192 8 Sequences and Series of Functions 197 8.1 Convergence of a Sequence of Functions.. Theorem 1.3.4 did not address the rule for “the limit of the product.” The reason is thatits proo
Trang 2Advanced Calculus
Theory and Practice
Trang 3ADVANCED CALCULUS: THEORY AND PRACTICE
John Srdjan Petrovic
COLLEGE GEOMETRY: A UNIFIED DEVELOPMENT
MATHEMATICAL AND EXPERIMENTAL MODELING OF PHYSICAL AND BIOLOGICAL PROCESSES
H T Banks and H T Tran
ORDINARY DIFFERENTIAL EQUATIONS: APPLICATIONS, MODELS, AND COMPUTING
Charles E Roberts, Jr.
REAL ANALYSIS AND FOUNDATIONS, THIRD EDITION
Steven G Krantz
Trang 4Advanced Calculus
Theory and Practice
John Srdjan Petrovic
Western Michigan University
Kalamazoo, USA
Trang 56000 Broken Sound Parkway NW, Suite 300
Boca Raton, FL 33487-2742
© 2014 by Taylor & Francis Group, LLC
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Trang 6Preface ix
1 Sequences and Their Limits 1
1.1 Computing the Limits: Part I 1
1.2 Definition of the Limit 4
1.3 Properties of Limits 8
1.4 Monotone Sequences 13
1.5 Number e 17
1.6 Cauchy Sequences 22
1.7 Limit Superior and Limit Inferior 25
1.8 Computing the Limits: Part II 29
2 Real Numbers 35 2.1 Axioms of the Set R 35
2.2 Consequences of the Completeness Axiom 39
2.3 Bolzano–Weierstrass Theorem 44
2.4 Some Thoughts about R 47
3 Continuity 51 3.1 Computing Limits of Functions 51
3.2 Review of Functions 56
3.3 Continuous Functions: A Geometric Viewpoint 58
3.4 Limits of Functions 61
3.5 Other Limits 66
3.5.1 One-Sided Limits 66
3.5.2 Limits at Infinity 68
3.5.3 Infinite Limits 70
3.6 Properties of Continuous Functions 72
3.7 Continuity of Elementary Functions 77
3.8 Uniform Continuity 81
3.9 Two Properties of Continuous Functions 85
4 Derivative 91 4.1 Computing the Derivatives 91
4.2 Derivative 93
4.3 Rules of Differentiation 99
4.4 Monotonicity: Local Extrema 103
4.5 Taylor’s Formula 109
Trang 74.6 L’Hˆopital’s Rule 113
5 Indefinite Integral 117 5.1 Computing Indefinite Integrals 117
5.2 Antiderivative 121
5.2.1 Rational Functions 124
5.2.2 Irrational Functions 127
5.2.3 Binomial Differentials 128
5.2.4 Some Trigonometric Integrals 131
6 Definite Integral 135 6.1 Computing Definite Integrals 135
6.2 Definite Integral 138
6.3 Integrable Functions 143
6.4 Riemann Sums 148
6.5 Properties of Definite Integrals 152
6.6 Fundamental Theorem of Calculus 155
6.7 Infinite and Improper Integrals 159
6.7.1 Infinite Integrals 159
6.7.2 Improper Integrals 163
7 Infinite Series 169 7.1 Review of Infinite Series 169
7.2 Definition of a Series 173
7.3 Series with Positive Terms 176
7.4 Root and Ratio Tests 181
7.4.1 Additional Tests for Convergence 183
7.5 Series with Arbitrary Terms 186
7.5.1 Additional Tests for Convergence 189
7.5.2 Rearrangement of a Series 192
8 Sequences and Series of Functions 197 8.1 Convergence of a Sequence of Functions 197
8.2 Uniformly Convergent Sequences of Functions 202
8.3 Function Series 208
8.3.1 Applications to Differential Equations 211
8.3.2 Continuous Nowhere Differentiable Function 214
8.4 Power Series 216
8.5 Power Series Expansions of Elementary Functions 221
9 Fourier Series 229 9.1 Introduction 229
9.2 Pointwise Convergence of Fourier Series 233
9.3 Uniform Convergence of Fourier Series 239
9.4 Ces`aro Summability 244
9.5 Mean Square Convergence of Fourier Series 250
Trang 89.6 Influence of Fourier Series 255
10 Functions of Several Variables 259 10.1 Subsets of Rn 259
10.2 Functions and Their Limits 264
10.3 Continuous Functions 269
10.4 Boundedness of Continuous Functions 273
10.5 Open Sets in Rn 278
10.6 Intermediate Value Theorem 285
10.7 Compact Sets 291
11 Derivatives 297 11.1 Computing Derivatives 297
11.2 Derivatives and Differentiability 300
11.3 Properties of the Derivative 306
11.4 Functions from Rn to Rm 310
11.5 Taylor’s Formula 314
11.6 Extreme Values 318
12 Implicit Functions and Optimization 325 12.1 Implicit Functions 325
12.2 Derivative as a Linear Map 330
12.3 Open Mapping Theorem 335
12.4 Implicit Function Theorem 339
12.5 Constrained Optimization 344
12.6 Second Derivative Test 351
12.6.1 Absolute Extrema 355
13 Integrals Depending on a Parameter 357 13.1 Uniform Convergence 357
13.2 Integral as a Function 361
13.3 Uniform Convergence of Improper Integrals 367
13.4 Integral as a Function 371
13.5 Some Important Integrals 377
14 Integration in Rn 387 14.1 Double Integrals over Rectangles 387
14.2 Double Integrals over Jordan Sets 393
14.3 Double Integrals as Iterated Integrals 397
14.4 Transformations of Jordan Sets in R2 402
14.5 Change of Variables in Double Integrals 407
14.6 Improper Integrals 413
14.7 Multiple Integrals 418
Trang 915 Fundamental Theorems 425
15.1 Curves in Rn 425
15.2 Line Integrals 430
15.3 Green’s Theorem 434
15.4 Surface Integrals 439
15.5 Divergence Theorem 446
15.6 Stokes’ Theorem 449
15.7 Differential Forms on Rn 453
15.8 Exact Differential Forms on Rn 457
16 Solutions and Answers to Selected Problems 465
Trang 10This text was written for a one-semester or a two-semester course in advanced calculus Ithas several goals: To expand the material covered in elementary calculus, and to present
it in a rigorous fashion; to improve problem-solving and proof-writing skills of students; tomake the reader aware of the historical development of calculus concepts and the reasonsbehind it; and to point out the connection between various topics
If one were to distill the principles of exposition applied throughout this book to onekey word, it would be motivation It can be seen in the selection of proofs, the order inwhich results are introduced, the inclusion of a substantial amount of history, and the largenumber of examples A professional mathematician typically has a well-developed sense ofwhether a particular problem or a concept is important, even without necessarily knowingwhy The topological definition of compactness (in terms of open covers) makes perfectsense, even before the Heine–Borel Theorem is formulated Nowadays, few mathematiciansknow that Heine wanted to reduce an infinite cover of the interval [a, b] to a finite one as away of proving that a continuous function on [a, b] is uniformly continuous (page 291) Yet,
to a student, such an insight can help make the idea less abstract
The exposition follows the idea that the learning goes from specific to general Perhaps
it was best formulated by Ralph Boas in [7]:
Suppose that you want to teach the “cat” concept to a very young child Do
you explain that a cat is a relatively small, primarily carnivorous mammal
with retractable claws, a distinctive sonic output, etc.? I’ll bet not You
probably show the kid a lot of different cats, saying “kitty” each time, until
it gets the idea To put it more generally, generalizations are best made
by abstraction from experience
John B Conway echoes the same idea in [19]:
To many, mathematics is a collection of theorems For me, mathematics
is a collection of examples; a theorem is a statement about a collection of
examples and the purpose of proving theorems is to classify and explain
the examples
In order to stick to these principles, the book contains almost 300 examples, which are used
to motivate and lead to important theorems As an illustration, the Extreme Value Theorem(Theorem 3.9.11) is formulated only after Examples 3.9.8–3.9.10 show what happens whenthe domain fails to be closed or bounded, or when it has both properties An equallyimportant role of examples is to help students develop the habit of scrutinizing theorems.Does the converse hold? Will it remain true if one of the hypotheses is relaxed? If not, what
is the counterexample?
Motivation as a driving force can also be detected in the proofs Instead of aiming for theshortest or the most elegant one, the book follows a simple plan: Start a proof with the ideathat should look reasonable to a student, and pursue it A good example is the “standard”proof of the Mean Value Theorem, which begins by introducing a strange-looking functionand then proves that it satisfies the hypotheses of Rolle’s Theorem The approach used inthis book is to note that the picture is a slanted version of the one seen in the proof of
Trang 11Rolle’s Theorem Then the task is to adjust the picture (to fit Rolle’s Theorem), i.e., todiscover the appropriate form of the function that will accomplish this.
The reader will notice the unorthodox order of the first two chapters The completeness
of the real line, while itself an “obvious” property to stipulate, makes much more sense whenused to prove important theorems about sequences Of course, these theorems themselvesare motivated by concrete sequences such as (1 + 1/n)n, etc
Finally, some significant effort was made for the book to include a historical tive A reader will get a glimpse into the development of calculus and its ideas from theage of Newton and Leibniz, all the way into the 20th century It has been documentedelsewhere that these details make learning mathematics a meaningful experience Knowingmore about the mathematicians helps get a timeline, and follow the evolution of a concept.Continuity of a function became a hot topic only after Arbogast showed that solutions ofpartial differential equations should be sought among piecewise continuous functions Theproper definition was then given by Cauchy, but even he struggled until the German schoolintroduced the ε− δ symbolism, which allowed understanding of the uniform continuity,and made it possible to create continuous but nowhere differentiable functions
prospec-It is helpful to know that some of the “natural” ideas were accepted until they proved
to be flawed, and that it took a long time and several tries to come up with a properformulation Piecewise-defined functions make a novice uncomfortable, so it is refreshing
to find out that they had the same effect on the best minds of the 18th century, and theywere not considered to be functions Nevertheless, they could not be outlawed once it wasdiscovered that they often appear as a sum of a Fourier series
To a student, various topics in calculus may seem unrelated, but in reality many ofthem have a common root, because they have been developed through attempts to solvesome specific problems For example, Cantor studied the sets on the real line, because hewas interested in Fourier series, and their sets of convergence Weierstrass discovered thatuniform convergence was the reason why the sum of a power series is a continuous functionand the sum of a trigonometric series need not be It also helps students understand thatmathematics has had (and still has) its share of disagreements and controversies, and that
it is a lively area full of opportunity to contribute to its advancement There is a long list
of references which will allow the instructor to direct students to further research of thehistory of a concept, or a result
An important goal of the book is to lead students toward the mastery of calculus niques, by strengthening those gained through elementary calculus, and by challenging them
tech-to acquire new ones Because of the former, early chapters start with a review of the priate topic This approach also has the advantage of pointing out specific results that werepreviously taken for granted, and that will be proved, thereby providing motivation for therest of the chapter The book also contains close to 100 worked out exercises that shouldhelp students, together with homework problems (the book contains more than 1,000 ofthose, some of them with solutions), develop problem-solving skills If pressed for time, theinstructor can leave much of this material to be read outside of the class A lucky one, withadvanced students, may decide to ignore the computational problems and focus on the moretheoretical ones
appro-The prerequisites for a course based on this book are a standard calculus sequence, ear algebra, and discrete mathematics or a “proofs” course, but these requirements can berelaxed Certainly, a student is expected to have had some exposure to ε− δ arguments,and a moderate use of quantifiers When it comes to linear algebra, it is not necessary whenstudying functions of one real variable It is used in the treatment of functions that depend
lin-on more than lin-one variable, and even there to the extent that a student has probably seenwhen learning the basics of multivariable calculus The only place where a solid foundation
in linear algebra is necessary is Chapter 12, which deals with the Implicit Function Theorem
Trang 12and the Inverse Function Theorem, and considers the derivative as a linear transformationbetween Euclidean spaces The course will present a serious (but not insurmountable) chal-lenge for a student who has had little experience with proofs in calculus and in general.Although some basic material can be found throughout the book, its purpose is to servemainly as a refresher This includes the use of quantifiers, which often make long statementsconcise and easier to handle.
The list of topics is more or less the usual one with a few exceptions That is, theinclusion of some material reflects the author’s personal preferences and biases Some obvi-ous examples are Sections 5.2.1–5.2.4 (various techniques of integration), 7.4.1 (some lesserknown test for convergence of series), and Chapter 13 (integrals depending on a parameter),
or at least large parts of them While these can be safely skipped by a more mainstreaminstructor, a student should at least be aware of their content (perhaps through indepen-dent study) With the advent of technology, some of the integration techniques may seemoutdated Yet, even a sophisticated computer algebra system failed to solve Exercise 5.2.12.Given that there is no universal test for the convergence of a series, it is not a bad idea
to have a reference text that has more than the usual few As for Chapter 13, in addition
to possessing its intrinsic beauty, it shows in a very obvious way some of the shortcomings
of the Riemann integral, and more than justifies the transition to the Lebesgue theory ofintegration
A less obvious standout is the chapter on Fourier series Many schools have a separatecourse, and have no need for this topic in advanced calculus Nevertheless, its significancefor the development of calculus cannot be overstated Section 9.6 is written as an attempt tojustify interest in the previously studied material, and to explain the birth of some modernmathematical disciplines
The book can be used both for a one-semester and a year-long course Most sections aredesigned in such a way that they can be covered in a 50-minute class For example, in a 15-week, 4-credit course, one could cover Chapters 1 through 9 by leaving Sections 1.1, 1.8, 2.4,3.1, 4.1, 5.1, 6.1, 7.1, 9.1, and 9.6, for students to read outside of the class, and by condensing
or completely omitting Sections 1.5, 3.7, 5.2.1–5.2.4, 7.4.1, 7.5.1, 8.3.1, and 8.3.2 If the classmeets only 3 hours per week or if students do not have much experience with ε− δ proofs, abetter plan is to cover the first 6 chapters and as much of Chapter 7 as time permits If theinstructor has his heart set on giving the series a fair shake, the recommended choice would
be Sections 1.2–1.4, 1.6, 1.7, 2.1–2.3, 3.2–3.6 (omitting 3.5.2 and 3.5.3), 3.8, 3.9, 4.2–4.6(avoiding the proofs in 4.6), 5.1, 5.2 (without 5.2.1–5.2.4), 6.2–6.6, 7.2–7.5 (skipping 7.4.1,7.5.1, 7.5.2) In a year-long course, the factors that may influence the choice of topics are thenumber of contact hours as well as the emphasis that the instructor wants to give Assumingthat the class meets 3 times a week, a “pure” approach would be to cover Chapters 1–7
in one semester (as described above) and in the second semester Sections 8.1–8.4 (skipping8.3.1 and 8.3.2), 9.2–9.5, 10.1–10.5, 11.2–11.6 (with 11.5 optional), 12.1–12.4, 13.1–13.4,14.1–14.5 (assigning 13.5 and 14.7 for independent reading), with as much of Chapter 15 astime permits On the other hand, if the course is aimed at advanced engineering students,the second semester should include Sections 8.3.1, 12.5, 12.6, as well as Chapter 15, at theexpense of Chapter 13 and Sections 14.4, 14.5 Of course, these are merely suggestions, andthe book offers enough flexibility for an instructor to determine what will make the cut
Acknowledgments
It is my pleasant duty to thank several people who have made significant contributions tothe quality of the book Shelley Speiss is a rare combination of artist and computer whiz.She has made all the illustrations in the book, on some occasions making immeasurableimprovement to my original sketches Daniel Sievewright read the whole text and even
Trang 13checked the calculations in many examples and exercises (I know that for a fact because
he discovered numerous errors, both of typographical and computational nature.) NathanPoirier also discovered numerous errors If there are still some left, they can be explainedonly by the fact that at the last moment I went behind their backs and made some changes
Dr Dennis Pence has been a gracious audience during the academic year in which the bookwas brewing I often exchanged ideas with him about how to approach a particular issue,and his enthusiasm for the history of mathematics was contagious
Western Michigan University has put its trust in me and allowed me to take a sabbaticalleave to work on this text This is my opportunity to thank them I hope that they will feeltheir expectations fulfilled
I am grateful to Robert Stern, senior editor of CRC, who has shown a lot of patienceand willingness to help throughout the publishing process Several anonymous reviewershave made valuable suggestions, and my thanks go to them
My family has been, as always, extremely supportive, and helped me sustain the energylevel needed to finish the project
This book is dedicated to all my calculus teachers, and in particular to my first one,Mihailo Arsenovi´c, who got me hooked on the subject
Trang 14Sequences and Their Limits
The intuitive idea of a limit is quite old In ancient Greece, the so-called method of tion was used by Archimedes, around 225 BC, to calculate the area under a parabola Inthe 17th century, Newton and Leibniz based much of the calculus they developed on theidea of taking limits In spite of the success that calculus brought to the natural sciences,
exhaus-it also drew heavy crexhaus-iticism due to the use of infinexhaus-itesimals (infinexhaus-itely small numbers) Itwas not until 1821, when Cauchy published Cours d’analyse, probably the most influentialtextbook in the history of analysis, that calculus achieved the rigor that is quite close tomodern standards
1.1 Computing the Limits: Part I
In this section we will review some of the rules for evaluating limits Some of the easiestproblems occur when anis a rational function, i.e., a quotient (or a ratio) of two polynomials
Exercise 1.1.1 an= 2n− 3
5n + 1.Solution When both polynomials have the same degree, we divide both the numeratorand the denominator by the highest power of n Here, it is just n We obtain
an=
2−n3
5 + 1n
Now we take the limit, and use the rules for limits (such as “the limits of the sum equalsthe sum of the limits”):
n→∞(5 + 1/n) =
lim
n→∞2− limn→∞3/nlim
n→∞5 + lim
n→∞1/n.Since limn→∞2 = 2, limn→∞5 = 5, limn→∞1/n = 0, and limn→∞3/n = 3 limn→∞1/n =
0, we obtain that limn→∞an= 2/5
Rule When both polynomials have the same degree, the limit is the ratio of the leadingcoefficients
Exercise 1.1.2 an= 3n + 5
n2− 4n + 7.Solution Again, we divide both the numerator and the denominator by the highest power
Trang 15This time, we obtain 0/1 = 0, so limn→∞an = 0.
Rule When the degree of the denominator is higher than the degree of the numerator, thelimit is 0
Exercise 1.1.3 an= n
3+ n2
n2+ 2n.Solution The highest power of n is n3 However, we have to be careful—it would beincorrect to use the rule “the limits of the quotient equals the quotient of the limits.” Thereason is that, after dividing by n3, the denominator 1/n + 2/n2 has limit 0 Nevertheless,the numerator is now 1 + 1/n, which has limit 1, and we can conclude that the sequence an
Solution Here we will use a2
− b2 = (a− b)(a + b) If we multiply and divide an by
Solution We will use the rule ln b− ln a = ln(b/a) Thus
an= lnn + 1
n = ln
1 + 1n
and limn→∞an = ln 1 = 0
Trang 16Remark 1.1.6 We have used the fact that lim ln 1 + 1
Our next problem will use the fact that if an = an, for some constant a≥ 0, then thelimit of an is either 0 (if a < 1), or 1 (if a = 1), or it is infinite (if a > 1)
Exercise 1.1.7 an= 3
n+ 4n+1
2· 3n− 4n.Solution Dividing by 4n yields
3n/4n+ 4n+1/4n
2· 3n/4n− 4n/4n = (3/4)
n+ 42(3/4)n− 1.Since limn→∞(3/4)n= 0 we see that limn→∞an=−4
Sometimes, we will need to use the Squeeze Theorem: if an ≤ bn ≤ cn for each n∈ Nand if lim an= lim cn= L, then lim bn= L
Exercise 1.1.8 an= sin(n
2)
√n Solution For any x∈ R, −1 ≤ sin x ≤ 1 This can be written as | sin x| ≤ 1 Consequently,
| sin(n2)| ≤ 1 for all n ∈ N and
0≤
sin(n√ 2)n
≤ √1n.
Since limn→∞1/√n = 0 we obtain that lim
n→∞an = 0
Exercise 1.1.9 an= 2
n
n!.Solution Notice that
·
22
·
23
· · ·
2n
<
21
·
2n
=
4n
Therefore 0 < 2n/n! < 4/n, and limn→∞4/n = 0, so we conclude that limn→∞an= 0.Did you know? The first use of the abbreviation lim (with a period at the end) was
by a Swiss mathematician, Simon L’ Huilier (1750–1840), in 1786 German mathematicianKarl Weierstrass (1815–1897), who is often cited as the “father of modern analysis,” used it(without a period) as early as 1841, but it did not appear in print until 1894 In the 1850s,
he began to write limx=c, and it appears that we owe the arrow (instead of the equality)
to two English mathematicians John Gaston Leathem (1871–1923) pioneered its use in
1905, and Godfrey Harold Hardy (1877–1947) made it popular through his 1908 textbook
A Course of Pure Mathematics
Weierstrass was supposed to study law and finance, but instead spent time studyingmathematics That is why he did not get a degree, and he started his career as a highschool teacher He spent about 15 years there, until his mathematical work brought himfame: an honorary doctoral degree from the University of K¨onigsberg, and a position ofprofessor at the University of Berlin, which was considered the leading university in theworld to study mathematics
Trang 17n2+ 1− n.
1.1.11 lim √3
n2+ 1−√3
n2+ n
1.1.17 lim
1
2 ·34· · · 2n2n− 1
1.2 Definition of the Limit
In the previous section we relied on some rules, such as that limn→∞1/n = 0, or the
“Squeeze Theorem.” They are all intuitively clear, but we will soon encounter some that arenot For example, in Section 1.5 we will study the sequence{an} defined by an = 1 +1
n
n
.Two most common errors are to conclude that its limit is 1 or ∞ The first (erroneous)argument typically notices that 1 + 1
n has limit 1, and 1 raised to any power is 1 Thesecond argues that since the exponent goes to infinity, and the base is bigger than 1, thelimit must be infinite In fact, the limit is neither 1 nor is it infinite As we find ourselves
in more and more complicated situations, our intuition becomes less and less reliable Theonly way to ensure that our results are correct is to define precisely every concept that weuse, and to furnish a proof for every assertion that we make
Having made this commitment, let us take a careful look at the idea of a limit We saidthat the sequence an= 1/n has the limit 0 What does it really mean? We might say that,
as n increases (without bound), an is getting closer and closer to 0 This last part is true,but it would remain true if 0 were replaced by−1: an is getting closer and closer to−1 Ofcourse,−1 would be a poor choice for the limit since anwill never be really close to−1 Forexample, none of the members of the sequence{an} will fall into the interval (−1.5, −0.5),with center at−1 and radius 0.5 (Figure 1.1)
What about−0.001 as a limit? Although closer than −1, it suffers from the same flaw.Namely, we can find an interval around −0.001 that contains no member of the sequence{an} (Example: (−0.0015, −0.0005)) We see that the crucial property of the limit is that
Trang 18Figure 1.1:−1 is not the limit of 1/n.
there cannot be such an interval around it More precisely, if L is the limit of {an}, thenany attempt to select such an interval centered at L must fail In other words, regardless ofhow small positive number ε is, the interval (L− ε, L + ε) must contain at least one member
of the sequence{an} The following examples illustrate this phenomenon
You may have noticed that a80 ∈ (−0.02, 0.02) as well Nevertheless, we are interested
in limits (as n increases without bound), so there is no urgency to choose the first member
of the sequence that belongs to this interval
Example 1.2.2 an=2n− 3
5n + 1, ε = 10
−4.Now L = 2/5, so we are looking for an that would be on a distance from 2/5 less than
10−4 Such is, for example, a68000 = (2· 68000 − 3)/(5 · 68000 + 1) = 0.39999, because
|a68000−2
5| = 0.00001 < 10−4
These examples illustrate the requirement that, for any ε > 0, the interval (L− ε, L + ε)must contain at least one member of the sequence an However, this is not sufficient Goingback to our original example an = 1/n, we see that this condition is satisfied by 1 Indeed,for any ε > 0, the interval (1− ε, 1 + ε) contains a1 Of course, for small values of ε (such
as 0.1) the interval (1− ε, 1 + ε) contains only a1 What we really want is that it containsalmost all an
Trang 19Example 1.2.4 an=2n− 3
5n + 1, ε = 10−4.Here|a68000−2
175(5n + 1) ≤ 5(5 17
· 68000 + 1)= 0.00001 < 10−4.
In view of these examples, we will require that, starting with some member of thesequence{an}, all coming after it must belong to this interval Here is the official definition:Definition 1.2.5 We say that a real number L is a limit of a sequence{an}, and we writelim an = L (or an → L), if for any ε > 0 there exists a positive integer N such that, for
n ≥ N, |an− L| < ε In such a situation we say that the sequence {an} is convergent.Otherwise, it is divergent
Remark 1.2.6 We will mostly avoid the more cumbersome notation that includes n→ ∞
So, instead of limn→∞an, we will write lim an; also, we will write an→ L instead of an → L,
as n→ ∞ The longer version will be used only for emphasis
Remark 1.2.7 Using Definition 1.2.5 will require finding a positive integer N It is useful
to have a symbol for the set of positive integers (also known as natural numbers) and wewill use N The fact that N is a positive integer can be written as N ∈ N Occasionally, wewill be dealing with the set of non-negative integers We will denote it by N0
Now we can indeed prove that lim 1/n = 0 We will do this in two stages Our first task
is, given ε, to come up with N ; the second part consists of proving that this choice of Nindeed works In order to find N we focus on the inequality|an− L| < ε In our situation
L = 0, an = 1/n (so |an| = 1/n), and the inequality is 1/n < ε Since this is the same
as n > 1/ε, we will select N so that N > 1/ε (Reason: any n that satisfies n ≥ N, willautomatically satisfy n > 1/ε.) Since 1/ε need not be an integer, we need a description of
an integer that is bigger than 1/ε A common strategy is to use the “floor” function ⌊x⌋(also called the greatest integer function) which gives the largest integer less than or equal
to x For example,⌊3.2⌋ = 3, ⌊−3.2⌋ = −4, ⌊6⌋ = 6 Using this function, N = ⌊1/ε⌋ + 1 is
an integer that is bigger than 1/ε Now that we have N , we can write the proof
Proof Let ε > 0 and define N =⌊1/ε⌋ + 1 Suppose that n ≥ N Since N > 1/ε, we havethat n > 1/ε, so 1/n < ε, and|an− 0| = |1
n| = 1
n < ε
Notice the structure of the proof: it is the same whenever we want to establish that thesuspected number is indeed the limit of the sequence Also, in the proof we do not need toexplain how we found N , only to demonstrate that it works This makes proofs shorter buthides the motivation In the proof above (unless you can go behind the scenes), it is notimmediately clear why we defined N =⌊1/ε⌋ + 1 In general, it is always a good idea whenreading a proof to try to see where a particular choice came from (Be warned, though: it
is far from easy!)
The task of finding N can be quite complicated
Exercise 1.2.8 Let an =n
2+ 3n− 22n2− 1 Prove that{an} is convergent.
Solution We “know” that the limit is 1/2, so we focus on |an− 1/2|
an−12 =n
2+ 3n− 22n2− 1 −
1
2 =2(n2+ 3n− 2) − (2n2
− 1)2(2n2− 1) =
3(2n− 1)2(2n2− 1).
Trang 20Since both the numerator and the denominator of the last fraction are positive, for any
n∈ N, we see that the inequality |an− 1/2| < ε becomes
3(2n− 1)2(2n2− 1) < ε.
In the previous example, at this point we “solved” the inequality for n Here, it would bevery hard, and in some examples it might be impossible A better plan is to try to find asimpler expression that is bigger than (or equal to) the left side For example, 2n− 1 < 2n,and 2n2− 1 ≥ 2n2− n2= n2 so
3(2n− 1)2(2n2− 1) <
3· 2n2n2 = 3
n.Now, we require that 3/n < ε, which leads to n > 3/ε and N =⌊3/ε⌋ + 1
Proof Let ε > 0 and define N =⌊3/ε⌋ + 1 Suppose that n ≥ N Since N > 3/ε, we havethat n > 3/ε so 3/n < ε Now
an−12