Multivariable Calculus: Concepts and Contexts (with Tools for Enriching Calculus, Interactive Video...

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Tools for Enriching™ Calculus CD-ROM (TEC) This CD-ROM helps you understand and visualize calculus by exploring concepts through
interactive modules and animations. TEC icons in the text direct you to the appropriate module for
exploration. TEC also has homework hints for specially marked exercises in each section.
Interactive Video Skillbuilder CD-ROM This CD-ROM contains video instruction for every major concept in the text. It lets you review
material that you may not have understood the first time you heard it, or if you missed class.
In order to help you evaluate your progress, each section contains a 10-question Web quiz
per section (the results of which can be emailed to the instructor) and a test for each chapter,
with answers.
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1
ARITHMETIC OPERATIONS
GEOMETRIC FORMULAS
Triangle Circle Sector of Circle
Sphere Cylinder Cone
Point-slope equation of line through with slope m:
Slope-intercept equation of line with slope m and y-intercept b:
CIRCLES
x h2 y k2 r 2
h, k
P1x1, y1
m y2 y1
x1 x2
P2x2, y2P1x1, y1
h
r
r
3 r 3
b
s r in radians C 2r 1 2 ab sin
A 1 2 r 2 A r 2 A 1
2 bh
a 0
a cb ca b
2a ax 2 bx c 0
n
1 2 3 k
kx nkyk nxyn1 yn
x yn x n nx n1y nn 1
2 x n2y2
x y3 x 3 3x 2y 3xy2 y3
x y3 x 3 3x 2y 3xy2 y3
x y2 x 2 2xy y2x y2 x 2 2xy y2
x 3 y3 x yx 2 xy y2
x 3 y3 x yx 2 xy y2
x 2 y2 x yx y
nx
s n xy s
n xs n y
x mn s n x m (sn x )mx 1n s
n x
xn 1
x m
x n x mnx mx n x mn
a
b
c
d
a
b
d
c
ad
bc
2
radians
0
π 2π x
s radians 180
T R I G O N O M E T R Y
FUNDAMENTAL IDENTITIES
1 tan2x
cos 2x cos2x sin2x 2 cos2x 1 1 2 sin2x
sin 2x 2 sin x cos x
tanx y tan x tan y
1 tan x tan y
tanx y tan x tan y
1 tan x tan y
cosx y cos x cos y sin x sin y
cosx y cos x cos y sin x sin y
sinx y sin x cos y cos x sin y
sinx y sin x cos y cos x sin y
c 2 a 2 b 2 2ab cos C
b 2 a 2 c 2 2ac cos B
a 2 b 2 c 2 2bc cos A
A
b
c
a
B
C
sin2 cos2 1cot 1
3
S P E C I A L F U N C T I O N S
POWER FUNCTIONS
(i i i)
INVERSE TRIGONOMETRIC FUNCTIONS

x
1
y
10
y=
x
x
y
0
x
y
0
π
2
_ π
2
y
0
x
4
S P E C I A L F U N C T I O N S
EXPONENTIAL AND LOGARITHMIC FUNCTIONS
1.
2.
3.
1 x y tanh1x &? tanh y x
cosh1x ln(x sx 2 1) y cosh1x &? cosh y x and y 0
sinh1x ln(x sx 2 1) y sinh1x &? sinh y x
coth x cosh x
sinh x tanh x
loga x
y loga x loga ye ln x x lnex x
logaxy loga x loga ya loga x x logaax x
ey x&?ln x y
ln e 1ln x loge x
ay x&?loga x y
y
1
0
ln x
EXPONENTIAL AND LOGARITHMIC FUNCTIONS
d
d
d
d
d
d
d
d
d
d
dx f tx f txtx
d
tx txf x f xtx
tx 2
d
dx f xtx f xtx txf x
d
d
d
d
dx c 0
D I F F E R E N T I A T I O N R U L E S
M U L T I V A R I A B L E
The cover photograph shows the Walt Disney Concert Hall in Los Angeles, designed and built 1992–2003 by Frank Gehry and Associates. It is a daring building, a layered composition of curved surfaces in the form of billowing sails with brushed stain- less steel cladding.
The highly complex structures that Frank Gehry designs would be impossible to build without the computer. The CATIA software that his architects and engineers use to produce the computer models is based on principles of calculus— fitting curves by matching tangent lines, making sure the curvature isn’t too large, and control- ling parametric surfaces. “Consequently,” says Gehry, “we have a lot of freedom. I can play with shapes.”
The process starts with Gehry’s initial sketches, which are translated into a succession of physical models. (Hundreds of different physical models were constructed during the design of the building, first with basic wooden blocks and then evolving into more sculptural forms.) Then an engineer uses a digitizer to record the coordinates of a series of points on a physical model. The digitized points are fed into a computer and the CATIA software is used to link these points with smooth curves. (It joins curves so that their tangent lines coincide.) The architect has considerable freedom in creating these curves, guided by displays of the curve, its derivative, and its curvature. Then the
Calculus and the Architecture of Curves
Images not available due to copyright restrictions
Image not available due to copyright restrictions
curves are connected to each other by a parametric surface, and again the architect can do so in many possible ways with the guidance of displays of the geometric characteristics of the surface.
The CATIA model is then used to produce another physical model, which, in turn, suggests modifications and leads to additional computer and physical models.
The CATIA program was developed in France by Dassault Systèmes, originally for designing airplanes, and was subsequently employed in the automotive industry. Frank Gehry, because of his complex sculptural shapes, is the first to use it in architecture. It helps him answer his question, “How wiggly can you get and still make a building?”
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Calculus Concepts and Contexts 3E
James Stewart M C M A S T E R U N I V E R S I T Y
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Credits continue on page A51.
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8.1 Sequences 557
8.2 Series 567
8.4 Other Convergence Tests 586
8.5 Power Series 594
8.7 Taylor and Maclaurin Series 605
Laboratory Project An Elusive Limit 617
8.8 The Binomial Series 617
Writing Project How Newton Discovered the Binomial Series 621
8.9 Applications of Taylor Polynomials 621
Applied Project Radiation from the Stars 630
Review 631
Vectors and the Geometry of Space 636
9.1 Three-Dimensional Coordinate Systems 637
9.2 Vectors 642
Discovery Project The Geometry of a Tetrahedron 665
9.5 Equations of Lines and Planes 666
Laboratory Project Putting 3D in Perspective 675
9
8
vii
Contents
9.6 Functions and Surfaces 676
9.7 Cylindrical and Spherical Coordinates 685
Laboratory Project Families of Surfaces 690
Review 690
Vector Functions 694
10.2 Derivatives and Integrals of Vector Functions 702
10.3 Arc Length and Curvature 708
10.4 Motion in Space: Velocity and Acceleration 716
Applied Project Kepler’s Laws 727
10.5 Parametric Surfaces 728
Partial Derivatives 738
11.2 Limits and Continuity 750
11.3 Partial Derivatives 756
11.5 The Chain Rule 780
11.6 Directional Derivatives and the Gradient Vector 788
11.7 Maximum and Minimum Values 801
Applied Project Designing a Dumpster 811
Discovery Project Quadratic Approximations and Critical Points 812
11.8 Lagrange Multipliers 813
Review 822
Multiple Integrals 828
12.2 Iterated Integrals 837
12.5 Applications of Double Integrals 857
12.6 Surface Area 867
12.7 Triple Integrals 872
12.8 Triple Integrals in Cylindrical and Spherical Coordinates 882
Applied Project Roller Derby 888
Discovery Project The Intersection of Three Cylinders 889
12.9 Change of Variables in Multiple Integrals 889
Review 898
Vector Calculus 904
13.4 Green’s Theorem 933
13.5 Curl and Divergence 940
13.6 Surface Integrals 948
13.7 Stokes’ Theorem 959
13.8 The Divergence Theorem 966
13.9 Summary 973
Appendixes A1
E A Few Proofs A3
H Polar Coordinates A6
I Complex Numbers A22
Index A53
x
When the first edition of this book appeared eight years ago, a heated debate about calculus reform was taking place. Such issues as the use of technology, the relevance of rigor, and the role of discovery versus that of drill were causing deep splits in mathematics departments. Since then the rhetoric has calmed down somewhat as reform- ers and traditionalists have realized that they have a common goal: to enable students to understand and appreciate calculus.
The first and second editions were intended to be a synthesis of reform and tradi- tional approaches to calculus instruction. In this third edition I continue to follow that path by emphasizing conceptual understanding through visual, numerical, and algebraic approaches.
What’s New in the Third Edition
By way of preparing to write the third edition of this text, I spent a year teaching calculus at the University of Toronto. I listened carefully to my students’ questions and my colleagues’ suggestions. And as I prepared each lecture I sometimes realized that an additional example was needed, or a sentence could be clarified, or a section could use a few more exercises of a certain type. In addition, I paid attention to the suggestions sent to me by many users and to the comments of the reviewers.
Many hundreds of improvements, large and small, have been incorporated into this edition. Here are some of them.
Many examples have been added or changed.
Extra steps have been provided in some of the existing examples.
The data in examples and exercises have been updated to be more timely.
More than 25% of the exercises in each chapter are new. Here are a few of my favorites:
Preface
8.4.36 593 8.6.37–38 605 8.9.22 628
10.1.37–38 701 11.4.38 779 11.5.36 787
New phrases and margin notes have been added to clarify the exposition.
A number of pieces of art have been redrawn.
I’ve also added new problems to the Focus on Problem Solving sections. See, for instance, Problems 14 and 16 on page 635.
Two new projects have been added. The project on page 617 shows how computer algebra systems use Taylor series to compute limits, and the project on page 675 shows how computer graphics programmers use clipping planes and hidden line rendering to portray three-dimensional objects on a two- dimensional screen.
The CD called Tools for Enriching Calculus (TEC) has been completely redesigned and now includes what we call Visuals, brief animations of various figures in the text. In addition there are now Visuals, Modules, and Homework Hints for the multivariable chapters. See the description on page xiii.
The symbol has been placed beside examples (an average of three per section) for which there are videos of instructors explaining the example in more detail. These videos are free to adopters. This material is also included on an Interactive VideoSkillbuilder CD. See the description of the Interactive Video Skillbuilder on page xiii.
Conscious of the need to control the size of the book, I’ve put new topics (as well as expanded coverage of some topics already in the book) on the revamped web site www.stewartcalculus.com rather than in the text itself. (See the list of additional topics in the description of the web site on page xvii.) As a result, the number of pages in the text is actually a bit less than in the second edition.
Features
Conceptual Exercises The most important way to foster conceptual understanding is through the problems that we assign. To that end I have devised various types of problems. Some exercise sets begin with requests to explain the meanings of the basic concepts of the section. (See, for instance, the first couple of exercises in Sections 8.2, 11.2, and 11.3. I often use them as a basis for classroom discussions.) Similarly, review sections begin with a Concept Check and a True-False Quiz. Other exercises test conceptual understanding through graphs or tables (see Exercises 8.7.2, 10.2.1–2, 10.3.27–33, 11.1.1–2, 11.1.9–14, 11.3.3–8, 11.6.1–2, 11.7.3–4, 12.1.5–10, 13.1.11–18, 13.2.15–16, and 13.3.1–2).
Graded Exercise Sets Each exercise set is carefully graded, progressing from basic conceptual exercises and skill-development problems to more challenging problems involving applications and proofs.
Real-World Data My assistants and I have spent a great deal of time looking in libraries, contacting companies and government agencies, and searching the Internet for interesting real- world data to introduce, motivate, and illustrate the concepts of calculus. As a result, many of the examples and exercises deal with functions defined by such numerical
PREFACE xi
xii PREFACE
data or graphs. For instance, functions of two variables are illustrated by a table of values of wave heights (Example 3 in Section 9.6) and by a table of values of the wind-chill index as a function of air temperature and wind speed (Example 1 in Sec- tion 11.1). Partial derivatives are introduced in Section 11.3 by examining a column in a table of values of the heat index (perceived air temperature) as a function of the actual temperature and the relative humidity. This example is pursued further in connection with linear approximations (Example 3 in Section 11.4). Directional derivatives are introduced in Section 11.6 by using a temperature contour map to estimate the rate of change of temperature at Reno in the direction of Las Vegas. Double integrals are used to estimate the average snowfall in Colorado on December 24, 1982 (Example 4 in Section 12.1). Vector fields are introduced in Section 13.1 by depictions of actual velocity vector fields showing San Francisco Bay wind patterns.
Projects One way of involving students and making them active learners is to have them work (perhaps in groups) on extended projects that give a feeling of substantial accom- plishment when completed. Applied Projects involve applications that are designed to appeal to the imagination of students. The project after Section 11.8 uses Lagrange multipliers to determine the masses of the three stages of a rocket so as to minimize the total mass while enabling the rocket to reach a desired velocity. Discovery Projects explore aspects of geometry: tetrahedra (after Section 9.4), hyperspheres (after Sec- tion 12.7), and intersections of three cylinders (after Section 12.8). The Laboratory Project on page 690 uses technology to discover how interesting the shapes of surfaces can be and how these shapes evolve as the parameters change in a family. The Writing Project on page 965 explores the historical and physical origins of Green’s Theorem and Stokes’ Theorem and the interactions of the three men involved. Many additional projects are provided in the Instructor’s Guide.
Problem Solving Students usually have difficulties with problems for which there is no single well- defined procedure for obtaining the answer. I think nobody has improved very much on George Polya’s four-stage problem-solving strategy and, accordingly, I have included a version of his problem-solving principles at the end of Chapter 1. They are applied, both explicitly and implicitly, throughout the book. After the other chapters I have placed sections called Focus on Problem Solving, which feature examples of how to tackle challenging calculus problems. In selecting the varied problems for these sections I kept in mind the following advice from David Hilbert: “A mathemati-cal problem should be difficult in order to entice us, yet not inaccessible lest it mock our efforts.” When I put these challenging problems on assignments and tests I grade them in a different way. Here I reward a student significantly for ideas toward a solution and for recognizing which problem-solving principles are relevant.
Technology The availability of technology makes it not less important but more important to understand clearly the concepts that underlie the images on the screen. But, when properly used, graphing calculators and computers are powerful tools for discovering and understanding those concepts. I assume that the student has access to either a graphing calculator or a computer algebra system. The icon ; indicates an exercise that definitely requires the use of such technology, but that is not to say that a graphing device can’t be used on the other exercises as well. The symbol is reserved for problems in which the full resources of a computer algebra system (like Derive, Maple, Mathematica, or the TI-89/92) are required. But technology doesn’t make pen- cil and paper obsolete. Hand calculation and sketches are often preferable to technol-
CAS
PREFACE xiii
ogy for illustrating and reinforcing some concepts. Both instructors and students need to develop the ability to decide where the hand or the machine is appropriate.
Tools for Enriching™ Calculus The CD-ROM called TEC is a companion to the text and is intended to enrich and complement its contents. Developed by Harvey Keynes, Dan Clegg, Hubert Hohn, and myself, TEC uses a discovery and exploratory approach. In sections of the book where technology is particularly appropriate, marginal icons direct students to TEC Visuals and Modules that provide a laboratory environment in which they can explore the topic in different ways and at different levels. Visuals are animations of figures in the text; Modules are more elaborate activities and include exercises. Instructors can choose to become involved at several different levels, ranging from simply encourag- ing students to use the Visuals and Modules for independent exploration, to assigning specific exercises from those included with each Module, or to creating additional exercises, labs, and projects that make use of the Visuals and Modules.
TEC also includes Homework Hints for representative exercises (usually odd- numbered) in every section of the text, indicated by printing the exercise number in red. These hints are usually presented in the form of questions and try to imitate an effective teaching assistant by functioning as a silent tutor. They are constructed so as not to reveal any more of the actual solution than is minimally necessary to make further progress.
Interactive Video Skilbuilder CD-ROM The Interactive Video Skillbuilder CD-ROM contains more than eight hours of video instruction. The problems worked during each video lesson are shown next to the viewing screen so that students can try working them before watching the solution. To help students evaluate their progress, each section contains a ten-question Web quiz (the results of which can be emailed to the instructor) and each chapter contains a chapter test, with answers to each problem.
Web Site: www.stewartcalculus.com This has been renovated and now includes the following.
Algebra Review
Lies My Calculator and Computer Told Me
History of Mathematics, with links to the better historical web sites
Additional Topics (complete with exercise sets): Trigonometric Integrals, Trigonometric Substitution, Strategy for Integration, Volumes by Cylindrical Shells, Strategy for Testing Series, Fourier Series, Formulas for the Remainder Term in Taylor Series, Linear Differential Equations, Second-Order Linear Differential Equations, Nonhomogeneous Linear Equations, Applications of Second-Order Differential Equations, Using Series to Solve Differential Equa- tions, Rotation of Axes
Links, for each chapter, to outside Web resources
Archived Problems (drill exercises that appeared in previous editions, together with their solutions)
Challenge Problems (some from the Focus on Problem Solving sections of prior editions)
Downloadable versions of CalcLabs for Derive and TI graphing calculators
Chapter 8 Infinite Sequences and Series
Tests for the convergence of series are considered briefly, with intuitive rather than formal justifications. Numerical estimates of sums of series are based on which test was used to prove convergence. The emphasis is on Taylor series and polynomials and their applications to physics. Error estimates include those from graphing devices.
Chapter 9 Vectors and the Geometry of Space
The dot product and cross product of vectors are given geometric definitions, moti- vated by work and torque, before the algebraic expressions are deduced. To facilitate the discussion of surfaces, functions of two variables and their graphs are introduced here.
Chapter 10 Vector Functions
The calculus of vector functions is used to prove Kepler’s First Law of planetary motion, with the proofs of the other laws left as a project. In keeping with the introduction of parametric curves in Chapter 1, parametric surfaces are introduced as soon as possible, namely, in this chapter. I think an early familiarity with such surfaces is desirable, especially with the capability of computers to produce their graphs. Then tangent planes and areas of parametric surfaces can be discussed in Sections 11.4 and 12.6.
Chapter 11 Partial Derivatives
Functions of two or more variables are studied from verbal, numerical, visual, and algebraic points of view. In particular, I introduce partial derivatives by looking at a specific column in a table of values of the heat index (perceived air temperature) as a function of the actual temperature and the relative humidity. Directional derivatives are estimated from contour maps of temperature, pressure, and snowfall.
Chapter 12 Multiple Integrals
Contour maps and the Midpoint Rule are used to estimate the average snowfall and average temperature in given regions. Double and triple integrals are used to compute probabilities, areas of parametric surfaces, volumes of hyperspheres, and the volume of intersection of three cylinders.
Chapter 13 Vector Fields
Vector fields are introduced through pictures of velocity fields showing San Francisco Bay wind patterns. The similarities among the Fundamental Theorem for line integrals, Green’s Theorem, Stokes’ Theorem, and the Divergence Theorem are emphasized.
Ancillaries
Multivariable Calculus: Concepts and Contexts, Third Edition, is supported by a complete set of ancillaries developed under my direction. Each piece has been designed to enhance student understanding and to facilitate creative instruction. The table on pages xv and xvi lists ancillaries available for instructors and students.
PREFACE xv
Contains Electronic Instructor’s Guide, Resource Integra- tion Guide, iLrn Testing, Instructions for iLrn Homework, and Power Point Lecture notes.
Tools for Enriching™ Calculus CD-ROM by James Stewart and Dan Clegg ISBN 0-534-40989-X
Completely revised and updated, TEC provides a laboratory environment in which students can explore selected topics. TEC also includes homework hints for representative exercises.
Instructor’s Guide by Douglas Shaw ISBN 0-534-41030-8
Each section of the main text is discussed from several view- points and contains suggested time to allot, points to stress, text discussion topics, core materials for lecture, workshop/ discussion suggestions, group work exercises in a form suitable for handout, with solutions, and suggested homework problems. An electronic version is available on the Instructor’s Resource CD-ROM.
Complete Solutions Manual, Multivariable by Dan Clegg ISBN 0-534-41012-X
Includes worked-out solutions to all exercises in the text.
Printed Test Bank By William Tomhave & Xueqi Zeng ISBN 0-534-41031-6
Contains multiple-choice and short-answer test items that key directly to the text.
ILrn Adopter’s Fulfillment Folder Kit ISBN 0-534-41033-2
Featuring full algorithmic generation of problems and free- response mathematics, iLrn allows you to customize exams and track student progress in an accessible, browser-based format, with results flowing automatically into your grade- book! This kit contains the generic Instructor’s Guide, Mathematics Instructors User’s Guide, Installation CD-ROM (for offline users), a text-specific content CD-ROM, a Quick
Start Guide, and a “How do I” quick introduction to widely used functions in iLrn.
Text-Specific Videos ISBN 0-534-41037-5
Text-specific videotape sets, available at no charge to adopters, consisting of one tape per text chapter. Each tape features a 10- to 20-minute problem-solving lesson for each section of the chapter. Covers both single- and multivariable calculus.
Transparencies, Multivariable by James Stewart ISBN 0-534-41015-4
Full-color, large-scale sheets of reproductions of material from the text.
Solutions Builder CD-ROM ISBN 0534410383
This CD is an electronic version of the complete solutions manual. It provides instructors with an efficient method for creating solution sets to homework or exams. Instructors can easily view, select, and save solution sets that can then be printed or posted.
Resources for Instructors and Students
Stewart Specialty Web Site: www.stewartcalculus.com
Contents: Algebra Review Additional Topics Drill exercises Problems Plus Web Links History of Mathematics Downloadable versions of CalcLabs for Derive and TI graphing calculators Maple Projects Mathematica Projects
iLrn Homework [http:// iLrn.com] ISBN 0-534-40988-1
iLrn Homework allows instructors to assign machine- gradable homework problems that help students identify where they need additional help. That assistance is available through worked-out solutions that guide students through the steps of problem solving, or via live online tutoring at vMentor. The tutors at this online service will skillfully guide students through a problem, using unique two-way audio and whiteboard features.
Electronic items Printed items
(cont.)
The Brooks/Cole Mathematics Resource Center Web Site http://mathematics.brookscole.com
When you adopt a Thomson–Brooks/Cole mathematics text, you and your students will have access to a variety of
teaching and learning resources. This Web site features everything from book-specific resources to newsgroups. It’s a great way to make teaching and learning an interactive and intriguing experience.
WebTutor Advantage™ on WebCT ISBN 0-534-41028-6
Lecture notes, discussion threads, and quizzes on WebCT.
WebTutor Advantage™ on Blackboard ISBN 0-534-41039-1
Lecture notes, discussion threads, and quizzes on Blackboard.
Student Resources
Tools for Enriching™ Calculus CD-ROM by James Stewart and Dan Clegg ISBN 0-534-40989-X
TEC provides a laboratory environment in which students can explore selected topics. TEC also includes homework hints for representative exercises.
Interactive Video SkillBuilder CD-ROM ISBN 0-534-41036-7
Think of it as portable office hours! The Interactive Video Skillbuilder CD-ROM contains more than eight hours of video instruction. The problems worked during each video lesson are shown next to the viewing screen so that students can try working them before watching the solution. To help students evaluate their progress, each section contains a ten-question Web quiz (the results of which can be emailed to the instructor) and each chapter contains a chapter test, with answers to each problem.
iLrn Student Resource Kit ISBN 0-534-39914-2
This helpful kit provides your students with a CD-ROM that contains the plug-ins needed to use the iLrn system and a Student Guide that offers additional assistance for students using iLrn.
Study Guide, Multivariable
by Robert Burton & Dennis Garity ISBN 0-534-41006-5
Contains key concepts, skills to master, a brief discussion of the ideas of the section, and worked-out examples with tips on how to find the solution.
Student Solutions Manual, Multivariable by Dan Clegg ISBN 0-534-41005-7
Provides completely worked-out solutions to all odd- numbered exercises within the text, giving students a way to check their answers and ensure that they took the correct steps to arrive at an answer.
CalcLabs with Maple, Multivariable by Philip Yasskin and Art Belmonte ISBN 0-534-41010-3
This comprehensive lab manual will help students learn to effectively use the technology tools available to them. Each lab contains clearly explained exercises and a variety of labs and projects to accompany the text.
Linear Algebra for Calculus by Konrad J. Heuvers, William P. Francis, John H. Kuisti, Deborah F. Lockhart, Daniel S. Moak, and Gene M. Ortner ISBN 0-534-25248-6
This comprehensive book, designed to supplement the calculus course, provides an introduction to and review of the basic ideas of linear algebra.
Electronic items Printed items
Acknowledgments
I am grateful to the following reviewers for sharing their knowledge and judgment with me. I have learned something from each of them.
William Ardis, Collin County Community College
Judith Broadwin, Jericho High School
Charles Bu, Wellesley University
Robert A. Chaffer, Central Michigan University
Joe W. Fisher, University of Cincinnati
Barry D. Hughes, University of Melbourne
Prem K. Kythe, University of New Orleans
Joyce Riseberg, Montgomery College
Richard Rochberg, Washington University
Denise Taunton Reid, Valdosta State University
Clifton Wingard, Delta State University
Teri E. Woodington, Colorado School of Mines
Second Edition Reviewers
Martina Bode, Northwestern University
Roxanne M. Byrne, University of Colorado at Denver
Larry Cannon, Utah State University
Deborah Troutman Cantrell, Chattanooga State Technical Community College
Barbara R. Fink, DeAnza College
Joe W. Fisher, University of Cincinnati
Richard L. Ford, California State University Chico
Gerrald Gustave Greivel, Colorado School of Mines
John R. Griggs, North Carolina State University
Barbara Bell Grover, Salt Lake Community College
John William Hagood, Northern Arizona University
Gary Steven Itzkowitz, Rowan University
Mohammad A. Kazemi, University of North Carolina, Charlotte
Kandace Alyson Kling, Portland Community College
Carrie L. Kyser, Clackamas Community College
Beth Turner Long, Pellissippi State Technical Community College
Andre Mathurin, Bellarmine College Prep
Richard Eugene Mercer, Wright State University
Laura J. Moore-Mueller, Green River Community College
Scott L. Mortensen, Dixie State College
Tejinder Singh Neelon, California State University San Marcos
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PREFACE xvii
xviii PREFACE
I also thank those who have responded to a survey about attitudes to calculus reform:
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PREFACE xix
In addition, I would like to thank George Bergman, Emile LeBlanc, Martin Erick- son, Stuart Goldenberg, Gerald Leibowitz, Larry Peterson, Charles Pugh, Marina Rat- ner, Peter Rosenthal, and Alan Weinstein for their suggestions; Dan Clegg for his
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xx PREFACE
research in libraries and on the Internet; Arnold Good for his treatment of optimiza- tion problems with implicit differentiation; Al Shenk and Dennis Zill for permission to use exercises from their calculus texts; COMAP for permission to use project material; George Bergman, David Bleecker, Dan Clegg, John Hagood, Victor Kaftal, Anthony Lam, Jamie Lawson, Ira Rosenholtz, Lowell Smylie, and Larry Wallen for ideas for exercises; Dan Drucker for the roller derby project; Tom Farmer, Fred Gass, John Ramsay, Larry Riddle, V. K. Srinivasan, and Philip Straffin for ideas for projects; and Jeff Cole and Dan Clegg for preparing the answer manuscript. I’m grateful to Jeff Cole for suggesting ways to improve the exercises. Dan Clegg acted as my assistant throughout; he proofread, made suggestions, and contributed some of the new exercises.
In addition, I thank those who have contributed to past editions: Ed Barbeau, Fred Brauer, Andy Bulman-Fleming, Tom DiCiccio, Garret Etgen, Chris Fisher, Gene Hecht, Harvey Keynes, Kevin Kreider, E. L. Koh, Zdislav Kovarik, David Leep, Lothar Redlin, Carl Riehm, Doug Shaw, and Saleem Watson.
I also thank Brian Betsill, Stephanie Kuhns, and Kathi Townes of TECH-arts for their production services, Tom Bonner for the cover image, and the following Brooks/ Cole staff: Janet Hill, editorial production project manager; Vernon Boes, art director; Karin Sandberg, Erin Mitchell, and Bryan Vann, marketing team; Earl Perry, technology project manager; Stacy Green, assistant editor; Katherine Cook, editorial assistant; Joohee Lee, permissions editor; Karen Hunt, print/media buyer; and Denise Davidson, cover designer. They have all done an outstanding job.
I have been very fortunate to have worked with some of the best mathematics editors in the business over the past two decades: Ron Munro, Harry Campbell, Craig Barth, Jeremy Hayhurst, Gary Ostedt, and now Bob Pirtle. Bob continues in that tra- dition of editors who, while offering sound advice and ample assistance, trust my instincts and allow me to write the books that I want to write.
JAMES STEWART
Calculus Concepts and Contexts 3E
Infinite Sequences and Series
Infinite sequences and series were introduced briefly in A Preview of Calculus in connection with Zeno’s paradoxes and the decimal representa- tion of numbers. Their importance in calculus stems from Newton’s idea of representing functions as sums of infinite series. For instance, in finding areas he often integrated a function by first expressing it as a series and then integrating each term of the series. We will pursue his idea in Section 8.7 in order to integrate such functions as
. (Recall that we have previously been unable to do this.) Many of the functions that arise in mathematical physics and chemistry, such as Bessel functions, are defined as sums of series, so it is important to be familiar with the basic concepts of convergence of infinite sequences and series.
Physicists also use series in another way, as we will see in Section 8.9. In studying fields as diverse as optics, special relativity, and electromag- netism, they analyze phenomena by replacing a function with the first few terms in the series that represents it.
ex2
Sequences
A sequence can be thought of as a list of numbers written in a definite order:
The number is called the first term, is the second term, and in general is the nth term. We will deal exclusively with infinite sequences and so each term will have a successor .
Notice that for every positive integer there is a corresponding number and so a sequence can be defined as a function whose domain is the set of positive integers. But we usually write instead of the function notation for the value of the function at the number .
NOTATION The sequence { , , , . . .} is also denoted by
EXAMPLE 1 Some sequences can be defined by giving a formula for the nth term. In the following examples we give three descriptions of the sequence: one by using the preceding notation, another by using the defining formula, and a third by writing out the terms of the sequence. Notice that doesn’t have to start at 1.
(a)
(b)
(c)
(d)
EXAMPLE 2 Find a formula for the general term of the sequence
assuming that the pattern of the first few terms continues.
SOLUTION We are given that
Notice that the numerators of these fractions start with 3 and increase by 1 whenever we go to the next term. The second term has numerator 4, the third term has numerator 5; in general, the th term will have numerator . The denominators are n 2n
a 5 7
3125 a 4

{0, 1, s2, s3, . . . , sn 3, . . .}an sn 3, n 3{sn 3}n3
2
3 ,
3
9 ,
4
27 ,
5
81 , . . . ,
1nn 1 3n1nn 1
3n 1
a3a2a1
8.1
557
the powers of 5, so has denominator . The signs of the terms are alternately positive and negative, so we need to multiply by a power of . In Example 1(b) the factor meant we started with a negative term. Here we want to start with a positive term and so we use or . Therefore,
EXAMPLE 3 Here are some sequences that don’t have a simple defining equation.
(a) The sequence , where is the population of the world as of January 1 in the year .
(b) If we let be the digit in the nth decimal place of the number , then is a well-defined sequence whose first few terms are
(c) The Fibonacci sequence is defined recursively by the conditions
Each term is the sum of the two preceding terms. The first few terms are
This sequence arose when the 13th-century Italian mathematician known as Fibonacci solved a problem concerning the breeding of rabbits (see Exercise 39).
. . . . . .
From Figure 1 or 2 it appears that the terms of the sequence are approaching 1 as becomes large. In fact, the difference
can be made as small as we like by taking sufficiently large. We indicate this by writing
In general, the notation
means that the terms of the sequence approach as becomes large. Notice that the following definition of the limit of a sequence is very similar to the definition of a limit of a function at infinity given in Section 2.5.
nLan
n, an 3, a32, a21, a1
an nn 1
fn
7, 1, 8, 2, 8, 1, 8, 2, 8, 4, 5, . . .
an ean
5 n
1 5 nan
0 11
7
Definition A sequence has the limit and we write
if we can make the terms as close to as we like by taking sufficiently large. If exists, we say the sequence converges (or is convergent). Otherwise, we say the sequence diverges (or is divergent).
Figure 3 illustrates Definition 1 by showing the graphs of two sequences that have the limit .
If you compare Definition 1 with Definition 2.5.4 you will see that the only difference between and is that is required to be an integer. Thus, we have the following theorem, which is illustrated by Figure 4.
Theorem If and when is an integer, then .
In particular, since we know from Section 2.5 that when we have
if
If becomes large as n becomes large, we use the notation
In this case the sequence is divergent, but in a special way. We say that diverges to .
The Limit Laws given in Section 2.3 also hold for the limits of sequences and their proofs are similar.
an an
FIGURE 4
y=ƒ
20 x
L
limn l an L nf n anlimx l f x L2
nlimx l f x Llimn l an L
0 n
an
L
FIGURE 3 Graphs of two sequences with lim an= L n `
L
an L
SECTION 8.1 SEQUENCES 559
A more precise definition of the limit of a sequence is given in Appendix D.
If and are convergent sequences and is a constant, then
The Squeeze Theorem can also be adapted for sequences as follows (see Figure 5).
If for and , then .
Another useful fact about limits of sequences is given by the following theorem, which follows from the Squeeze Theorem because .
Theorem If , then .
EXAMPLE 4 Find .
SOLUTION The method is similar to the one we used in Section 2.5: Divide numerator and denominator by the highest power of that occurs in the denominator and then use the Limit Laws.
Here we used Equation 3 with .
EXAMPLE 5 Calculate .lim n l
ln n
lim n l
an p [lim
lim n l
an
an lim n l
an lim n l
Limit Laws for Convergent Sequences
Squeeze Theorem for Sequences
This shows that the guess we made earlier from Figures 1 and 2 was correct.
FIGURE 5 The sequence bn is squeezed between the sequences an and cn.
0 n
cn
an
bn
SOLUTION Notice that both numerator and denominator approach infinity as . We can’t apply l’Hospital’s Rule directly because it applies not to sequences but to functions of a real variable. However, we can apply l’Hospital’s Rule to the related function and obtain
Therefore, by Theorem 2 we have
EXAMPLE 6 Determine whether the sequence is convergent or divergent.
SOLUTION If we write out the terms of the sequence, we obtain
The graph of this sequence is shown in Figure 6. Since the terms oscillate between 1 and infinitely often, does not approach any number. Thus, does not exist; that is, the sequence is divergent.
EXAMPLE 7 Evaluate if it exists.
SOLUTION
EXAMPLE 8 Discuss the convergence of the sequence , where .
SOLUTION Both numerator and denominator approach infinity as but here we have no corresponding function for use with l’Hospital’s Rule ( is not defined when is not an integer). Let’s write out a few terms to get a feeling for what happens to as gets large:
It appears from these expressions and the graph in Figure 8 that the terms are decreasing and perhaps approach 0. To confirm this, observe from Equation 5 that
an 1
a3 1 2 3
3 3 3 a2
lim n l
1, 1, 1, 1, 1, 1, 1, . . .
an 1n
lim n l
n l
FIGURE 6
The graph of the sequence in Example 7 is shown in Figure 7 and supports the answer.
FIGURE 7
0 n
an
1
1
_1
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator. So
We know that as . Therefore, as by the Squeeze Theorem.
EXAMPLE 9 For what values of is the sequence convergent?
SOLUTION We know from Section 2.5 and the graphs of the exponential functions in Section 1.5 that for and for . There- fore, putting and using Theorem 2, we have
For the cases and we have
and
If , then , so
and therefore by Theorem 4. If , then diverges as in Example 6. Figure 9 shows the graphs for various values of . (The case is shown in Figure 6.)
The results of Example 9 are summarized for future use as follows.
The sequence is convergent if and divergent for all other values of .
Definition A sequence is called increasing if for all , that is, It is called decreasing if for all . A sequence is monotonic if it is either increasing or decreasing.
n 1an an1a1 a2 a3 . n 1an an1an
lim n l
r n 0
r>1
r=1
FIGURE 9 The sequence an=rn
r 1r r n r 1lim n l r n 0
lim n l
r n 0
lim n l
0 0lim n l
1n lim n l
if 0 r 1
a r 0 a 1limx l ax 0a 1limx l ax
r n r
n l an l 0n l 1n l 0
0 an 1
562 CHAPTER 8 INFINITE SEQUENCES AND SERIES
CREATING GRAPHS OF SEQUENCES Some computer algebra systems have special commands that enable us to create sequences and graph them directly. With most graphing calculators, however, sequences can be graphed by using parametric equations. For instance, the sequence in Example 8 can be graphed by entering the parametric equations
and graphing in dot mode starting with , setting the -step equal to . The
result is shown in Figure 8. 1tt 1
x t y t!t t
FIGURE 8
and so for all .
EXAMPLE 11 Show that the sequence is decreasing.
SOLUTION 1 We must show that , that is,
This inequality is equivalent to the one we get by cross-multiplication:
Since , we know that the inequality is true. Therefore, and so is decreasing.
SOLUTION 2 Consider the function :
Thus, is decreasing on and so . Therefore, is decreasing.
Definition A sequence is bounded above if there is a number such that
It is bounded below if there is a number such that
If it is bounded above and below, then is a bounded sequence.
For instance, the sequence is bounded below but not above. The sequence is bounded because for all .n0 an 1an nn 1
an 0an n
m
Man
whenever x2 1f x x 2 1 2x 2
x 2 12 1 x 2
x 2 12 0
f x x
x 2 1
1 n2 n&?
n3 n2 n 1 n3 2n2 2n&?
n 1n2 1 nn 12 1 &? n 1
n 12 1
n 5 SECTION 8.1 SEQUENCES 563
The right side is smaller because it has a larger denominator.
We know that not every bounded sequence is convergent [for instance, the sequence satisfies but is divergent from Example 6] and not every monotonic sequence is convergent . But if a sequence is both bounded and monotonic, then it must be convergent. This fact is stated without proof as Theorem 7, but intuitively you can understand why it is true by looking at Fig- ure 10. If is increasing and for all , then the terms are forced to crowd together and approach some number .
Monotonic Sequence Theorem Every bounded, monotonic sequence is convergent.
EXAMPLE 12 Investigate the sequence defined by the recurrence relation
SOLUTION We begin by computing the first several terms:
These initial terms suggest that the sequence is increasing and the terms are approaching 6. To confirm that the sequence is increasing, we use mathematical induction to show that for all . This is true for because
. If we assume that it is true for , then we have
so
and
Thus
We have deduced that is true for . Therefore, the inequality is true for all by induction.
Next we verify that is bounded by showing that for all . (Since the sequence is increasing, we already know that it has a lower bound: for all .) We know that , so the assertion is true for . Suppose it is true for
. Then
so
and
Thus
This shows, by mathematical induction, that for all .nan 6
ak1 6
ak 6 12
an a1 2 nan 6an
n n k 1an1 an
ak2 ak1
ak1 6 ak 6
ak1 ak
a9 5.984375 a8 5.96875 a7 5.9375
a6 5.875 a5 5.75 a4 1 2 5 6 5.5
a3 1 2 4 6 5 a2 1
2 2 6 4 a1 2
for n 1, 2 , 3, . . .an1 1 2 an 6a1 2
an
7
20 n
FIGURE 10
Mathematical induction is often used in dealing with recursive sequences. See page 87 for a discussion of the Principle of Mathematical Induction.
Since the sequence is increasing and bounded, the Monotonic Sequence Theorem guarantees that it has a limit. The theorem doesn’t tell us what the value of the limit is. But now that we know exists, we can use the given recurrence relation to write
Since , it follows that too (as , also). So we have
Solving this equation for , we get , as we predicted.L 6L
L 1 2 L 6
n 1 l n l an1 l Lan l L
lim n l
12 an 6 1 2 ( lim
n l an 6) 1
2 L 6

; 29–34 Use a graph of the sequence to decide whether the sequence is convergent or divergent. If the sequence is convergent, guess the value of the limit from the graph and then prove your guess. (See the margin note on page 562 for advice on graphing sequences.)
29. 30.
31. 32.
33. 34.

35. If $1000 is invested at 6% interest, compounded annually, then after years the investment is worth dollars. (a) Find the first five terms of the sequence . (b) Is the sequence convergent or divergent? Explain.
36. Find the first 40 terms of the sequence defined by
and . Do the same if . Make a conjecture about this type of sequence.
a1 25a1 11
3an 1
an
n3
n!
n 1
an ln n2
n 0, 1, 0, 0, 1, 0, 0, 0, 1, . . . 25.
an sin 2n
2n21. 1. (a) What is a sequence?
(b) What does it mean to say that ? (c) What does it mean to say that ?
2. (a) What is a convergent sequence? Give two examples. (b) What is a divergent sequence? Give two examples.
List the first six terms of the sequence defined by
Does the sequence appear to have a limit? If so, find it.
4. List the first nine terms of the sequence . Does this sequence appear to have a limit? If so, find it. If not, explain why.
5–8 Find a formula for the general term of the sequence, assuming that the pattern of the first few terms continues.
6.
7. 8.
9–28 Determine whether the sequence converges or diverges. If it converges, find the limit.
10.
an cos2n e n e n
e 2n 1 an
5, 1, 5, 1, 5, 1, . . .2, 7, 12, 17, . . .
{ 1 4 , 29 , 3
an
Exercises8.1
is increasing and for all . Deduce that is convergent and find its limit.
48. Show that the sequence defined by
satisfies and is decreasing. Deduce that the sequence is convergent and find its limit.
.
estimating the value of to five decimal places.
51. The size of an undisturbed fish population has been modeled by the formula
where is the fish population after years and and are positive constants that depend on the species and its environment. Suppose that the population in year 0 is . (a) Show that if is convergent, then the only possible
values for its limit are 0 and . (b) Show that . (c) Use part (b) to show that if , then ;
in other words, the population dies out. (d) Now assume that . Show that if , then
is increasing and . Show also that if , then is decreasing and . Deduce that if , then .
52. A sequence is defined recursively by
Find the first eight terms of the sequence . What do you notice about the odd terms and the even terms? By consider- ing the odd and even terms separately, show that is convergent and deduce that
This gives the continued fraction expansion
s2 1 1
an1 1 1
1 an a1 1
limn l pn b aa b pn b a pnp 0 b a
0 pn b a pn p 0 b aa b
limn l pn 0a b pn1 bapn
b a pn
f L Llimn l an L fan1 f an
a3 f a2 f f aa2 f aa1 a
0.8n 0.000001 n
0 an 2
an1 1
3 an a1 2
an nan 3(a) Determine whether the sequence defined as follows is convergent or divergent:
(b) What happens if the first term is ?
38. (a) If , what is the value of ? (b) A sequence is defined by
Find the first ten terms of the sequence correct to five decimal places. Does it appear that the sequence is convergent? If so, estimate the value of the limit to three decimal places.
(c) Assuming that the sequence in part (b) has a limit, use part (a) to find its exact value. Compare with your estimate from part (b).
39. (a) Fibonacci posed the following problem: Suppose that rabbits live forever and that every month each pair produces a new pair which becomes productive at age 2 months. If we start with one newborn pair, how many pairs of rabbits will we have in the month? Show that the answer is , where is the Fibonacci sequence defined in Example 3(c).
(b) Let and show that . Assuming that is convergent, find its limit.
40. Find the limit of the sequence
41–44 Determine whether the sequence is increasing, decreasing, or not monotonic. Is the sequence bounded?
42.

Suppose you know that is a decreasing sequence and all its terms lie between the numbers 5 and 8. Explain why the sequence has a limit. What can you say about the value of the limit?
46. A sequence is given by , . (a) By induction or otherwise, show that is increasing
and bounded above by 3. Apply the Monotonic Sequence Theorem to show that exists.
(b) Find .
an1 3 1
a1 s2an
an 2n 3
3n 4 an
an an1 1 1an2an fn1fn
fn fn
a1 1 an1 11 an for n 1
an limn l an1limn l an L
a1 2
37.
SECTION 8.2 SERIES 567
Logistic Sequences
A sequence that arises in ecology as a model for population growth is defined by the logistic difference equation
where measures the size of the population of the generation of a single species. To keep the numbers manageable, is a fraction of the maximal size of the population, so
. Notice that the form of this equation is similar to the logistic differential equation in Section 7.5. The discrete model—with sequences instead of continuous functions—is preferable for modeling insect populations, where mating and death occur in a periodic fashion.
An ecologist is interested in predicting the size of the population as time goes on, and asks these questions: Will it stabilize at a limiting value? Will it change in a cyclical fashion? Or will it exhibit random behavior?
Write a program to compute the first terms of this sequence starting with an initial population . Use this program to do the following.
1. Calculate 20 or 30 terms of the sequence for and for two values of such that . Graph the sequences. Do they appear to converge? Repeat for a different
value of between 0 and 1. Does the limit depend on the choice of ? Does it depend on the choice of ?
2. Calculate terms of the sequence for a value of between 3 and 3.4 and plot them. What do you notice about the behavior of the terms?
3. Experiment with values of between 3.4 and 3.5. What happens to the terms?
4. For values of between 3.6 and 4, compute and plot at least 100 terms and comment on the behavior of the sequence. What happens if you change by 0.001? This type of behavior is called chaotic and is exhibited by insect populations under certain conditions.
p0
k
k
k
2
0 pn 1 pn
CAS
L A B O R A T O R Y P R O J E C T
Series
If we try to add the terms of an infinite sequence we get an expression of the form
which is called an infinite series (or just a series) and is denoted, for short, by the symbol
But does it make sense to talk about the sum of infinitely many terms? It would be impossible to find a finite sum for the series
because if we start adding the terms we get the cumulative sums 1, 3, 6, 10, 15, 21, . . . and, after the term, we get , which becomes very large as increases.
nnn 12nth
anor
an n1
8.2
However, if we start to add the terms of the series
we get , , , , , , . . . , , . . . . The table shows that as we add more and more terms, these partial sums become closer and closer to 1. (See also Figure 11 in A Preview of Calculus, page 7.) In fact, by adding sufficiently many terms of the series we can make the partial sums as close as we like to 1. So it seems reasonable to say that the sum of this infinite series is 1 and to write
We use a similar idea to determine whether or not a general series (1) has a sum. We consider the partial sums
and, in general,
These partial sums form a new sequence , which may or may not have a limit. If exists (as a finite number), then, as in the preceding example, we call it
the sum of the infinite series .
Definition Given a series , let denote its th partial sum:
If the sequence is convergent and exists as a real number, then the series is called convergent and we write
The number is called the sum of the series. If the sequence is divergent, then the series is called divergent.

an
sn n
n sn
an
i1 ai
s3 a1 a2 a3
n Sum of first n terms
1 0.50000000 2 0.75000000 3 0.87500000 4 0.93750000 5 0.96875000 6 0.98437500 7 0.99218750
10 0.99902344 15 0.99996948 20 0.99999905 25 0.99999997
Compare with the improper integral
To find this integral we integrate from 1 to and then let . For a series, we sum from 1 to
and then let .n l n t l
t
y
t l y
1 f x dx
EXAMPLE 1 An important example of an infinite series is the geometric series
Each term is obtained from the preceding one by multiplying it by the common ratio . (We have already considered the special case where and .)
If , then . Since doesn’t exist, the geometric series diverges in this case.
If , we have
If , we know from (8.1.6) that as , so
Thus, when the geometric series is convergent and its sum is . If or , the sequence is divergent by (8.1.6) and so, by Equation 3,
does not exist. Therefore, the geometric series diverges in those cases.
We summarize the results of Example 1 as follows.
The geometric series
If , the geometric series is divergent.
EXAMPLE 2 Find the sum of the geometric series
SOLUTION The first term is and the common ratio is . Since , the series is convergent by (4) and its sum is
5 10
4
lim n l
rn a
1 r
sn a1 rn
r 1
lim n l snsn a a a na l r 1 r 1
2a 1 2
n1 arn1
Figure 1 provides a geometric demonstration of the result in Example 1. If the triangles are constructed as shown and is the sum of the series, then, by similar triangles,
s
a
a
aa
a
ara-ar
ar
ar@
ar#
ar@
s
In words: The sum of a convergent geometric series is
first term
EXAMPLE 3 Is the series convergent or divergent?
SOLUTION Let’s rewrite the nth term of the series in the form :
We recognize this series as a geometric series with and . Since , the series diverges by (4).
EXAMPLE 4 Write the number . . . as a ratio of integers.
SOLUTION
After the first term we have a geometric series with and . Therefore
EXAMPLE 5 Find the sum of the series , where .
SOLUTION Notice that this series starts with and so the first term is . (With series, we adopt the convention that even when .) Thus

n0 xn 1 x x 2 x 3 x 4
x 0x 0 1 x 0 1n 0
x 1
2.3171717. . . 2.3 17

n
1 5.000000 2 1.666667 3 3.888889 4 2.407407 5 3.395062 6 2.736626 7 3.175583 8 2.882945 9 3.078037
10 2.947975
sn
What do we really mean when we say that the sum of the series in Example 2 is ? Of course, we can’t literally add an infinite number of terms, one by one. But, according to Definition 2, the total sum is the limit of the sequence of partial sums. So, by taking the sum of sufficiently many terms, we can get as close as we like to the number . The table shows the first ten partial sums and the graph in Figure 2 shows how the sequence of partial sums approaches .3
sn
3
3
Another way to identify and is to write out the first few terms:
4 16 3
ra

EXAMPLE 6 Show that the series is convergent, and find its sum.
SOLUTION This is not a geometric series, so we go back to the definition of a convergent series and compute the partial sums.
We can simplify this expression if we use the partial fraction decomposition
(see Section 5.7). Thus, we have
and so
is divergent.
SOLUTION For this particular series it’s convenient to consider the partial sums , , , and show that they become large.
1 1 2
3 1 4 ) ( 1
5 1 8 ) ( 1
3 1 4 ) ( 1
3 1 4 ) 1
1 1
n 1
1 1
2 1
Notice that the terms cancel in pairs. This is an example of a telescoping sum: Because of all the cancellations, the sum collapses (like a pirate’s collapsing telescope) into just two terms.
Figure 3 illustrates Example 6 by showing the graphs of the sequence of terms
and the sequence of partial sums. Notice that and
. See Exercises 46 and 47 for two geometric interpretations of Example 6. sn l 1
an l 0 sn an 1[nn 1]
FIGURE 3
Similarly, , , and in general
This shows that as and so is divergent. Therefore, the harmonic series diverges.
Theorem If the series is convergent, then .
Proof Let . Then . Since is convergent, the sequence is convergent. Let . Since as , we also have . Therefore
NOTE 1 With any series we associate two sequences: the sequence of its partial sums and the sequence of its terms. If is convergent, then the limit of the sequence is (the sum of the series) and, as Theorem 6 asserts, the limit of the sequence is 0.
| NOTE 2 The converse of Theorem 6 is not true in general. If , we cannot conclude that is convergent. Observe that for the harmonic series we have as , but we showed in Example 7 that is divergent.
The Test for Divergence If does not exist or if , then the
series is divergent.
The Test for Divergence follows from Theorem 6 because, if the series is not divergent, then it is convergent, and so .
EXAMPLE 8 Show that the series diverges.
SOLUTION
So the series diverges by the Test for Divergence.
NOTE 3 If we find that , we know that is divergent. If we find that , we know nothing about the convergence or divergence of . Remember the warning in Note 2: If , the series might converge or it might diverge.
anlim n l an 0 anlim n l an 0
anlim n l an 0
lim n l
5 0

an7
1nn l an 1n l 0 1n an
lim n l an 0
an ssn
sn lim n l
sn1
lim n l sn1 s n l n 1 l lim n l sn ssn
anan sn sn1sn a1 a2 an
lim n l
s2n 1 n
5 2
The method used in Example 7 for showing that the harmonic series diverges is due to the French scholar Nicole Oresme (1323–1382).
Theorem If and are convergent series, then so are the series (where is a constant), , and , and
(i) (ii)
(iii)
These properties of convergent series follow from the corresponding Limit Laws for Convergent Sequences in Section 8.1. For instance, here is how part (ii) of Theo- rem 8 is proved:
Let
and, using Equation 5.2.9, we have
Therefore, is convergent and its sum is
EXAMPLE 9 Find the sum of the series .
SOLUTION The series is a geometric series with and , so
In Example 6 we found that
So, by Theorem 8, the given series is convergent and
3 1 1 4

n l
NOTE 4 A finite number of terms doesn’t affect the convergence or divergence of a series. For instance, suppose that we were able to show that the series
is convergent. Since
it follows that the entire series is convergent. Similarly, if it is known that the series converges, then the full series
is also convergent.

13. 14.

17–26 Determine whether the series is convergent or divergent. If it is convergent, find its sum.
17. 18.
19. 20.
21. 22.
23. 24.

27–30 Determine whether the series is convergent or divergent by expressing as a telescoping sum (as in Example 6). If it is convergent, find its sum.
28.
31–34 Express the number as a ratio of integers.
32.

2 n
3 n

5 n1
n1 5( 2
3 )n11. (a) What is the difference between a sequence and a series? (b) What is a convergent series? What is a divergent series?
2. Explain what it means to say that .
; 3–8 Find at least 10 partial sums of the series. Graph both the sequence of terms and the sequence of partial sums on the same screen. Does it appear that the series is convergent or divergent? If it is convergent, find the sum. If it is divergent, explain why.
4.
(a) Determine whether is convergent. (b) Determine whether is convergent.
10. (a) Explain the difference between
(b) Explain the difference between
11–16 Determine whether the geometric series is convergent or divergent. If it is convergent, find its sum.
11.
5 10 3

Exercises8.2
(b) Calculate the total time that the ball travels. (c) Suppose that each time the ball strikes the surface
with velocity it rebounds with velocity , where . How long will it take for the ball to come
to rest?
What is the value of if ?
; 46. Graph the curves , , for on a common screen. By finding the areas between
successive curves, give a geometric demonstration of the fact, shown in Example 6, that
47. The figure shows two circles and of radius 1 that touch at . is a common tangent line; is the circle that touches , , and ; is the circle that touches , , and ; is the circle that touches , , and . This procedure can be continued indefinitely and produces an infinite sequence of circles . Find an expression for the diameter of and thus provide another geometric demonstration of Example 6.
48. A right triangle is given with and . is drawn perpendicular to , is drawn perpendicu-
lar to , , and this process is continued indefinitely as shown in the figure. Find the total length of all the perpendiculars
in terms of and .

0 k 1 kvv
35–37 Find the values of for which the series converges. Find the sum of the series for those values of .
36.
37.

38. We have seen that the harmonic series is a divergent series whose terms approach 0. Show that
is another series with this property.
39–40 Use the partial fraction command on your CAS to find a convenient expression for the partial sum, and then use this expression to find the sum of the series. Check your answer by using the CAS to sum the series directly.
39. 40.
find and .
42. If the partial sum of a series is , find and .
43. When money is spent on goods and services, those that receive the money also spend some of it. The people receiv- ing some of the twice-spent money will spend some of that, and so on. Economists call this chain reaction the multiplier effect. In a hypothetical isolated community, the local government begins the process by spending dollars. Suppose that each recipient of spent money spends and saves
of the money that he or she receives. The values and s are called the marginal propensity to consume and the marginal propensity to save and, of course, . (a) Let be the total spending that has been generated after
transactions. Find an equation for . (b) Show that , where . The number
is called the multiplier. What is the multiplier if the marginal propensity to consume is ?
Note: The federal government uses this principle to justify deficit spending. Banks use this principle to justify lend- ing a large percentage of the money that they receive in deposits.
44. A certain ball has the property that each time it falls from a height onto a hard, level surface, it rebounds to a height
, where . Suppose that the ball is dropped from an initial height of meters. (a) Assuming that the ball continues to bounce indefinitely,
find the total distance that it travels. (Use the fact that the ball falls in .)t seconds1
2 tt 2 meters
H 0 r 1rh
Sn
n1 anan
sn n 1
n2 n2
2n
(b) The Sierpinski carpet is a two-dimensional counterpart of the Cantor set. It is constructed by removing the center one-ninth of a square of side 1, then removing the centers of the eight smaller remaining squares, and so on. (The figure shows the first three steps of the construction.) Show that the sum of the areas of the removed squares is 1. This implies that the Sierpinski carpet has area 0.
56. (a) A sequence is defined recursively by the equation for , where and can be
any real numbers. Experiment with various values of and and use your calculator to guess the limit of the sequence.
(b) Find in terms of and by expressing in terms of and summing a series.
57. Consider the series
(a) Find the partial sums and . Do you recognize the denominators? Use the pattern to guess a formula for .
(b) Use mathematical induction to prove your guess. (c) Show that the given infinite series is convergent, and
find its sum.
58. In the figure there are infinitely many circles approaching the vertices of an equilateral triangle, each circle touching other circles and sides of the triangle. If the triangle has sides of length 1, find the total area occupied by the circles.
sn
an
What is wrong with the following calculation?
(Guido Ubaldus thought that this proved the existence of God because “something has been created out of nothing.”)
50. Suppose that is known to be a convergent series. Prove that is a divergent series.
51. If is convergent and is divergent, show that the series is divergent. [Hint: Argue by contradiction.]
52. If and are both divergent, is necessar- ily divergent?
Suppose that a series has positive terms and its partial sums satisfy the inequality for all . Explain why must be convergent.
54. The Fibonacci sequence was defined in Section 8.1 by the equations
Show that each of the following statements is true.
(a)
(b)
(c)
The Cantor set, named after the German mathematician Georg Cantor (1845–1918), is constructed as follows. We start with the closed interval and remove the open interval . That leaves the two intervals and and we remove the open middle third of each. Four intervals remain and again we remove the open middle third of each of them. We continue this procedure indefinitely, at each step removing the open middle third of every interval that remains from the preceding step. The Cantor set consists of the numbers that remain in after all those intervals have been removed. (a) Show that the total length of all the intervals that are
removed is 1. Despite that, the Cantor set contains infinitely many numbers. Give examples of some numbers in the Cantor set.
[0, 1]
3, 23 ) [0, 1]

1
an
1 1 1 1 1 1 1
1 1 1 1 1 1
1 1 1 1 1 1
0 0 0 0
SECTION 8.3 THE INTEGRAL AND COMPARISON TESTS; EST IMATING SUMS 577
The Integral and Comparison Tests; Estimating Sums
In general, it is difficult to find the exact sum of a series. We were able to accomplish this for geometric series and the series because in each of those cases we could find a simple formula for the partial sum . But usually it isn’t easy to compute . Therefore, in this section and the next we develop tests that enable us to determine whether a series is convergent or divergent without explicitly finding its sum. In some cases, however, our methods will enable us to find good estimates of the sum.
In this section we deal only with series with positive terms, so the partial sums are increasing. In view of the Monotonic Sequence Theorem, to decide whether a series is convergent or divergent, we need to determine whether the partial sums are bounded or not.
Testing with an Integral
Let’s investigate the series whose terms are the reciprocals of the squares of the positive integers:
There’s no simple formula for the sum of the first terms, but the computer- generated table of values given in the margin suggests that the partial sums are approaching a number near 1.64 as and so it looks as if the series is convergent.
We can confirm this impression with a geometric argument. Figure 1 shows the curve and rectangles that lie below the curve. The base of each rectangle is an interval of length 1; the height is equal to the value of the function at the right endpoint of the interval. So the sum of the areas of the rectangles is
If we exclude the first rectangle, the total area of the remaining rectangles is smaller than the area under the curve for , which is the value of the integral . In Section 5.10 we discovered that this improper integral is convergent and has value 1. So the picture shows that all the partial sums are less than
1
1 1x 2 dx x 1y 1x 2
FIGURE 1
y= 1
n l
8.3
n
100 1.6350 500 1.6429
1000 1.6439 5000 1.6447
i 2
Thus, the partial sums are bounded and the series converges. The sum of the series (the limit of the partial sums) is also less than 2:
[The exact sum of this series was found by the Swiss mathematician Leonhard Euler (1707–1783) to be , but the proof of this fact is beyond the scope of this book.]
Now let’s look at the series
The table of values of suggests that the partial sums aren’t approaching a finite number, so we suspect that the given series may be divergent. Again we use a picture for confirmation. Figure 2 shows the curve , but this time we use rectangles whose tops lie above the curve.
The base of each rectangle is an interval of length 1. The height is equal to the value of the function at the left endpoint of the interval. So the sum of the areas of all the rectangles is
This total area is greater than the area under the curve for , which is equal to the integral . But we know from Section 5.10 that this improper integral is divergent. In other words, the area under the curve is infinite. So the sum of the series must be infinite, that is, the series is divergent.
The same sort of geometric reasoning that we used for these two series can be used to prove the following test.
The Integral Test Suppose is a continuous, positive, decreasing function on and let . Then the series is convergent if and only if
the improper integral is convergent. In other words:
(a) If is convergent, then is convergent.
(b) If is divergent, then is divergent.
n1 any
x
1
s1
1
s2
1
s3
1
s4
1
s5
area= 1
n
100 18.5896 500 43.2834
1000 61.8010 5000 139.9681
1
si
NOTE When we use the Integral Test it is not necessary to start the series or the integral at . For instance, in testing the series
Also, it is not necessary that be always decreasing. What is important is that be ultimately decreasing, that is, decreasing for larger than some number . Then
is convergent, so is convergent by Note 4 of Section 8.2.
EXAMPLE 1 Determine whether the series converges or diverges.
SOLUTION The function is positive and continuous for because the logarithm function is continuous. But it is not obvious whether or not is decreasing, so we compute its derivative:
Thus, when , that is, . It follows that is decreasing when and so we can apply the Integral Test:
Since this improper integral is divergent, the series is also divergent by the Integral Test.
EXAMPLE 2 For what values of is the series convergent?
SOLUTION If , then . If , then . In either case , so the given series diverges by the Test for Divergence [see (8.2.7)].
If , then the function is clearly continuous, positive, and decreasing on . We found in Chapter 5 [see (5.10.2)] that
It follows from the Integral Test that the series converges if and diverges if . (For , this series is the harmonic series discussed in Example 7 in Section 8.2.)
The series in Example 2 is called the p-series. It is important in the rest of this chapter, so we summarize the results of Example 2 for future reference as follows.
The -series is convergent if and divergent if .p 1p 1
n1
y
1
1
xp dx converges if p 1 and diverges if p 1
1, f x 1xpp 0

2 1
f x x1x ln x
x 2 1 ln x
x 2

n4
n 1
In order to use the Integral Test we need to be able to evaluate and therefore we have to be able to find an antiderivative of . Frequently this is difficult or impossible, so we need other tests for convergence too.
f
For instance, the series
is convergent because it is a p-series with . But the series
is divergent because it is a p-series with .
Testing by Comparing
The series
reminds us of the series , which is a geometric series with and and is therefore convergent. Because the series (2) is so similar to a convergent series, we have the feeling that it too must be convergent. Indeed, it is. The inequality
shows that our given series (2) has smaller terms than those of the geometric series and therefore all its partial sums are also smaller than 1 (the sum of the geometric series). This means that its partial sums form a bounded increasing sequence, which is convergent. It also follows that the sum of the series is less than the sum of the geometric series:
Similar reasoning can be used to prove the following test, which applies only to series whose terms are positive. The first part says that if we have a series whose terms are smaller than those of a known convergent series, then our series is also convergent. The second part says that if we start with a series whose terms are larger than those of a known divergent series, then it too is divergent.
The Comparison Test Suppose that and are series with positive terms.
(a) If is convergent and for all , then is also convergent.
(b) If is divergent and for all , then is also divergent.
In using the Comparison Test we must, of course, have some known series for the purpose of comparison. Most of the time we use one of these series:
A -series [ converges if and diverges if ; see (1)] A geometric series [ converges if and diverges if ;
see (8.2.4)] r 1 r 1 arn1
p 1p 1 1npp
bn

Standard Series for Use with the Comparison Test
EXAMPLE 3 Determine whether the series converges or diverges.
SOLUTION For large the dominant term in the denominator is , so we compare the given series with the series . Observe that
because the left side has a bigger denominator. (In the notation of the Comparison Test, is the left side and is the right side.) We know that
is convergent ( -series with ). Therefore
is convergent by part (a) of the Comparison Test.
Although the condition or in the Comparison Test is given for all , we need verify only that it holds for , where is some fixed integer, because
the convergence of a series is not affected by a finite number of terms. This is illustrated in the next example.
EXAMPLE 4 Test the series for convergence or divergence.
SOLUTION We used the Integral Test to test this series in Example 1, but we can also test it by comparing it with the harmonic series. Observe that for and so
We know that is divergent ( -series with ). Thus, the given series is divergent by the Comparison Test.
NOTE The terms of the series being tested must be smaller than those of a convergent series or larger than those of a divergent series. If the terms are larger than the terms of a convergent series or smaller than those of a divergent series, then the Com- parison Test doesn’t apply. Consider, for instance, the series
The inequality
is useless as far as the Comparison Test is concerned because is convergent and . Nonetheless, we have the feeling that ought to be 12n 1an bn
bn ( 1 2 )n

2n 2 4n 3
convergent because it is very similar to the convergent geometric series . In such cases the following test can be used.
The Limit Comparison Test Suppose that and are series with positive terms. If
where c is a finite number and , then either both series converge or both diverge.
Although we won’t prove the Limit Comparison Test, it seems reasonable because for large .
EXAMPLE 5 Test the series for convergence or divergence.
SOLUTION We use the Limit Comparison Test with
and obtain
Since this limit exists and is a convergent geometric series, the given series converges by the Limit Comparison Test.
Estimating the Sum of a Series
Suppose we have been able to use the Integral Test to show that a series is convergent and we now want to find an approximation to the sum of the series. Of course, any partial sum is an approximation to because . But how good is such an approximation? To find out, we need to estimate the size of the remainder
The remainder is the error made when , the sum of the first terms, is used as an approximation to the total sum.
We use the same notation and ideas as in the Integral Test, assuming that is decreasing on . Comparing the areas of the rectangles with the area under
for in Figure 3, we see that
Similarly, we see from Figure 4 that
Rn an1 an2 y
n1 f x dx
Rn an1 an2 y
n f x dx
f
nsnRn
limn l sn sssn
2n
bn 1
2nan 1
2n 1
FIGURE 3
0 x
So we have proved the following error estimate.
Remainder Estimate for the Integral Test Suppose , where is a continuous, positive, decreasing function for and is convergent. If
, then
EXAMPLE 6 (a) Approximate the sum of the series by using the sum of the first 10 terms. Estimate the error involved in this approximation. (b) How many terms are required to ensure that the sum is accurate to within ?
SOLUTION In both parts (a) and (b) we need to know . With , which satisfies the conditions of the Integral Test, we have
(a)
According to the remainder estimate in (3), we have
So the size of the error is at most .
(b) Accuracy to within means that we have to find a va

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