Chapter 6: Numerical Series

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

Chapter 7: Numerical Series

Chapter 10: Power Series

6.3 Testing Series With Positive Terms

Testing for Convergence

Math 113 Exam 3 Practice

CHAPTER 10 INFINITE SEQUENCES AND SERIES

INFINITE SEQUENCES AND SERIES

Infinite Sequences and Series

Part I: Covers Sequence through Series Comparison Tests

MAT1026 Calculus II Basic Convergence Tests for Series

SCORE. Exam 2. MA 114 Exam 2 Fall 2016

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

Math 106 Fall 2014 Exam 3.2 December 10, 2014

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

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

SCORE. Exam 2. MA 114 Exam 2 Fall 2016

9.3 The INTEGRAL TEST; p-series

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

SUMMARY OF SEQUENCES AND SERIES

Math 106 Fall 2014 Exam 3.1 December 10, 2014

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

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.

Practice Test Problems for Test IV, with Solutions

INFINITE SEQUENCES AND SERIES

Math 113 Exam 4 Practice

AP Calculus Chapter 9: Infinite Series

5 Sequences and Series

Chapter 6 Infinite Series

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 +

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.

JANE PROFESSOR WW Prob Lib1 Summer 2000

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 III. Chapter Alternating Series

The Interval of Convergence for a Power Series Examples

10.6 ALTERNATING SERIES

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

7 Sequences of real numbers

SCORE. Exam 2. MA 114 Exam 2 Fall 2017

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

11.6 Absolute Convergence and the Ratio and Root Tests

MTH 142 Exam 3 Spr 2011 Practice Problem Solutions 1

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

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

Section 11.8: Power Series

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

MATH 2300 review problems for Exam 2

Section 5.5. Infinite Series: The Ratio Test

10.5 Positive Term Series: Comparison Tests Contemporary Calculus 1

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

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

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

MTH 246 TEST 3 April 4, 2014

n=1 a n is the sequence (s n ) n 1 n=1 a n converges to s. We write a n = s, n=1 n=1 a n

Strategy for Testing Series

Math 113 Exam 3 Practice

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

ENGI Series Page 6-01

Math 163 REVIEW EXAM 3: SOLUTIONS

Ma 530 Infinite Series I

Sequences. Notation. Convergence of a Sequence

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

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.

Math 132, Fall 2009 Exam 2: Solutions

Section 11.6 Absolute and Conditional Convergence, Root and Ratio Tests

Not for reproduction

Solutions to Homework 7

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

n 3 ln n n ln n is convergent by p-series for p = 2 > 1. n2 Therefore we can apply Limit Comparison Test to determine lutely convergent.

5.6 Absolute Convergence and The Ratio and Root Tests

INFINITE SERIES PROBLEMS-SOLUTIONS. 3 n and 1. converges by the Comparison Test. and. ( 8 ) 2 n. 4 n + 2. n n = 4 lim 1

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

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

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

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

1 Lecture 2: Sequence, Series and power series (8/14/2012)

A sequence of numbers is a function whose domain is the positive integers. We can see that the sequence

MATH 166 TEST 3 REVIEW SHEET

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

Math 116 Practice for Exam 3

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

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

The Comparison Tests. Examples. math 131 infinite series, part iii: comparison tests 18

MA131 - Analysis 1. Workbook 9 Series III

Quiz No. 1. ln n n. 1. Define: an infinite sequence A function whose domain is N 2. Define: a convergent sequence A sequence that has a limit

THE INTEGRAL TEST AND ESTIMATES OF SUMS

Additional Notes on Power Series

We are mainly going to be concerned with power series in x, such as. (x)} converges - that is, lims N n

Fall 2018 Exam 2 PIN: 17 INSTRUCTIONS

E. Incorrect! Plug n = 1, 2, 3, & 4 into the general term formula. n =, then the first four terms are found by

P-SERIES AND INTEGRAL TEST

Sequences A sequence of numbers is a function whose domain is the positive integers. We can see that the sequence

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

Sequences, Series, and All That

Sequences. A Sequence is a list of numbers written in order.

(A sequence also can be thought of as the list of function values attained for a function f :ℵ X, where f (n) = x n for n 1.) x 1 x N +k x N +4 x 3

a 3, a 4, ... are the terms of the sequence. The number a n is the nth term of the sequence, and the entire sequence is denoted by a n

x a x a Lecture 2 Series (See Chapter 1 in Boas)

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)

Transcription:

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 (which is what most studets thik all Calculus is comprised of) ad those that do appear do so at the ed of the chapter. The sese of discoectedess is heighteed by the fact that most Calculus studets have ot see series sice freshma year (ad, the, oly briefly). For the purposes of AP, this topic is broke ito four basic subtopics: Numerical sequeces ad series Radius ad iterval of covergece for a Power Series Taylor polyomials Creatig a ew series from a old oe This is the order most texts use, ad all topics are i oe chapter, which ca be overwhelmig. The fial subtopic is where the derivatives ad itegrals reappear, built o the theoretical foudatios laid by the first three. We are optig to break the topic of sequeces ad series ito two chapters. This chapter covers covergece ad divergece of umerical sequeces ad series. The other chapter will cover the material related to Power Series ad occurs at the ed of the curriculum. What we will cosider i Calculus is the issue of whether a sequece or series coverges or ot. There are several processes to test if a series is coverget or ot. There are seve Tests to prove covergece or divergece of a umerical series.. The Divergece (th Term) Test 2. The Compariso Test 3. The Limit Compariso Test 4. The Itegral Test 5. The Ratio Test 6. The th Root Test 7. The Alteratig Series Test 328

Each may be easier or more difficult depedig o the series beig tested. Obviously, we will ot cover #4 i this chapter, sice we have ot leared Itegratio yet. The rest rely o Limits-at-Ifiity, ad they provide a good cotext withi which to practice that material. Most of these Tests apply to series comprised of positive values. The Alteratig Series Test applies to series i which the terms alterate sigs. Though each sectio itroduces a ew test, the homework i each sectio will apply to all the tests that were itroduced previously. 329

6.: Sequece ad Series Algebra Review Let s have a quick review from Algebra 2: A sequece is simply a list of umbers. Sequeces are oly iterestig to a mathematicia whe there is a patter withi the umbers. For example:, 3, 5, 7, 4, 7, 0, -7, -4,, 3, 9, 27, 4, 2,, 2, 4,... I Algebra, a sequece is writte as,, 2, 3, 5, 8, 3, a, a 2, a 3, a 4, a 5,..., a,... The geeral terms i the sequece have a subscript idicatig the cardiality--that is, which place they are i the sequece. a is the first umber i the sequece, a 2 is the secod, ad so o. a is the th umber. Note that was always a coutig umber. I calculus ad physics, this will chage i that the subscript may be tied to the expoet ad the series will geerally be writte a 0, a, a 2, a 3, a 4,..., a,... Sequece-- a fuctio with the Natural Numbers as the domai. Arithmetic Sequece-- A sequece i which oe term equals a costat added to the precedig term, i.e. a + = a + ( )d. Geometric Sequece-- A sequece i which oe term equals a costat multiplied to the precedig term, i.e. a = a r. Series--the sum of a sequece. Much of what we did with sequeces i Algebra 2H ivolved fidig a or S (the sum of the first umbers i the sequece) directly without havig to fid all the umbers i the sequece. The patter of the sequece leads to a equatio where 330

a is determied by, ad that patter is determied by the type of sequece it is. It was basically a cotext for arithmetic. OBJECTIVES Idetify Sequeces ad Series Fid Partial Sums of a give Series. Fid the terms, partial sums, ifiite sums, or i a geometric sequece. Ex Fid the ext three terms ad the geeral term i each of these sequeces ad classify them as a arithmetic sequece, a geometric sequece or either. a), 3, 5, 7, b) 4, 7, 0, -7, -4, c) 4, 2,, 2,,... d),, 2, 3, 5, 8, 3, 4 e) + 4 + 9 + 6 + 25 +... (f), 2, 3, 4, 5,... a) 9,, 3,,+ 2( ). It is arithmetic because we add 2 each time. b) -2, -28, -35,, 4 7( ). It is arithmetic because we add -7 each time. c) 8, 6, 32,..., 2 2 each time. i. It is geometric because we multiply by 2 d) 2, 34, 55. This is either arithmetic or geometric. We do ot multiply by or add the same amout each step. It is called the Fiboacci Sequece. e) 36, 49, 64,..., 2. This is a p-series with p=2. f) 6, 7, 8,..., +. This is the Alteratig Harmoic Series. 33

Partial Sum--Def: the sum of the first terms of a sequece. Series are geerally writte with sigma otatio: a k k= where a k is the kth term i the sequece ad we are substitutig the itegers through i for k, gettig the terms ad addig them up. stads for sum. There is a symbol for the product of a sequece ( Π), but we will ot go ito that here. Ex 2 Fid k 2 4 k=0 4 ( k 2 ) = ( 0 2 )+ ( 2 )+ ( 2 2 )+ ( 3 2 )+ 4 2 k=0 =+ 0 3 8 5 = 25 Ex 3 Fid 5 k=2 3 k 2 5 3 = 3 k=2 k 2 5 k=2 k 2 = 3 2 + 2 3 + 2 4 + 2 5 2 = 3 4 + 9 + 6 + 25 =.39 There are formulas for the th partial sum of a arithmetic ad a geometric series that was taught i Algebra 2H. But these are ot of much use where we are goig. 332

The oe formula from freshma year that will be useful is for the sum of a Ifiite Geometric Series. I some cases, a ifiite series might have a total, if the later terms get ifiitely small. Much of the series work i Calculus is about fidig whether a ifiite series is coverget (has a total) or diverget (has total keeps gettig bigger idefiitely). Much of what we will be doig is fidig out if a series has a fiite total. The geometric series is oe of the few series that has a rule that allow us to actually fid the total. Geometric ad Arithmetic series are the oes we ecoutered i Algebra 2. There are several other kids, some of which are importat for our study of series. Arithmetic Series: a + = a + ( )d Geometric Series: a = a r p-series: Alteratig series: Harmoic Series: p ( ) a or ( cos π)a or ay other series where the =0 =0 sigs of the terms alterate betwee + ad -. =+ 2 + 3 + + (Note that this is a p-series where p=) 4 Alteratig Harmoic Series: ( ) + = 2 + 3 4 + 333

Geometric Ifiite Sum S = a r, if ad oly if r <. If r, the series is diverget. Ex 4 Fid the ifiite sum for the geometric sequece with a =024 ad r = 2. S = a r S = 024 2 = 682 2 3 Ex 5 Fid the ifiite sum for the geometric sequece with a = 24 ad r = 5 4. Sice r = 5 4 diverget. >, the formula will yield a icorrect aswer. This series is Geometric ad Arithmetic series are the oes we ecoutered i Algebra 2. There are several other kids, some of which are importat for our study of series. We will look at them more closely later. 334

6. Homework Fid the ext four umbers i each sequece ad classify it as arithmetic, geometric or either. Fid the ext four umbers i each sequece ad classify it as arithmetic, geometric or either.. 2, 4, 8, 6, 32,! 2. { 2, 7, 2, 7,!} 3., 2 3, 4 9, 8 27,! 5. { 2, 4, 6, 8, 0,! } 4. { 0, 2, 0, 2, 0,!} Fid the sum. 6. 5 4 7. 3k 5 k k= k= 5 k ( 2k +) 8. 9. 2 k k= 7 k=3 5 k 2 3 0. k=0 2. 7 3 k=0 k k+2 7. 2 + k 2 k=3 3. 3 3 2 k=0 k 4. =0 2 + 7 335

6.2: Covergece ad Divergece Vocabulary Ifiite Sequece Def: a sequece with a ifiite umber of terms. Coverget Sequece --Def: a ifiite sequece where a approaches a particular value whe goes to ifiity. Diverget Sequece --Def: a ifiite sequece where a does ot approach a particular value whe goes to ifiity. I other words, A sequece is coverget if ad oly if lim a = c. A sequece is diverget if ad oly if lim a c. Ex Which of the followig sequeces coverge? I. 4 3 II. e 2 2 III. e 2 e 2 I each case, we just eed to check the Limit at Ifiity. That is, we eed to check the ed behavior. I. lim 4 3 = 4 ; therefore, this sequece coverges. 3 II. III. lim e2 2 lim = ; therefore, this oe diverges. e 2 L'H e 2 = lim e2 2 e 2 2 =; therefore, this coverges. So, oly I ad III coverge. This is a typical AP questio. The key is CRITICAL READING. Sometimes, the questio is about Sequeces, sometimes it is about Series (which we will cosider 336

later it he chapter). Sometimes the questio is about covergece ad sometimes it is about divergece. Ex 2 Which of the followig sequeces diverge? I. 3 2 7 3 II. 7l 2 III. 4 4 2 3 I. lim 3 2 = 0 7 3 ; therefore, this sequece does ot diverges. II. lim 7l L'H 2 = lim 7 2 = 0 ; therefore, this oe does ot diverge. III. lim 4 4 2 3 = ; therefore, this diverges. So, oly III diverges. Vocabulary Ifiite Sum--Def: the sum of all the terms of a sequece. This is ot possible for a arithmetic series, but might be for a geometric series. Coverget Series--Def: a ifiite series that has a total. Diverget Series--Def: a ifiite series that does ot have a total. I other words, A series is coverget if ad oly if =c. a A series is diverget if ad oly if a c. Whe testig a umerical series for divergece or covergece, the two easiest tests are: 337

Divergece Test (th term test): If Lim a 0 If diverges. a Lim a = 0 o coclusio The Divergece Test basically says that if the terms at the ed are ot 0, the sum will just keep gettig larger ad larger (i.e., diverges). While easy to use, each has a disadvatage. The Divergece Test tells you if a fuctio has terms approachig 0. If the terms at the upper ed of the series do ot approach 0, the series caot coverge. But, there are may series with ed-terms approachig 0 that still diverge. So half the test is icoclusive. OBJECTIVE Determie the covergece or divergece of a sequece. Determie the divergece of a series. Use the Alteratig Series Test to check for covergece or divergece. Ex 3 What coclusio ca be draw from the Divergece Test? I. II. 2 2 + III. I. II. III. lim 2 = 0 ; so the Divergece Test is icoclusive for I. Note that this is ot eough to decide if it coverges or ot. 2 Lim + = 2 Lim 2 + = ; therefore, II diverges. Lim = 0 ; therefore, the Divergece Test is icoclusive for III. I ad III pass the Divergece Test 338

A test which is similar to the Divergece Test, i process, is the Alteratig Series test: The Liebitz Alteratig Series Test If a series is Alteratig Series ad is a a +, the it is coverget if ad oly if Lim a = 0. This meas the o-alteratig part of the series must be decreasig ad pass the Divergece Test (that is, the sequece coverges to 0). Ex 4 Is the Alteratig Harmoic Series coverget or diverget? + Remember that the Alteratig Harmoic Series is. ( i) a = ) + ii) lim = ad a + = ( ) +2 + = + > +, so a a + ( ) + = lim = 0 + So coverges. 339

Ex 5 Is cos π coverget or diverget?! Remember that the Alteratig Harmoic Series is. i) a = ii) cos( π)! =! ad a + = a =! > ( +)!, so a a + ( ) + lim = lim = 0 ( ) cos π + ( +)! = ( +)! So cos( π)! coverges. 340

6.2 Homework Set A Determie if the sequece coverges or diverges.. { ( ) } 2. 3+ 2 2 + 2 3. + 4. 2 3 + 5. 7. L 2! ( + 2)! 6. { 2 } Determie which series might be coverget by the Divergece Test. 8. 9. 4 3 2 5 + 0. 3 8. 2 =0 2. =0 4 + 5 Test these series for covergece or divergece. 3. ( ) 4. 3 ( ) + 4 2 + 5. ( ) 6. + ( ) L 34

L 7. ( ) 8. cos( π) 3 4 9. ( ) 20. 4 + cos( π) 5 4 2. ( 3) 22. ( ) 2 4 + 6.2 Homework Set B Determie if the sequece coverges or diverges.. 2 2 + 3 5 2. { 2 + } { } 4. e 3. ta ( 2 3 ) 5. { cos} 6. { 6 2 + 4} 3 7. + 3 49 2 + 8. + Determie which series might be coverget by the Divergece Test. 9. 5 4 3 0. 3 e 2 =0. 80 2 2. =0 3. 2 4. =0 si 3 =0 ta ( 4) 342

6.3: The Itegral Test Whe testig a umerical series for divergece or covergece, oe of the easiest tests is: Itegral Test: a ad x dx either both coverge or both diverge. The reaso the Itegral Test works is best explaied visually. The series ca be visualized as a sum of Riema Rectagles with width = ad height = the term values. If the itegral diverges (i.e., the sum is ifiite), the rectagles ca be draw left-had ad their sum is larger tha a ifiite umber. If the itegral coverges (i.e., the sum is fiite), the rectagles ca be draw right-had ad their sum is smaller tha a fiite umber. I both situatios, the fuctio must be decreasig. While easy to use, there is a disadvatage: The Itegral Test oly works o fuctios that ca be itegrated. Not all fuctios ca be itegrated. 343

Ex Use the Itegral Test to determie which of these series coverge. I. II. III. 2 + =2 l 2 I. II. x dx x 2 + dx = l x = de ; therefore, I diverges. = ta π x = 2 π 4 = π ; therefore, II coverges. 4 III. xl 2 x dx = 2 coverges. l 2 u du 2 = = 0 ( l2) = l2 ; therefore, III u l 2 II ad III coverge. Two otes about this example. First, I. proves that the Harmoic Series always diverges. Secod, III was a series that started at = 2 istead of =. This was to avoid the problem of the first term beig trasfiite. Ex 2 For what values of p does the p-series coverge? p Lim p = 0 for all p>0, so p must be positive. x p dx = x p+ p +. This itegral will be fiite as log as p>, because this will leave x i the deomiator so the limit ca go to 0. Therefore, p coverges for all p>. We will use this fact later i this chapter, whe applyig the Compariso Tests. 344

Ex 3 Determie whether is coverget or diverget. + 2 0 0 x + x 2 dx = lim b = 2 lim b b 0 x + x 2 dx x=b x=0 du u 0 b = 2 lim l + x2 b l = 2 lim l + b2 b = Sice this itegral diverges, the series diverges. + 2 0 345

6.3 Homework Set A Apply the Itegral Test to each of these series to determie divergece or covergece. If the series is coverget, state the improper itegral value.. 2 + 2. 3 2 =2 L 2 3. 4. 3 + + 4 5. + 8 + 27 + 64 + 25 +... 6. 5 2 3 7. 8. 2 + 2 + 9. 0. e 5 + 2 2 4 6.3 Homework Set B Apply the Itegral Test to each of these series to determie divergece or covergece. If the series is coverget, state the improper itegral value.. 2 e 3 2. 3 2 + 6 3. 6 5 4. + 6 =0 2 4 + 3 5. l 6. si =2 7. 8. + 2 4 346

6.4: The Direct ad Limit Compariso Tests The Direct ad Limit Compariso Tests are much less algebraic tha the Divergece ad Itegral Tests. They ivolve comparig a series to some series that we already kow coverges or diverges. So what do we kow? Geometric Series: p-series: a r --coverges to S = a =0 r if r < --diverges if r --coverges for p> p --diverges for p Harmoic Series: = + 2 + 3 + 4 + Diverges The Direct Compariso Test Whe comparig a give series to a kow series b, a ) if a b ad b coverges, the a coverges. 2) if a b ad b diverges, the a diverges. The use of this test is very verbal. 347

The Limit Compariso Test Whe comparig a give series a to a kow series b,. If lim a b = ay positive Real umber, the both coverge or both diverge. 2. If lim a = 0, the a coverges if b b 3. If lim a = de, the a diverges if b b coverges. diverges. Key Idea I: What series to compare a to depeds o where the variable is. If is i the base, compare to the p-series. If is i the expoet, compare to the Geometric Series. If is i both the base ad expoet (but ot ), try either. If, use the Nth Root Test (sectio 5-4). REMINDER: Factual Kowledge for Comparisos Geometric series:. =0 a r coverges if r <ad diverges if r 2. If a r coverges, it coverges to =0 a 0 r p-series:. =0 p coverges if p >ad diverges if p 348

Key Idea II: Which test to use depeds o the combiatio of the relative sizes ad whether the kow series coverges or diverges. b coverges b diverges a b Direct Compariso Test Limit Compariso Test a b Limit Compariso Test Direct Compariso Test Key Idea III: If a is ratioal, we wat to compare to the ed behavior model of the correspodig fuctio. OBJECTIVE Use the Compariso Tests to check for covergece or divergece. Ex Does coverge? 3 + 2 We ca make a direct compariso betwee ad 3 + 2. 3 + 2 < because 3 3 + 2 has a larger deomiator. Sice coverges (it is a p-series with p>) ad coverges. 3 + 2 3 3 3 + 2 is smaller tha 3, the 349

Ex 2 Does coverge? 2 =3 We ca make a direct compariso betwee ad 2. 2 > because 2 has a smaller deomiator. Sice diverges (it is the Harmoic Series--a p-series with p=) ad 2 is bigger tha, the diverges. 2 =3 Ex 3 Does coverge? 2 4 If we ca try to make a direct compariso betwee we fid that 2 4 > (because 2 2 4 ad 2 4, has a larger deomiator) but coverges (it is a p-series with p>). So the Direct Compariso Test 2 does ot apply. 2 Lim 2 4 2 = Lim 2 2 4 = > 0 This limit is greater tha 0 ad coverges also. coverges. Therefore, 2 2 4 350

Ex 4 Does coverge? 2 2 > 2 because 2 has a smaller deomiator. But 2 coverges (it is a Geometric Series with r<), so Direct Compariso will ot work. Lim 2 2 = Lim L 2 = ' H Lim 2 l2 2 l2 = 2 This limit is greater tha 0 ad coverges also. coverges. Therefore, 2 2 Ex 5 Does coverge? 2 + 3 We ca make a direct compariso betwee ad 2 + 3. 2 + 3 < 2 because 2 + 3 has a larger deomiator. Sice 2 coverges ad 2 + 3 is smaller tha 2, the coverges. 2 + 3 2 35

Ex 6 Does coverge? 3 + 2 4 We ca use the Limit Compariso Test betwee ad 3 + 2 4. 3 4 Lim ( 3 + 2) 4 3 4 = Lim 3 4 ( 3 + 2) 4 = diverges, therefore, 3 4 diverges. 3 + 2 4 352

6.4 Homework Set A Test these series for covergece or divergece.. 2. 2 + + 5 2 + 3 3. + 4. 2 3 2 5. 7. 3 + 6. 3 3+ cos 8. 3 ( +)2 + 9. 2 5 0. si 3 + + 6.4 Homework Set B Test these series for covergece or divergece.. si 2 2. 3 + 3. 3 =2 3! 4. 3 2 + 2 5. 4 2 2 + 7 5 6. ( + 2) 4 4 + 2 ( + 3 ) 2 7. + 2si 8. 4 9. + 4 2 + 4 0. 2 + 4 353

6.5: Absolute ad Coditioal Covergece The four tests i the previous two sectios oly apply to series of positive values. There is a separate test for alteratig series called the Liebitz Alteratig Series Test. The Alteratig Series Test (AST) If a series is Alteratig Series ad is a a +, the it is coverget if ad oly if Lim a = 0. This test was itroduced i Sectio 6.2. Vocabulary Absolute Covergece Def: Whe a alteratig series ad its absolute value are both coverget. Coditioal Covergece Def: Whe a alteratig series is coverget but its absolute value are diverget. OBJECTIVE Use the Alteratig Series Test to check for covergece or divergece. Ex Is cos π coverget or diverget? 3 Remember that the Alteratig Harmoic Series is. i) a = cos( π) = 3 3 ad a + = a > 3 + 3, so ( ) = 3 cos π + = + ( +) 3 354

ii) Lim + 3 So = Lim = 0 3 cos π coverges. 3 Ex 2 Is cos π absolutely or coditioally coverget? 3 Ex proved that cos π is coverget by the AST. To demostrate 3 coditioal vs. absolute covergece, we test the series without the alterator: 3 cos π = 3 3 is a p-series with p >, therefore, sice both the alteratig ad oalteratig versios of the series coverge, cos( π) 3 coverges absolutely. ( ) + ( + 2) Ex 3 Is coverget or diverget? If coverget, is it absolutely or ( +) coditioally coverget? i) a = ( + + 2) = + 2 ( + ) 2 + ad a + = + + ( +) + 2 ( +) + + 2 2 + > + 3 2 + 3 + 2, so a a + = + 3 2 + 3 + 2 355

ii) Lim ( ) + + 2 + = Lim + 2 2 + = 0 Ex 4 Is So ( ) + ( + 2) + coverges. To test for absolute vs. coditioal covergece, we eed to test + 2. + We ca do the Limit Compariso Test agaist the ed behavior model. + 2 ( +) + 2 Lim = Lim ( +) = Lim 2 + 2 2 + = This limit is greater tha 0, therefore, both do the same thig--amely, they diverge. So, ( ) + ( + 2) + is coditioally coverget. ( ) + coverget or diverget? If coverget, is it absolutely or 2 coditioally coverget? ( i) a = ) + ii) +2 2 = 2 ad a + = = 2 + 2 2 2 > 2 2, so a a + Lim ( ) + = 2 Lim 2 = 0 So ( ) + coverges. Furthermore, 2 coverges because it is a Geometric series with r = 2. So, 2 ( ) + coverges absolutely. 2 356

6.5 Homework Set A Test these series for covergece or divergece. If it is coverget, determie if it has absolute or coditioal covergece.. ( 3) 2. ( ) 2 4 + 3. ( ) 4. + ( ) L L 5. ( ) 6. cosπ 3 4 7. ( ) + 8. 4 2 + cosπ 3 4 6.6 Homework Set B a) Test these series for covergece or divergece. b) If it is coverget, determie if it has absolute or coditioal covergece.. ( 3) + 2. 2 4 ( ) + 5 3. =2 4 + 4. 5. + 6. + cos( π)! 7. si π 2 8. 6 9. arcta 0.. arcta π cos π 2. 2 si π 2 e 357

6.6: The Ratio ad Nth Root Tests The Ratio Test is oe of the more easily used tests ad it works o most series, especially those that ivolve factorials ad those that are a combiatio of geometric ad p-series. Ufortuately, it is sometimes icoclusive, especially with alteratig series. Cauchy Ratio Test: If lim a + a If lim a + a If lim a + a < it coverges; > it diverges; = the test is icoclusive. The Nth Root Test is best for series with is both the base ad the expoet. The Nth Root Test: If lim a If lim a If lim a < it coverges; > it diverges; = the test is icoclusive. OBJECTIVE Use the Ratio ad Nth Root Tests to check for covergece or divergece. Sice the Cauchy Ratio Test ivolves Limits at Ifiity, It might be a good idea to review the Hierarchy of fuctios: 358

The Hierarchy of Fuctios. Logs grow the slowest. 2. Polyomials, Ratioals ad Radicals grow faster tha logs ad the degree of the EBM determies which algebraic fuctio grows fastest. For example, y = x 2 grows more slowly tha y = x 2. 3. The trig iverses fall i betwee the algebraic fuctios at the value of their respective horizotal asymptotes. 4. Expoetial fuctios grow faster tha the others. (I BC Calculus, we will see the factorial fuctio, y =!, grows the fastest.) 5. The fastest growig fuctio i the combiatio determies the ed behavior, just as the highest degree term did amog the algebraics. Ex Does coverge?! Lim a + a = Lim = Lim = Lim = Lim = 0 ( +)!!! ( + )!! ( + )! + 359

Sice this limit is <, coverges.! Ex 2 Does coverge? 2 Lim a + a = Lim + 2 + 2 2 = Lim + 2 + = Lim + 2 = 2 Sice this limit is <, coverges. 2 Ex 3 Does + 2 coverge? + Lim a + a = Lim = Lim = Lim ( +) + 2 + ( +) + + 2 ( +) + 3 ( + ) ( + 2) ( + ) + 2 2 + 3 2 + 4 + 4 = So the Ratio Test is icoclusive. The easiest test would be the Limit Compariso Test: 360

+ 2 Compare to ( +). + 2 ( +) + 2) lim = lim ( +) = Sice the Limit =,both series coverge or both diverge. therefore, diverges, + 2 diverges also. + Ex 4 Does coverge? l ( + 2) This series has to a power; therefore, the Root Test is appropriate. Lim l = ( + 2) Lim Sice this limit is <, l + 2 = 0 coverges. l ( + 2) Ex 5 Does + coverge? e This series has to a power; therefore, the Root Test is appropriate. Lim + e = Lim + e Lim ( + )e = e > = + Sice this limit is >, diverges. e 36

6.6 Homework Set A Test these series for covergece or divergece.. 2 2. 2 3. ( 3) 3 4. ( 3) 5. ( )! 6. 5 + ( 2)! 7. cos π 3 8.! 0 ( +)4 2+ 9. 2 + 2 2 + 0. 3 +3. For which of the followig series is the Ratio Test icoclusive. What test would you use istead? (a) (b) 3 ( 3) (c) (d) 2 + 2 6.6 Homework Set B Test these series for covergece or divergece.. 4 2. 4 4 =2 4 3. e! 4. 3 5 7... 2 + 5.! l =2 6. 3 7. 3 8. =0 =0! 4 7 0... 3 + 362

9. 4 + 5 0. 2 3.! e 2! 2. ( 3) 3 3. For which of the followig series is the Ratio Test icoclusive? What test would you use istead? (a) 2 (b) 2 + 2 (c) ( 3 (d) )! =2 4 2 + 2 3 3+ 363

6.7 Geeral Series Summary A series is coverget if ad oly if A series is diverget if ad oly if = c. a a c. Previously, we ivestigated the seve tests that will help determie if a give series coverges or ot. Those tests were:. Divergece Test (th term test): If Lim a 0 it diverges. If Lim a = 0 o coclusio 2. Itegral Test: If f(x) is a decreasig fuctio, the a ad x dx either both coverge or both diverge. 3. Cauchy Ratio Test: If Lim a + a If Lim a + a If Lim a + a < it coverges; > it diverges; = the test is icoclusive. 4. The Alteratig Series Test: A Alteratig Series is coverget if ad oly if ad i) a a +, ii) Lim a = 0. 364

5. The Nth Root Test: If Lim a < it coverges; If Lim a > it diverges; If Lim a = the test is icoclusive. 6. The Direct Compariso Test: Whe comparig a give series a to a kow series b, i) if a b ad b coverges, the a coverges. ii) if a b ad b diverges, the a diverges. 7. The Limit Compariso Test: Whe comparig a give series a to a kow series b, i. If Lim a b > 0 the both coverge or both diverge. ii. iii. If If Lim a = 0, the a coverges if b b Lim a =, the a diverges if b b coverges. diverges. OBJECTIVE Fid whether a give umerical series coverges or diverges. 365

oes lim a = 0? Yes No Diverget Series, by the Divergece Test What kid of series is it? Alteratig Series Apply the AST Apply the th Root Test Geometric Series Apply the Direct Compariso Test p-series Is it easy to itegrate? Yes Apply the Itegral Test No Apply the Limit Compariso Test Factorials or Combiatio Is it easy to itegrate? Yes Apply the Itegral Test No Apply the Ratio Test 366

6.7 Homework ) Which of the followig sequeces coverge? 3 I. 5 cos 7 4 II. π III.! e 2 (A) I oly (B) II oly (C) I ad III oly (D) II ad III oly (E) III oly 2) Which of the followig sequeces diverge? I. l 5 II. 5cos3 e III.! ( + 2 )! (A) II oly (B) III oly (C) I ad II oly (D) I ad III oly (E) I, II, ad III 3) Which of the followig series diverge? I. 9 8 + II. l5 III. cos 2π =3 (A) I oly (B) III oly (C) I ad II oly =2 2 3 + ( 2e) (D) II ad III oly (E) I, II, ad III 4) What are all values of k for which the ifiite series k 5 coverges? (A) k < 5 (B) k 5 (C) k < (D) k (E) k = 0 367

5) If f x = si x =0 2, the f π 6 = (A) 2 (B) 4 3 (C) 2 (D) 3 2 (E) diverget 6) Which of the followig series coverge? I. 2 + 3 II. 2 + 3 + 6 + III. e 2 + 4 (A) I oly (B) III oly (C) I ad III oly (D) II ad III oly (E) I ad II oly 7) Which of the followig series are diverget? I. ( ) II. =0 3 + e III. =0 ( + 5)!! 3 (A) I oly (B) II oly (C) III oly (D) I ad II oly (E) II ad III oly 8) Which of the followig series are coditioally coverget? I. ( ) II. =0 e III. =0 ( + 5)! cos( π) (A) I oly (B) II oly (C) III oly (D) I ad II oly (E) I ad III oly 368

9) Which of the followig series are diverget? I. II. 3 + =0 ( + 2)! III.! =0! 3 (A) I ad III oly (B) II oly (C) III oly (D) I ad II oly (E) II ad III oly 0) What are all values of p for which the ifiite series 5 + p diverges? (A) p > 2 (B) p 2 (C) p < 2 (D) p 2 (E) All real values of p 369

Numerical Series Test Calculator Allowed ) Which of the followig sequeces coverge? I. ( ) 3 L II. a L 2 = ( ) III. 3 3 + 2 2 + (A) I oly (B) III oly (C) I ad II oly (D) I ad III oly (E) I, II, ad III 2) Which of the followig Series diverge? I. 5 6 II. 3 5 III. 74 5 (A) I oly (B) III oly (C) I ad II oly (D) II ad III oly (E) I, II, ad III 3) Which of the followig are series diverget? I. e l5 2 II. ( III. 3 + + 5)! =3cos( 2π) =2 ( 2e) (A) I oly (B) III oly (C) I ad III oly (D) II ad III oly (E) I, II, ad III 4) What are all values of p for which the ifiite series 5 + p coverges? (A) p > 2 (B) p 2 (C) p < 2 (D) p 2 (E) All real values of p 370

5) Which of the followig are series coverget? I. 5 II. 3 + 3 III. ( 3 + 7) 4 + 2 =2 (A) I oly (B) III oly (C) I ad II oly (D) I ad III oly (E) II ad III oly 6) Which of the followig series diverge? I. II. 2 + III. cos 3 + (A) I oly (B) II oly (C) I ad II oly (D) I ad III oly (E) I, II ad III 7) If a coverges, the which of the followig is true? I. a diverges II. III. a coverges absolutely 4a coverges (A) I oly (B) II oly (C) III oly (D) II ad III oly (E) I, II ad III 37

8) Which of the followig series are coditioally coverget? I. ( ) II. L =2 ( ) III. ta ( 2) 2 (A) I oly (B) I ad III oly (C) II ad III oly (D) I ad III oly (E) I, II, ad III 9) si π 6 = (A) (B) 2 (C) 3 2 (D) 3 2 3 2 (E) diverget 0. Which of the followig three tests will establish that the series coverges? 2 5 + I. Direct Compariso Test with 5 2 II. Limit Compariso Test with 3 2 III. Direct Compariso Test with 2 (A) I oly (B) II oly (C) I ad II oly (D) II ad III oly (E) I, II, ad III 372

Numerical Series Test Calculator Allowed. Use the Itegral Test to determie if 2 e 3 is coverget or diverget. 2. Use the Ratio Test to determie if 2 3! is coverget or diverget. 3. Determie whether the series coverges or diverges. ( + 4) 2 373

6. Homework. Geometric 64, 28, 256 2. Arithmetic 22, 27, 32 3. Geometric 6 8, 32 243, 64 729 4. Neither 2, 0, 2 5. Neither 2, 4, 6 6. 8. -7 9. 248 0. 37 60 532 287 7. 0. 45 2. 5 2 3. No sum 4. 4 5 6.2 Homework. Diverget 2. Coverget 3. Coverget 4. Coverget 5. Coverget 6. Coverget 7. Coverget 8. Might Coverge 9. Might Coverge 0. Diverget. Diverget 2. Might Coverge 3. Coverget 4. Coverget 5. Coverget 6. Diverget 7. Coverget 8. Coverget 9. Coverget 20. Coverget 2. Diverget 22. Diverget 6.3 Homework. Coverget, 2 2. Coverget, l2 3. Diverget 374

4. Coverget, 2 5. Coverget, 2 6. Coverget, 7 6 7. Coverget, π 4 8. Diverget 9. Coverget,.0002 0. Coverget, 2e 6.4 Homework. Coverges by LCT 2. Coverges by LCT 3. Diverges by LCT 4. Coverges by the Ratio Test 5. Diverges by the Divergece Test 6. Coverges by the Ratio Test 7. Coverges by DCT 8. Diverges by LCT 9. Diverges by LCT 0. Diverges by LCT 6.5 Homework. Diverges by Alt. Series Test 2. Diverges by Alt. Series Test 3. Coditioally Coverget by AST 4. Diverges by Alt. Series Test 5. Coditioally Coverget by AST 6. Coditioally Cov. by AST 7. Absolutely Coverget by AST 8. Absolutely Coverget 6.6 Homework. Diverget by Divergece Test 2. Coverges by the Itegral Test 3. Diverget by Ratio Test 4. Coverges by Ratio Test 5. Diverget by Divergece Test 6. Coverges by Ratio Test 7. Coverges by Ratio Test 8. Coverges by Ratio Test 9. Coverges by Root Test 0. Diverges by Root Test 375

a. Ratio Test icoclusive b. Ratio Test works c. Ratio Test works d. Ratio Test icoclusive 6.7 Homework. B 2. A 3. D 4. A 5. B 6. D 7. E 8. C 9. E 0. B 376