral test to decide whether a series

In this section, we show how to use the integral test to decide whether a series Itegral Test Itegral Test Example Itegral Test Example p-series Compariso Test Example Example 2 Example 3 Example 4 Example 5 Exa Itegral Test I this sectio, we show how to use the itegral test to decide

More information

11.6 Absolute Convergence and the Ratio and Root Tests

11.6 Absolute Convergence and the Ratio and Root Tests .6 Absolute Covergece ad the Ratio ad Root Tests The most commo way to test for covergece is to igore ay positive or egative sigs i a series, ad simply test the correspodig series of positive terms. Does

More information

Chapter 6 Infinite Series

Chapter 6 Infinite Series Chapter 6 Ifiite Series I the previous chapter we cosidered itegrals which were improper i the sese that the iterval of itegratio was ubouded. I this chapter we are goig to discuss a topic which is somewhat

More information

(a) (b) All real numbers. (c) All real numbers. (d) None. to show the. (a) 3. (b) [ 7, 1) (c) ( 7, 1) (d) At x = 7. (a) (b)

(a) (b) All real numbers. (c) All real numbers. (d) None. to show the. (a) 3. (b) [ 7, 1) (c) ( 7, 1) (d) At x = 7. (a) (b) Chapter 0 Review 597. E; a ( + )( + ) + + S S + S + + + + + + S lim + l. D; a diverges by the Itegral l k Test sice d lim [(l ) ], so k l ( ) does ot coverge absolutely. But it coverges by the Alteratig

More information

Convergence: nth-term Test, Comparing Non-negative Series, Ratio Test

Convergence: nth-term Test, Comparing Non-negative Series, Ratio Test Covergece: th-term Test, Comparig No-egative Series, Ratio Test Power Series ad Covergece We have writte statemets like: l + x = x x2 + x3 2 3 + x + But we have ot talked i depth about what values of x

More information

Math 113, Calculus II Winter 2007 Final Exam Solutions

Math 113, Calculus II Winter 2007 Final Exam Solutions Math, Calculus II Witer 7 Fial Exam Solutios (5 poits) Use the limit defiitio of the defiite itegral ad the sum formulas to compute x x + dx The check your aswer usig the Evaluatio Theorem Solutio: I this

More information

Solutions to Practice Midterms. Practice Midterm 1

Solutions to Practice Midterms. Practice Midterm 1 Solutios to Practice Midterms Practice Midterm. a False. Couterexample: a =, b = b False. Couterexample: a =, b = c False. Couterexample: c = Y cos. = cos. + 5 = 0 sice both its exist. + 5 cos π. 5 + 5

More information

Chapter 6: Numerical Series

Chapter 6: Numerical Series Chapter 6: Numerical Series 327 Chapter 6 Overview: Sequeces ad Numerical Series I most texts, the topic of sequeces ad series appears, at first, to be a side topic. There are almost o derivatives or itegrals

More information

Calculus 2 - D. Yuen Final Exam Review (Version 11/22/2017. Please report any possible typos.)

Calculus 2 - D. Yuen Final Exam Review (Version 11/22/2017. Please report any possible typos.) Calculus - D Yue Fial Eam Review (Versio //7 Please report ay possible typos) NOTE: The review otes are oly o topics ot covered o previous eams See previous review sheets for summary of previous topics

More information

Chapter 6 Overview: Sequences and Numerical Series. For the purposes of AP, this topic is broken into four basic subtopics:

Chapter 6 Overview: Sequences and Numerical Series. For the purposes of AP, this topic is broken into four basic subtopics: Chapter 6 Overview: Sequeces ad Numerical Series I most texts, the topic of sequeces ad series appears, at first, to be a side topic. There are almost o derivatives or itegrals (which is what most studets

More information

Chapter 7: Numerical Series

Chapter 7: Numerical Series Chapter 7: Numerical Series Chapter 7 Overview: Sequeces ad Numerical Series I most texts, the topic of sequeces ad series appears, at first, to be a side topic. There are almost o derivatives or itegrals

More information

AP Calculus Chapter 9: Infinite Series

AP Calculus Chapter 9: Infinite Series AP Calculus Chapter 9: Ifiite Series 9. Sequeces a, a 2, a 3, a 4, a 5,... Sequece: A fuctio whose domai is the set of positive itegers = 2 3 4 a = a a 2 a 3 a 4 terms of the sequece Begi with the patter

More information

Math 116 Practice for Exam 3

Math 116 Practice for Exam 3 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

More information

The Ratio Test. THEOREM 9.17 Ratio Test Let a n be a series with nonzero terms. 1. a. n converges absolutely if lim. n 1

The Ratio Test. THEOREM 9.17 Ratio Test Let a n be a series with nonzero terms. 1. a. n converges absolutely if lim. n 1 460_0906.qxd //04 :8 PM Page 69 SECTION 9.6 The Ratio ad Root Tests 69 Sectio 9.6 EXPLORATION Writig a Series Oe of the followig coditios guaratees that a series will diverge, two coditios guaratee that

More information

Physics 116A Solutions to Homework Set #1 Winter Boas, problem Use equation 1.8 to find a fraction describing

Physics 116A Solutions to Homework Set #1 Winter Boas, problem Use equation 1.8 to find a fraction describing Physics 6A Solutios to Homework Set # Witer 0. Boas, problem. 8 Use equatio.8 to fid a fractio describig 0.694444444... Start with the formula S = a, ad otice that we ca remove ay umber of r fiite decimals

More information

INFINITE SEQUENCES AND SERIES

INFINITE SEQUENCES AND SERIES 11 INFINITE SEQUENCES AND SERIES INFINITE SEQUENCES AND SERIES 11.4 The Compariso Tests I this sectio, we will lear: How to fid the value of a series by comparig it with a kow series. COMPARISON TESTS

More information

Math 106 Fall 2014 Exam 3.2 December 10, 2014

Math 106 Fall 2014 Exam 3.2 December 10, 2014 Math 06 Fall 04 Exam 3 December 0, 04 Determie if the series is coverget or diverget by makig a compariso (DCT or LCT) with a suitable b Fill i the blaks with your aswer For Coverget or Diverget write

More information

MTH 133 Solutions to Exam 2 Nov. 18th 2015

MTH 133 Solutions to Exam 2 Nov. 18th 2015 Name: Sectio: Recitatio Istructor: READ THE FOLLOWING INSTRUCTIONS. Do ot ope your exam util told to do so. No calculators, cell phoes or ay other electroic devices ca be used o this exam. Clear your desk

More information

f(x) dx as we do. 2x dx x also diverges. Solution: We compute 2x dx lim

f(x) dx as we do. 2x dx x also diverges. Solution: We compute 2x dx lim Math 3, Sectio 2. (25 poits) Why we defie f(x) dx as we do. (a) Show that the improper itegral diverges. Hece the improper itegral x 2 + x 2 + b also diverges. Solutio: We compute x 2 + = lim b x 2 + =

More information

Math 113 (Calculus 2) Section 12 Exam 4

Math 113 (Calculus 2) Section 12 Exam 4 Name: Row: Math Calculus ) Sectio Exam 4 8 0 November 00 Istructios:. Work o scratch paper will ot be graded.. For questio ad questios 0 through 5, show all your work i the space provided. Full credit

More information

Alternating Series. 1 n 0 2 n n THEOREM 9.14 Alternating Series Test Let a n > 0. The alternating series. 1 n a n.

Alternating Series. 1 n 0 2 n n THEOREM 9.14 Alternating Series Test Let a n > 0. The alternating series. 1 n a n. 0_0905.qxd //0 :7 PM Page SECTION 9.5 Alteratig Series Sectio 9.5 Alteratig Series Use the Alteratig Series Test to determie whether a ifiite series coverges. Use the Alteratig Series Remaider to approximate

More information

( 1) n (4x + 1) n. n=0

( 1) n (4x + 1) n. n=0 Problem 1 (10.6, #). Fid the radius of covergece for the series: ( 1) (4x + 1). For what values of x does the series coverge absolutely, ad for what values of x does the series coverge coditioally? Solutio.

More information

MH1101 AY1617 Sem 2. Question 1. NOT TESTED THIS TIME

MH1101 AY1617 Sem 2. Question 1. NOT TESTED THIS TIME MH AY67 Sem Questio. NOT TESTED THIS TIME ( marks Let R be the regio bouded by the curve y 4x x 3 ad the x axis i the first quadrat (see figure below. Usig the cylidrical shell method, fid the volume of

More information

ENGI Series Page 6-01

ENGI Series Page 6-01 ENGI 3425 6 Series Page 6-01 6. Series Cotets: 6.01 Sequeces; geeral term, limits, covergece 6.02 Series; summatio otatio, covergece, divergece test 6.03 Stadard Series; telescopig series, geometric series,

More information

1 Introduction to Sequences and Series, Part V

1 Introduction to Sequences and Series, Part V MTH 22 Calculus II Essex Couty College Divisio of Mathematics ad Physics Lecture Notes #8 Sakai Web Project Material Itroductio to Sequeces ad Series, Part V. The compariso test that we used prior, relies

More information

5.6 Absolute Convergence and The Ratio and Root Tests

5.6 Absolute Convergence and The Ratio and Root Tests 5.6 Absolute Covergece ad The Ratio ad Root Tests Bria E. Veitch 5.6 Absolute Covergece ad The Ratio ad Root Tests Recall from our previous sectio that diverged but ( ) coverged. Both of these sequeces

More information

Math 106 Fall 2014 Exam 3.1 December 10, 2014

Math 106 Fall 2014 Exam 3.1 December 10, 2014 Math 06 Fall 0 Exam 3 December 0, 0 Determie if the series is coverget or diverget by makig a compariso DCT or LCT) with a suitable b Fill i the blaks with your aswer For Coverget or Diverget write Coverget

More information

Mathematics 116 HWK 21 Solutions 8.2 p580

Mathematics 116 HWK 21 Solutions 8.2 p580 Mathematics 6 HWK Solutios 8. p580 A abbreviatio: iff is a abbreviatio for if ad oly if. Geometric Series: Several of these problems use what we worked out i class cocerig the geometric series, which I

More information

10.6 ALTERNATING SERIES

10.6 ALTERNATING SERIES 0.6 Alteratig Series Cotemporary Calculus 0.6 ALTERNATING SERIES I the last two sectios we cosidered tests for the covergece of series whose terms were all positive. I this sectio we examie series whose

More information

MAT1026 Calculus II Basic Convergence Tests for Series

MAT1026 Calculus II Basic Convergence Tests for Series MAT026 Calculus II Basic Covergece Tests for Series Egi MERMUT 202.03.08 Dokuz Eylül Uiversity Faculty of Sciece Departmet of Mathematics İzmir/TURKEY Cotets Mootoe Covergece Theorem 2 2 Series of Real

More information

Section 11.6 Absolute and Conditional Convergence, Root and Ratio Tests

Section 11.6 Absolute and Conditional Convergence, Root and Ratio Tests Sectio.6 Absolute ad Coditioal Covergece, Root ad Ratio Tests I this chapter we have see several examples of covergece tests that oly apply to series whose terms are oegative. I this sectio, we will lear

More information

BC: Q401.CH9A Convergent and Divergent Series (LESSON 1)

BC: Q401.CH9A Convergent and Divergent Series (LESSON 1) BC: Q40.CH9A Coverget ad Diverget Series (LESSON ) INTRODUCTION Sequece Notatio: a, a 3, a,, a, Defiitio: A sequece is a fuctio f whose domai is the set of positive itegers. Defiitio: A ifiite series (or

More information

An alternating series is a series where the signs alternate. Generally (but not always) there is a factor of the form ( 1) n + 1

An alternating series is a series where the signs alternate. Generally (but not always) there is a factor of the form ( 1) n + 1 Calculus II - Problem Solvig Drill 20: Alteratig Series, Ratio ad Root Tests Questio No. of 0 Istructios: () Read the problem ad aswer choices carefully (2) Work the problems o paper as eeded (3) Pick

More information

Part I: Covers Sequence through Series Comparison Tests

Part I: Covers Sequence through Series Comparison Tests Part I: Covers Sequece through Series Compariso Tests. Give a example of each of the followig: (a) A geometric sequece: (b) A alteratig sequece: (c) A sequece that is bouded, but ot coverget: (d) A sequece

More information

Solutions to Final Exam Review Problems

Solutions to Final Exam Review Problems . Let f(x) 4+x. Solutios to Fial Exam Review Problems Math 5C, Witer 2007 (a) Fid the Maclauri series for f(x), ad compute its radius of covergece. Solutio. f(x) 4( ( x/4)) ( x/4) ( ) 4 4 + x. Sice the

More information

Sec 8.4. Alternating Series Test. A. Before Class Video Examples. Example 1: Determine whether the following series is convergent or divergent.

Sec 8.4. Alternating Series Test. A. Before Class Video Examples. Example 1: Determine whether the following series is convergent or divergent. Sec 8.4 Alteratig Series Test A. Before Class Video Examples Example 1: Determie whether the followig series is coverget or diverget. a) ( 1)+1 =1 b) ( 1) 2 1 =1 Example 2: Determie whether the followig

More information

9.3 The INTEGRAL TEST; p-series

9.3 The INTEGRAL TEST; p-series Lecture 9.3 & 9.4 Math 0B Nguye of 6 Istructor s Versio 9.3 The INTEGRAL TEST; p-series I this ad the followig sectio, you will study several covergece tests that apply to series with positive terms. Note

More information

Review for Test 3 Math 1552, Integral Calculus Sections 8.8,

Review for Test 3 Math 1552, Integral Calculus Sections 8.8, Review for Test 3 Math 55, Itegral Calculus Sectios 8.8, 0.-0.5. Termiology review: complete the followig statemets. (a) A geometric series has the geeral form k=0 rk.theseriescovergeswhe r is less tha

More information

62. Power series Definition 16. (Power series) Given a sequence {c n }, the series. c n x n = c 0 + c 1 x + c 2 x 2 + c 3 x 3 +

62. Power series Definition 16. (Power series) Given a sequence {c n }, the series. c n x n = c 0 + c 1 x + c 2 x 2 + c 3 x 3 + 62. Power series Defiitio 16. (Power series) Give a sequece {c }, the series c x = c 0 + c 1 x + c 2 x 2 + c 3 x 3 + is called a power series i the variable x. The umbers c are called the coefficiets of

More information

d) If the sequence of partial sums converges to a limit L, we say that the series converges and its

d) If the sequence of partial sums converges to a limit L, we say that the series converges and its Ifiite Series. Defiitios & covergece Defiitio... Let {a } be a sequece of real umbers. a) A expressio of the form a + a +... + a +... is called a ifiite series. b) The umber a is called as the th term

More information

11.6 Absolute Convrg. (Ratio & Root Tests) & 11.7 Strategy for Testing Series

11.6 Absolute Convrg. (Ratio & Root Tests) & 11.7 Strategy for Testing Series 11.6 Absolute Covrg. (Ratio & Root Tests) & 11.7 Strategy for Testig Series http://screecast.com/t/ri3unwu84 Give ay series Σ a, we ca cosider the correspodig series 1 a a a a 1 2 3 whose terms are the

More information

Read carefully the instructions on the answer book and make sure that the particulars required are entered on each answer book.

Read carefully the instructions on the answer book and make sure that the particulars required are entered on each answer book. THE UNIVERSITY OF WARWICK FIRST YEAR EXAMINATION: Jauary 2009 Aalysis I Time Allowed:.5 hours Read carefully the istructios o the aswer book ad make sure that the particulars required are etered o each

More information

MTH 246 TEST 3 April 4, 2014

MTH 246 TEST 3 April 4, 2014 MTH 26 TEST April, 20 (PLEASE PRINT YOUR NAME!!) Name:. (6 poits each) Evaluate lim! a for the give sequece fa g. (a) a = 2 2 5 2 5 (b) a = 2 7 2. (6 poits) Fid the sum of the telescopig series p p 2.

More information

1. C only. 3. none of them. 4. B only. 5. B and C. 6. all of them. 7. A and C. 8. A and B correct

1. C only. 3. none of them. 4. B only. 5. B and C. 6. all of them. 7. A and C. 8. A and B correct M408D (54690/54695/54700), Midterm # Solutios Note: Solutios to the multile-choice questios for each sectio are listed below. Due to radomizatio betwee sectios, exlaatios to a versio of each of the multile-choice

More information

Are the following series absolutely convergent? n=1. n 3. n=1 n. ( 1) n. n=1 n=1

Are the following series absolutely convergent? n=1. n 3. n=1 n. ( 1) n. n=1 n=1 Absolute covergece Defiitio A series P a is called absolutely coverget if the series of absolute values P a is coverget. If the terms of the series a are positive, absolute covergece is the same as covergece.

More information

Sequences. Notation. Convergence of a Sequence

Sequences. Notation. Convergence of a Sequence Sequeces A sequece is essetially just a list. Defiitio (Sequece of Real Numbers). A sequece of real umbers is a fuctio Z (, ) R for some real umber. Do t let the descriptio of the domai cofuse you; it

More information

Math 142, Final Exam. 5/2/11.

Math 142, Final Exam. 5/2/11. Math 4, Fial Exam 5// No otes, calculator, or text There are poits total Partial credit may be give Write your full ame i the upper right corer of page Number the pages i the upper right corer Do problem

More information

Math 116 Practice for Exam 3

Math 116 Practice for Exam 3 Math 6 Practice for Eam 3 Geerated April 4, 26 Name: SOLUTIONS Istructor: Sectio Number:. This eam has questios. Note that the problems are ot of equal difficulty, so you may wat to skip over ad retur

More information

JANE PROFESSOR WW Prob Lib1 Summer 2000

JANE PROFESSOR WW Prob Lib1 Summer 2000 JANE PROFESSOR WW Prob Lib Summer 000 Sample WeBWorK problems. WeBWorK assigmet Series6CompTests due /6/06 at :00 AM..( pt) Test each of the followig series for covergece by either the Compariso Test or

More information

Calculus with Analytic Geometry 2

Calculus with Analytic Geometry 2 Calculus with Aalytic Geometry Fial Eam Study Guide ad Sample Problems Solutios The date for the fial eam is December, 7, 4-6:3p.m. BU Note. The fial eam will cosist of eercises, ad some theoretical questios,

More information

Notice that this test does not say anything about divergence of an alternating series.

Notice that this test does not say anything about divergence of an alternating series. MATH 572H Sprig 20 Worksheet 7 Topics: absolute ad coditioal covergece; power series. Defiitio. A series b is called absolutely coverget if the series b is coverget. If the series b coverges, while b diverges,

More information

Strategy for Testing Series

Strategy for Testing Series Strategy for Testig Series We ow have several ways of testig a series for covergece or divergece; the problem is to decide which test to use o which series. I this respect testig series is similar to itegratig

More information

Math F15 Rahman

Math F15 Rahman Math 2-009 F5 Rahma Week 0.9 Covergece of Taylor Series Sice we have so may examples for these sectios ad it s usually a simple matter of recallig the formula ad pluggig i for it, I ll simply provide the

More information

Series III. Chapter Alternating Series

Series III. Chapter Alternating Series Chapter 9 Series III With the exceptio of the Null Sequece Test, all the tests for series covergece ad divergece that we have cosidered so far have dealt oly with series of oegative terms. Series with

More information

Ma 530 Infinite Series I

Ma 530 Infinite Series I Ma 50 Ifiite Series I Please ote that i additio to the material below this lecture icorporated material from the Visual Calculus web site. The material o sequeces is at Visual Sequeces. (To use this li

More information

Math 19B Final. Study Aid. June 6, 2011

Math 19B Final. Study Aid. June 6, 2011 Math 9B Fial Study Aid Jue 6, 20 Geeral advise. Get plety of practice. There s a lot of material i this sectio - try to do as may examples ad as much of the homework as possible to get some practice. Just

More information

Roberto s Notes on Series Chapter 2: Convergence tests Section 7. Alternating series

Roberto s Notes on Series Chapter 2: Convergence tests Section 7. Alternating series Roberto s Notes o Series Chapter 2: Covergece tests Sectio 7 Alteratig series What you eed to kow already: All basic covergece tests for evetually positive series. What you ca lear here: A test for series

More information

MTH 122 Calculus II Essex County College Division of Mathematics and Physics 1 Lecture Notes #20 Sakai Web Project Material

MTH 122 Calculus II Essex County College Division of Mathematics and Physics 1 Lecture Notes #20 Sakai Web Project Material MTH 1 Calculus II Essex Couty College Divisio of Mathematics ad Physics 1 Lecture Notes #0 Sakai Web Project Material 1 Power Series 1 A power series is a series of the form a x = a 0 + a 1 x + a x + a

More information

Taylor Series (BC Only)

Taylor Series (BC Only) Studet Study Sessio Taylor Series (BC Oly) Taylor series provide a way to fid a polyomial look-alike to a o-polyomial fuctio. This is doe by a specific formula show below (which should be memorized): Taylor

More information

REVIEW 1, MATH n=1 is convergent. (b) Determine whether a n is convergent.

REVIEW 1, MATH n=1 is convergent. (b) Determine whether a n is convergent. REVIEW, MATH 00. Let a = +. a) Determie whether the sequece a ) is coverget. b) Determie whether a is coverget.. Determie whether the series is coverget or diverget. If it is coverget, fid its sum. a)

More information

Midterm Exam #2. Please staple this cover and honor pledge atop your solutions.

Midterm Exam #2. Please staple this cover and honor pledge atop your solutions. Math 50B Itegral Calculus April, 07 Midterm Exam # Name: Aswer Key David Arold Istructios. (00 poits) This exam is ope otes, ope book. This icludes ay supplemetary texts or olie documets. You are ot allowed

More information

MATH 2300 review problems for Exam 2

MATH 2300 review problems for Exam 2 MATH 2300 review problems for Exam 2. A metal plate of costat desity ρ (i gm/cm 2 ) has a shape bouded by the curve y = x, the x-axis, ad the lie x =. (a) Fid the mass of the plate. Iclude uits. Mass =

More information

Section 11.8: Power Series

Section 11.8: Power Series Sectio 11.8: Power Series 1. Power Series I this sectio, we cosider geeralizig the cocept of a series. Recall that a series is a ifiite sum of umbers a. We ca talk about whether or ot it coverges ad i

More information

Calculus BC and BCD Drill on Sequences and Series!!! By Susan E. Cantey Walnut Hills H.S. 2006

Calculus BC and BCD Drill on Sequences and Series!!! By Susan E. Cantey Walnut Hills H.S. 2006 Calculus BC ad BCD Drill o Sequeces ad Series!!! By Susa E. Catey Walut Hills H.S. 2006 Sequeces ad Series I m goig to ask you questios about sequeces ad series ad drill you o some thigs that eed to be

More information

Section 1.4. Power Series

Section 1.4. Power Series Sectio.4. Power Series De itio. The fuctio de ed by f (x) (x a) () c 0 + c (x a) + c 2 (x a) 2 + c (x a) + ::: is called a power series cetered at x a with coe ciet sequece f g :The domai of this fuctio

More information

Solutions to quizzes Math Spring 2007

Solutions to quizzes Math Spring 2007 to quizzes Math 4- Sprig 7 Name: Sectio:. Quiz a) x + x dx b) l x dx a) x + dx x x / + x / dx (/3)x 3/ + x / + c. b) Set u l x, dv dx. The du /x ad v x. By Itegratio by Parts, x(/x)dx x l x x + c. l x

More information

MTH 133 Solutions to Exam 2 November 16th, Without fully opening the exam, check that you have pages 1 through 12.

MTH 133 Solutions to Exam 2 November 16th, Without fully opening the exam, check that you have pages 1 through 12. Name: Sectio: Recitatio Istructor: INSTRUCTIONS Fill i your ame, etc. o this first page. Without fully opeig the exam, check that you have pages through. Show all your work o the stadard respose questios.

More information

Power Series: A power series about the center, x = 0, is a function of x of the form

Power Series: A power series about the center, x = 0, is a function of x of the form You are familiar with polyomial fuctios, polyomial that has ifiitely may terms. 2 p ( ) a0 a a 2 a. A power series is just a Power Series: A power series about the ceter, = 0, is a fuctio of of the form

More information

Math 21, Winter 2018 Schaeffer/Solis Stanford University Solutions for 20 series from Lecture 16 notes (Schaeffer)

Math 21, Winter 2018 Schaeffer/Solis Stanford University Solutions for 20 series from Lecture 16 notes (Schaeffer) Math, Witer 08 Schaeffer/Solis Staford Uiversity Solutios for 0 series from Lecture 6 otes (Schaeffer) a. r 4 +3 The series has algebraic terms (polyomials, ratioal fuctios, ad radicals, oly), so the test

More information

n n 2 n n + 1 +

n n 2 n n + 1 + Istructor: Marius Ioescu 1. Let a =. (5pts) (a) Prove that for every ε > 0 there is N 1 such that a +1 a < ε if N. Solutio: Let ε > 0. The a +1 a < ε is equivalet with + 1 < ε. Simplifyig, this iequality

More information

MATH 2300 review problems for Exam 2

MATH 2300 review problems for Exam 2 MATH 2300 review problems for Exam 2. A metal plate of costat desity ρ (i gm/cm 2 ) has a shape bouded by the curve y = x, the x-axis, ad the lie x =. Fid the mass of the plate. Iclude uits. (b) Fid the

More information

Seunghee Ye Ma 8: Week 5 Oct 28

Seunghee Ye Ma 8: Week 5 Oct 28 Week 5 Summary I Sectio, we go over the Mea Value Theorem ad its applicatios. I Sectio 2, we will recap what we have covered so far this term. Topics Page Mea Value Theorem. Applicatios of the Mea Value

More information