# University of California Los Angeles Compute the Values Calculus Questions

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3. In this exercise, we will prove Darboux’s Theorem: Suppose f : [a; b] ! R is continuouson [a; b] and dierentiable on (a; b). Let a < x1 < x2 < b and suppose there is a L 2 Rsuch that f0(x1) < L < f0(x2) (or the other way around). Then, there exists x 2 (x1; x2)such that f0(x) = L.In particular, whilst the derivative may not be continuous it does satisfy the IntermediateValue Property.(a) Consider the function g(x) := f(x)􀀀Lx. Show that g is dierentiable on (a; b) andg0(x1) < 0 < g0(x2).(b) Argue that g must have a minimum at some point x 2 (x1; x2).(c) Use part (b) to complete the proof.13. Assume f : [a; b] ! R is Riemann integrable on [a; b].(a) Show that if one value of f(x) is changed at some point x 2 [a; b], then f is stillintegrable and integrates to the same value as before.(b) Show that the observation in (a) remains if we change only a nite number of valuesof f.(c) Find an example to show that altering f on an innite number of points may causeit cause the resulting function to no longer be Riemann integrable.

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Kenneth A. Ross
Elementary Analysis
The Theory of Calculus
Second Edition
In collaboration with Jorge M. López, University of
Puerto Rico, Rı́o Piedras
123
Kenneth A. Ross
Department of Mathematics
University of Oregon
Eugene, OR, USA
ISSN 0172-6056
ISBN 978-1-4614-6270-5
ISBN 978-1-4614-6271-2 (eBook)
DOI 10.1007/978-1-4614-6271-2
Springer New York Heidelberg Dordrecht London
Library of Congress Control Number: 2013950414
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Preface
Preface to the First Edition A study of this book, and especially the exercises, should give the reader a thorough understanding
of a few basic concepts in analysis such as continuity, convergence
of sequences and series of numbers, and convergence of sequences
and series of functions. An ability to read and write proofs will
be stressed. A precise knowledge of deﬁnitions is essential. The beginner should memorize them; such memorization will help lead to
understanding.
Chapter 1 sets the scene and, except for the completeness axiom,
should be more or less familiar. Accordingly, readers and instructors
are urged to move quickly through this chapter and refer back to it
when necessary. The most critical sections in the book are §§7–12 in
Chap. 2. If these sections are thoroughly digested and understood,
the remainder of the book should be smooth sailing.
The ﬁrst four chapters form a unit for a short course on analysis.
I cover these four chapters (except for the enrichment sections and
§20) in about 38 class periods; this includes time for quizzes and
examinations. For such a short course, my philosophy is that the
students are relatively comfortable with derivatives and integrals but
do not really understand sequences and series, much less sequences
and series of functions, so Chaps. 1–4 focus on these topics. On two
v
vi
Preface
or three occasions, I draw on the Fundamental Theorem of Calculus
or the Mean Value Theorem, which appears later in the book, but of
course these important theorems are at least discussed in a standard
calculus class.
In the early sections, especially in Chap. 2, the proofs are very
detailed with careful references for even the most elementary facts.
Most sophisticated readers ﬁnd excessive details and references a
hindrance (they break the ﬂow of the proof and tend to obscure the
main ideas) and would prefer to check the items mentally as they
proceed. Accordingly, in later chapters, the proofs will be somewhat
less detailed, and references for the simplest facts will often be omitted. This should help prepare the reader for more advanced books
which frequently give very brief arguments.
Mastery of the basic concepts in this book should make the
analysis in such areas as complex variables, diﬀerential equations,
numerical analysis, and statistics more meaningful. The book can
also serve as a foundation for an in-depth study of real analysis
given in books such as [4, 33, 34, 53, 62, 65] listed in the bibliography.
Readers planning to teach calculus will also beneﬁt from a careful
study of analysis. Even after studying this book (or writing it), it will
not be easy to handle questions such as “What is a number?” but
at least this book should help give a clearer picture of the subtleties
The enrichment sections contain discussions of some topics that I
think are important or interesting. Sometimes the topic is dealt with
lightly, and suggestions for further reading are given. Though these
sections are not particularly designed for classroom use, I hope that
some readers will use them to broaden their horizons and see how
this material ﬁts in the general scheme of things.
I have beneﬁtted from numerous helpful suggestions from my colleagues Robert Freeman, William Kantor, Richard Koch, and John
Leahy and from Timothy Hall, Gimli Khazad, and Jorge López. I
grammar and taste. Of course, remaining errors in grammar and
mathematics are the responsibility of the author.
Several users have supplied me with corrections and suggestions
that I’ve incorporated in subsequent printings. I thank them all,
Preface
vii
including Robert Messer of Albion College, who caught a subtle error
in the proof of Theorem 12.1.
Preface to the Second Edition After 32 years, it seemed time
to revise this book. Since the ﬁrst edition was so successful, I have
retained the format and material from the ﬁrst edition. The numbering of theorems, examples, and exercises in each section will be
the same, and new material will be added to some of the sections.
Every rule has an exception, and this rule is no exception. In §11,
a theorem (Theorem 11.2) has been added, which allows the simpliﬁcation of four almost-identical proofs in the section: Examples 3
and 4, Theorem 11.7 (formerly Corollary 11.4), and Theorem 11.8
(formerly Theorem 11.7).
Where appropriate, the presentation has been improved. See especially the proof of the Chain Rule 28.4, the shorter proof of Abel’s
Theorem 26.6, and the shorter treatment of decimal expansions in
§16. Also, a few examples have been added, a few exercises have been
modiﬁed or added, and a couple of exercises have been deleted.
Here are the main additions to this revision. The proof of the
irrationality of e in §16 is now accompanied by an elegant proof that
π is also irrational. Even though this is an “enrichment” section,
it is especially recommended for those who teach or will teach precollege mathematics. The Baire Category Theorem and interesting
consequences have been added to the enrichment §21. Section 31, on
Taylor’s Theorem, has been overhauled. It now includes a discussion
of Newton’s method for approximating zeros of functions, as well
as its cousin, the secant method. Proofs are provided for theorems
that guarantee when these approximation methods work. Section 35
on Riemann-Stieltjes integrals has been improved and expanded.
A new section, §38, contains an example of a continuous nowherediﬀerentiable function and a theorem that shows “most” continuous
functions are nowhere diﬀerentiable. Also, each of §§22, 32, and 33
has been modestly enhanced.
It is a pleasure to thank many people who have helped over
the years since the ﬁrst edition appeared in 1980. This includes
David M. Bloom, Robert B. Burckel, Kai Lai Chung, Mark Dalthorp
(grandson), M. K. Das (India), Richard Dowds, Ray Hoobler,
viii
Preface
Richard M. Koch, Lisa J. Madsen, Pablo V. Negrón Marrero
(Puerto Rico), Rajiv Monsurate (India), Theodore W. Palmer, Jürg
Rätz (Switzerland), Peter Renz, Karl Stromberg, and Jesús Sueiras
(Puerto Rico).
Special thanks go to my collaborator, Jorge M. López, who provided a huge amount of help and support with the revision. Working
with him was also a lot of fun. My plan to revise the book was supported from the beginning by my wife, Ruth Madsen Ross. Finally,
I thank my editor at Springer, Kaitlin Leach, who was attentive to
my needs whenever they arose.
Especially for the Student: Don’t be dismayed if you run into
material that doesn’t make sense, for whatever reason. It happens
to all of us. Just tentatively accept the result as true, set it aside as
Index or Symbols Index if some terminology or notation is puzzling.
Contents
Preface
1 Introduction
1
The Set N of Natural Numbers .
2
The Set Q of Rational Numbers
3
The Set R of Real Numbers . .
4
The Completeness Axiom . . . .
5
The Symbols +∞ and −∞ . . .
6
* A Development of R . . . . . .
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2 Sequences
7
Limits of Sequences . . . . . . . . . . . . . . .
8
A Discussion about Proofs . . . . . . . . . . .
9
Limit Theorems for Sequences . . . . . . . . .
10
Monotone Sequences and Cauchy Sequences .
11
Subsequences . . . . . . . . . . . . . . . . . . .
12
lim sup’s and lim inf’s . . . . . . . . . . . . . .
13
* Some Topological Concepts in Metric Spaces
14
Series . . . . . . . . . . . . . . . . . . . . . . .
15
Alternating Series and Integral Tests . . . . .
16
* Decimal Expansions of Real Numbers . . . .
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1
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13
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28
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33
33
39
45
56
66
78
83
95
105
109
ix
x
Contents
3 Continuity
17
Continuous Functions . . . . . . . . . . .
18
Properties of Continuous Functions . . .
19
Uniform Continuity . . . . . . . . . . . .
20
Limits of Functions . . . . . . . . . . . .
21
* More on Metric Spaces: Continuity . .
22
* More on Metric Spaces: Connectedness
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123
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178
4 Sequences and Series of Functions
23
Power Series . . . . . . . . . . . . . . . . . . .
24
Uniform Convergence . . . . . . . . . . . . . .
25
More on Uniform Convergence . . . . . . . . .
26
Diﬀerentiation and Integration of Power Series
27
* Weierstrass’s Approximation Theorem . . . .
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187
187
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200
208
216
5 Diﬀerentiation
28
Basic Properties of the Derivative
29
The Mean Value Theorem . . . .
30
* L’Hospital’s Rule . . . . . . . .
31
Taylor’s Theorem . . . . . . . . .
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280
291
298
331
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6 Integration
32
The Riemann Integral . . . . . . . .
33
Properties of the Riemann Integral
34
Fundamental Theorem of Calculus .
35
* Riemann-Stieltjes Integrals . . . .
36
* Improper Integrals . . . . . . . . .
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7 Capstone
339
37
* A Discussion of Exponents and Logarithms . . . . 339
38
* Continuous Nowhere-Diﬀerentiable Functions . . . 347
Appendix on Set Notation
365
367
A Guide to the References
394
Contents
xi
References
397
Symbols Index
403
Index
405
1
C H A P T E R

.
Introduction
The underlying space for all the analysis in this book is the set of
real numbers. In this chapter we set down some basic properties of
this set. These properties will serve as our axioms in the sense that
it is possible to derive all the properties of the real numbers using
only these axioms. However, we will avoid getting bogged down in
this endeavor. Some readers may wish to refer to the appendix on
set notation.
§1 The Set N of Natural Numbers
We denote the set {1, 2, 3, . . .} of all positive integers by N. Each
positive integer n has a successor, namely n + 1. Thus the successor
of 2 is 3, and 37 is the successor of 36. You will probably agree that
the following properties of N are obvious; at least the ﬁrst four are.
N1. 1 belongs to N.
N2. If n belongs to N, then its successor n + 1 belongs to N.
N3. 1 is not the successor of any element in N.
K.A. Ross, Elementary Analysis: The Theory of Calculus,
Undergraduate Texts in Mathematics, DOI 10.1007/978-1-4614-6271-2 1,
1
2
1. Introduction
N4. If n and m in N have the same successor, then n = m.
N5. A subset of N which contains 1, and which contains n + 1
whenever it contains n, must equal N.
Properties N1 through N5 are known as the Peano Axioms or
Peano Postulates. It turns out most familiar properties of N can be
proved based on these ﬁve axioms; see  or .
Let’s focus our attention on axiom N5, the one axiom that may
not be obvious. Here is what the axiom is saying. Consider a subset
S of N as described in N5. Then 1 belongs to S. Since S contains
n + 1 whenever it contains n, it follows that S contains 2 = 1 + 1.
Again, since S contains n + 1 whenever it contains n, it follows that
S contains 3 = 2 + 1. Once again, since S contains n + 1 whenever it
contains n, it follows that S contains 4 = 3+1. We could continue this
monotonous line of reasoning to conclude S contains any number in
N. Thus it seems reasonable to conclude S = N. It is this reasonable
conclusion that is asserted by axiom N5.
Here is another way to view axiom N5. Assume axiom N5 is false.
Then N contains a set S such that
(i) 1 ∈ S,
(ii) If n ∈ S, then n + 1 ∈ S,
and yet S = N. Consider the smallest member of the set {n ∈ N :
n ∈ S}, call it n0 . Since (i) holds, it is clear n0 = 1. So n0 is a
successor to some number in N, namely n0 − 1. We have n0 − 1 ∈ S
since n0 is the smallest member of {n ∈ N : n ∈ S}. By (ii), the
successor of n0 − 1, namely n0 , is also in S, which is a contradiction.
This discussion may be plausible, but we emphasize that we have not
proved axiom N5 using the successor notion and axioms N1 through
N4, because we implicitly used two unproven facts. We assumed
every nonempty subset of N contains a least element and we assumed
that if n0 = 1 then n0 is the successor to some number in N.
Axiom N5 is the basis of mathematical induction. Let P1 , P2 ,
P3 , . . . be a list of statements or propositions that may or may
not be true. The principle of mathematical induction asserts all the
statements P1 , P2 , P3 , . . . are true provided
(I1 ) P1 is true,
(I2 ) Pn+1 is true whenever Pn is true.
§1. The Set N of Natural Numbers
3
We will refer to (I1 ), i.e., the fact that P1 is true, as the basis for
induction and we will refer to (I2 ) as the induction step. For a sound
proof based on mathematical induction, properties (I1 ) and (I2 ) must
both be veriﬁed. In practice, (I1 ) will be easy to check.
Example 1
Prove 1 + 2 + · · · + n = 12 n(n + 1) for positive integers n.
Solution
Our nth proposition is
1
Pn : “1 + 2 + · · · + n = n(n + 1).”
2
Thus P1 asserts 1 = 12 · 1(1 + 1), P2 asserts 1 + 2 = 12 · 2(2 + 1), P37
asserts 1 + 2 + · · · + 37 = 12 · 37(37 + 1) = 703, etc. In particular, P1
is a true assertion which serves as our basis for induction.
For the induction step, suppose Pn is true. That is, we suppose
1 + 2 + · · · + n = 12 n(n + 1)
is true. Since we wish to prove Pn+1 from this, we add n + 1 to both
sides to obtain
1 + 2 + · · · + n + (n + 1) = 12 n(n + 1) + (n + 1)
= 12 [n(n + 1) + 2(n + 1)] = 12 (n + 1)(n + 2)
= 12 (n + 1)((n + 1) + 1).
Thus Pn+1 holds if Pn holds. By the principle of mathematical
induction, we conclude Pn is true for all n.
We emphasize that prior to the last sentence of our solution we
did not prove “Pn+1 is true.” We merely proved an implication: “if Pn
is true, then Pn+1 is true.” In a sense we proved an inﬁnite number
of assertions, namely: P1 is true; if P1 is true then P2 is true; if P2
is true then P3 is true; if P3 is true then P4 is true; etc. Then we
applied mathematical induction to conclude P1 is true, P2 is true, P3
is true, P4 is true, etc. We also confess that formulas like the one just
proved are easier to prove than to discover. It can be a tricky matter
to guess such a result. Sometimes results such as this are discovered
by trial and error.
4
1. Introduction
Example 2
All numbers of the form 5n − 4n − 1 are divisible by 16.
Solution
More precisely, we show 5n − 4n − 1 is divisible by 16 for each n in
N. Our nth proposition is
Pn : “5n − 4n − 1
is divisible by
16.”
The basis for induction P1 is clearly true, since 51 − 4 · 1 − 1 = 0.
Proposition P2 is also true because 52 − 4 · 2 − 1 = 16, but note
we didn’t need to check this case before proceeding to the induction
step. For the induction step, suppose Pn is true. To verify Pn+1 , the
trick is to write
5n+1 − 4(n + 1) − 1 = 5(5n − 4n − 1) + 16n.
Since 5n − 4n − 1 is a multiple of 16 by the induction hypothesis, it
follows that 5n+1 − 4(n + 1) − 1 is also a multiple of 16. In fact, if
5n − 4n − 1 = 16m, then 5n+1 − 4(n + 1) − 1 = 16 · (5m + n). We have
shown Pn implies Pn+1 , so the induction step holds. An application
of mathematical induction completes the proof.
Example 3
Show | sin nx| ≤ n| sin x| for all positive integers n and all real
numbers x.
Solution
Our nth proposition is
Pn : “| sin nx| ≤ n| sin x|
for all real numbers x.”
The basis for induction is again clear. Suppose Pn is true. We apply
the addition formula for sine to obtain
| sin(n + 1)x| = | sin(nx + x)| = | sin nx cos x + cos nx sin x|.
Now we apply the Triangle Inequality and properties of the absolute
value [see Theorems 3.7 and 3.5] to obtain
| sin(n + 1)x| ≤ | sin nx| · | cos x| + | cos nx| · | sin x|.
Since | cos y| ≤ 1 for all y we see that
| sin(n + 1)x| ≤ | sin nx| + | sin x|.
Exercises
5
Now we apply the induction hypothesis Pn to obtain
| sin(n + 1)x| ≤ n| sin x| + | sin x| = (n + 1)| sin x|.
Thus Pn+1 holds. Finally, the result holds for all n by mathematical
induction.
Exercises
1.1 Prove 12 + 22 + · · ·+ n2 = 16 n(n + 1)(2n + 1) for all positive integers n.
1.2 Prove 3 + 11 + · · · + (8n − 5) = 4n2 − n for all positive integers n.
1.3 Prove 13 + 23 + · · · + n3 = (1 + 2 + · · · + n)2 for all positive integers n.
1.4 (a) Guess a formula for 1 + 3 + · · · + (2n − 1) by evaluating the sum
for n = 1, 2, 3, and 4. [For n = 1, the sum is simply 1.]
(b) Prove your formula using mathematical induction.
1.5 Prove 1 +
1
2
n
+
1
4
+ ···+
1
2n
=2−
1
2n
for all positive integers n.
n
1.6 Prove (11) − 4 is divisible by 7 when n is a positive integer.
1.7 Prove 7n − 6n − 1 is divisible by 36 for all positive integers n.
1.8 The principle of mathematical induction can be extended as follows.
A list Pm , Pm+1 , . . . of propositions is true provided (i) Pm is true,
(ii) Pn+1 is true whenever Pn is true and n ≥ m.
(a) Prove n2 > n + 1 for all integers n ≥ 2.
(b) Prove n! > n2 for all integers n ≥ 4. [Recall n! = n(n − 1) · · · 2 · 1;
for example, 5! = 5 · 4 · 3 · 2 · 1 = 120.]
1.9 (a) Decide for which integers the inequality 2n > n2 is true.
(b) Prove your claim in (a) by mathematical induction.
1.10 Prove (2n + 1) + (2n + 3) + (2n + 5) + · · · + (4n − 1) = 3n2 for all
positive integers n.
1.11 For each n ∈ N, let Pn denote the assertion “n2 + 5n + 1 is an even
integer.”
(a) Prove Pn+1 is true whenever Pn is true.
(b) For which n is Pn actually true? What is the moral of this
exercise?
6
1. Introduction
1.12 For n ∈ N, let n! [read “n factorial”] denote the product 1 · 2 · 3 · · · n.
Also let 0! = 1 and deﬁne

n
n!
=
for k = 0, 1, . . . , n.
(1.1)
k
k!(n − k)!
The binomial theorem asserts that
n
n
n
n
n
an +
an−1 b+
an−2 b2 + · · · +
abn−1 +
bn
0
1
2
n−1
n
1
=an +nan−1 b+ n(n−1)an−2 b2 + · · · +nabn−1 +bn .
2
(a+b)n =
(a) Verify the binomial theorem for n = 1, 2, and 3.
n n+1
(b) Show nk + k−1
= k for k = 1, 2, . . . , n.
(c) Prove the binomial theorem using mathematical induction and
part (b).
§2 The Set Q of Rational Numbers
Small children ﬁrst learn to add and to multiply positive integers.
After subtraction is introduced, the need to expand the number system to include 0 and negative integers becomes apparent. At this
point the world of numbers is enlarged to include the set Z of all
integers. Thus we have Z = {0, 1, −1, 2, −2, . . .}.
Soon the space Z also becomes inadequate when division is introduced. The solution is to enlarge the world of numbers to include
all fractions. Accordingly, we study the space Q of all rational numbers, i.e., numbers of the form m
n where m, n ∈ Z and n = 0. Note
that Q contains all terminating decimals such as 1.492 = 1,492
1,000 . The
connection between decimals and real numbers is discussed in 10.3
on page 58 and in §16. The space Q is a highly satisfactory algebraic system in which the basic operations addition, multiplication,
subtraction and division can be fully studied. No system is perfect,
however, and Q is inadequate in some ways. In this section we will
consider the defects of Q. In the next section we will stress the good
features of Q and then move on to the system of real numbers.
The set Q of rational numbers is a very nice algebraic system until
one tries to solve equations like x2 = 2. It turns out that no rational
§2. The Set Q of Rational Numbers
7
FIGURE 2.1
number satisﬁes this equation, and yet there are good reasons to
believe some kind of number satisﬁes this equation. Consider, for
example, a square with sides having length one; see Fig. 2.1. If d is
the length of the diagonal, then from geometry we know 12 +12 = d2 ,
there
i.e., d2 = 2. Apparently

√ is a positive length whose square is 2,
which we write as 2. But 2 cannot
be a rational number, as we will

show in Example 2. Of course, 2 can be approximated by rational
numbers. There are rational numbers whose squares are close to 2;
for example, (1.4142)2 = 1.99996164 and (1.4143)2 = 2.00024449.
It is evident that there are lots of rational numbers and yet there
are “gaps” in Q. Here is another way to view this situation. Consider
the graph of the polynomial x2 −2 in Fig. 2.2. Does the graph of x2 −2
cross the x-axis? We are inclined to say it does, because when we
draw the x-axis we include “all” the points. We allow no “gaps.” But
notice that the graph of x2 − 2 slips by all the rational numbers on
the x-axis. The x-axis is our picture of the number line, and the set
of rational numbers again appears to have signiﬁcant “gaps.”
There are even more exotic numbers such as π and e that are not
rational numbers, but which come up naturally in mathematics. The
number π is basic to the study of circles and spheres, and e arises in
problems of exponential
growth.

We return to 2. This is an example of what is called an algebraic
number because it satisﬁes the equation x2 − 2 = 0.
8
1. Introduction
FIGURE 2.2
2.1 Definition.
A number is called an algebraic number if it satisﬁes a polynomial
equation
cn xn + cn−1 xn−1 + · · · + c1 x + c0 = 0
where the coeﬃcients c0 , c1 , . . . , cn are integers, cn = 0 and n ≥ 1.
Rational numbers are always algebraic numbers. In fact, if r = m
n
is a rational number [m, n ∈ Z and n = 0], then√ it satisﬁes
the

3
,
, etc. [or
equation nx − m = 0. Numbers deﬁned in terms of
fractional exponents, if you prefer] and ordinary algebraic operations
on the rational numbers are invariably algebraic numbers.
Example 1

√ √

3
3
4−2 3
4
,
3,
17,
2
+
5
and
are algebraic numbers. In fact,
17
√7
4
3 is a solution of x2 − 3 = 0, and
17 is a solution of 17x − 4 = 0,

3
3 − 17 = 0. The expression a =
17 is a solution
of
x
2 + 3 5 means

a2 = 2 + 3 5 or a2 − 2 = 3 5 so that (a2 − 2)3 = 5. Therefore we
§2. The Set Q of Rational Numbers
9

have a6 − 6a4 + 12a2 − 13 = 0, which shows a = 2 + 3 5 satisﬁes
the polynomial equation x6 − 6×4 + 12×2 − 13 = 0.

4−2 3
2 = 4 − 2 3,
to
7b
Similarly, the expression b =
7

hence 2 3 = 4−7b2 , hence 12 = (4−7b2 )2 , hence 49b4 −56b2 +4 = 0.
Thus b satisﬁes the polynomial equation 49×4 − 56×2 + 4 = 0.
The next theorem may be familiar from elementary algebra. It is
the theorem that justiﬁes the following remarks: the only possible rational solutions of x3 −7×2 +2x−12 = 0 are ±1, ±2, ±3, ±4, ±6, ±12,
so the only possible (rational) monomial factors of x3 − 7×2 + 2x − 12
are x − 1, x + 1, x − 2, x + 2, x − 3, x + 3, x − 4, x + 4, x − 6, x + 6,
x − 12, x + 12. We won’t pursue these algebraic problems; we merely
make these observations in the hope they will be familiar.
The next theorem also allows one to prove algebraic numbers that
do not√look like rational numbers are usually not √
rational
√ numbers.

Thus 4 is obviously a rational number, while 2, 3, 5, etc.
turn out to be nonrational. See the examples following the theorem.
Also, compare Exercise 2.7. Recall that an integer k is a factor of an
integer m or divides m if m
k is also an integer.
If the next theorem seems complicated, ﬁrst read the special case
in Corollary 2.3 and Examples 2–5.
2.2 Rational Zeros Theorem.
Suppose c0 , c1 , . . . , cn are integers and r is a rational number
satisfying the polynomial equation
cn xn + cn−1 xn−1 + · · · + c1 x + c0 = 0
(1)
where n ≥ 1, cn = 0 and c0 = 0. Let r = dc where c, d are integers having no common factors and d = 0. Then c divides c0 and d
divides cn .
In other words, the only rational candidates for solutions of (1)
have the form dc where c divides c0 and d divides cn .
Proof
We are given
c n
c n−1
c
cn
+ cn−1
+ · · · + c1
+ c0 = 0.
d
d
d
10
1. Introduction
We multiply through by dn and obtain
cn cn + cn−1 cn−1 d + cn−2 cn−2 d2 + · · · + c2 c2 dn−2 + c1 cdn−1 + c0 dn = 0.
(2)
If we solve for c0 dn , we obtain
c0 dn = −c[cn cn−1 +cn−1 cn−2 d+cn−2 cn−3 d2 +· · · +c2 cdn−2 +c1 dn−1 ].
It follows that c divides c0 dn . But c and dn have no common factors,
so c divides c0 . This follows from the basic fact that if an integer
c divides a product ab of integers, and if c and b have no common
factors, then c divides a. See, for example, Theorem 1.10 in .
Now we solve (2) for cn cn and obtain
cn cn = −d[cn−1 cn−1 +cn−2 cn−2 d+· · · +c2 c2 dn−3 +c1 cdn−2 +c0 dn−1 ].
Thus d divides cn cn . Since cn and d have no common factors,
d divides cn .
2.3 Corollary.
Consider the polynomial equation
xn + cn−1 xn−1 + · · · + c1 x + c0 = 0,
where the coeﬃcients c0 , c1 , . . . , cn−1 are integers and c0 = 0.1 Any
rational solution of this equation must be an integer that divides c0 .
Proof
In the Rational Zeros Theorem 2.2, the denominator of r must divide
the coeﬃcient of xn , which is 1 in this case. Thus r is an integer and
it divides c0 .
Example
2

2 is not a rational number.
Proof
By Corollary 2.3, the only rational numbers that could possibly be
solutions of x2 − 2 = 0 are ±1, ±2. [Here n = 2, c2 = 1, c1 = 0,
c0 = −2. So the rational solutions have the form dc where c divides
1
Polynomials like this, where the highest power has coeﬃcient 1, are called monic
polynomials.
§2. The Set Q of Rational Numbers
11
c0 = −2 and d divides c2 = 1.] One can substitute each of the four
numbers ±1, ±2 into the equation x2 − 2 = 0 to√quickly eliminate
them as possible solutions of the equation. Since 2 is a solution of
x2 − 2 = 0, it cannot be a rational number.
Example
3

17 is not a rational number.
Proof
The only possible rational solutions of x2 − 17 = 0 are ±1, ±17, and
none of these numbers are solutions.
Example
4

3
6 is not a rational number.
Proof
The only possible rational solutions of x3 −6 = 0 are ±1, ±2, ±3, ±6.
It is easy to verify that none of these eight numbers satisﬁes the
equation x3 − 6 = 0.
Example
5

a = 2 + 3 5 is not a rational number.
Proof
In Example 1 we showed a is a solution of x6 − 6×4 + 12×2 − 13 = 0.
By Corollary 2.3, the only possible rational solutions are ±1, ±13.
When x = 1 or −1, the left hand side of the equation is −6 and
when x = 13 or −13, the left hand side of the equation turns out to
equal 4,657,458. This last computation could be avoided by using a
little common sense. Either observe a is “obviously” bigger than 1
and less than 13, or observe
136 − 6 · 134 + 12 · 132 − 13 = 13(135 − 6 · 133 + 12 · 13 − 1) = 0
since the term in parentheses cannot be zero: it is one less than some
multiple of 13.
Example 6
b=

4−2 3
7
is not a rational number.
12
1. Introduction
Proof
In Example 1 we showed b is a solution of 49×4 − 56×2 + 4 = 0. By
Theorem 2.2, the only possible rational solutions are
±1, ±1/7, ±1/49, ±2, ±2/7, ±2/49, ±4, ±4/7, ±4/49.
To complete our proof, all we need to do is substitute these 18 candidates into the equation 49×4 − 56×2 + 4 = 0. This prospect is
so discouraging, however, that we choose to ﬁnd a more clever approach. In Example 1, we also showed 12 = (4 − 7b2)2 . Now if b were
rational, then 4 − 7b2 would also be rational [Exercise 2.6], so the
equation 12 = x2 would have a rational solution. But the only possible rational solutions to x2 − 12 = 0 are ±1, ±2, ±3, ±4, ±6, ±12,
and these all can be eliminated by mentally substituting them into
the equation. We conclude 4 − 7b2 cannot be rational, so b cannot
be rational.
As a practical matter, many or all of the rational candidates given
by the Rational Zeros Theorem can be eliminated by approximating
the quantity in question. It is nearly obvious that the values in Examples 2 through 5 are not integers, while all the rational candidates
are. The number b in Example 6 is approximately 0.2767; the nearest
rational candidate is +2/7 which is approximately 0.2857.
It should be noted that not all irrational-looking expressions are
actually irrational. See Exercise 2.7.
2.4 Remark.
While admiring the eﬃcient Rational Zeros Theorem for ﬁnding
rational zeros of polynomials with integer coeﬃcients, you might
wonder how one would ﬁnd other zeros of these polynomials, or zeros of other functions. In §31, we will discuss the most well-known
method, called Newton’s method, and its cousin, the secant method.
That discussion can be read now; only the proof of the theorem uses
material from §31.
Exercises
√ √ √ √

3, 5, 7, 24, and 31 are not rational numbers.
√ √

2.2 Show 3 2, 7 5 and 4 13 are not rational numbers.
2.1 Show
§3. The Set R of Real Numbers
13

2 + 2 is not a rational number.

3
2.4 Show 5 − 3 is not a rational number.

2.5 Show [3 + 2]2/3 is not a rational number.
2.3 Show
2.6 In connection with Example 6, discuss why 4 − 7b2 is rational if b is
rational.
2.7 Show the following
irrational-looking expressions
are actually rational

numbers: (a) 4 + 2 3 − 3, and (b) 6 + 4 2 − 2.
2.8 Find all rational solutions of the equation x8 −4×5 +13×3 −7x +1 = 0.
§3 The Set R of Real Numbers
The set Q is probably the largest system of numbers with which
you really feel comfortable. There are some subtleties but you have
learned to cope with them. For example, Q is not simply the set of
symbols m/n, where m, n ∈ Z, n = 0, since we regard some pairs of
diﬀerent looking fractions as equal. For example, 24 and 36 represent
the same element of Q. A rigorous development of Q based on Z,
which in turn is based on N, would require us to introduce the notion
of equivalence classes. In this book we assume a familiarity with and
understanding of Q as an algebraic system. However, in order to
clarify exactly what we need to know about Q, we set down some of
its basic axioms and properties.
The basic algebraic operations in Q are addition and multiplication. Given a pair a, b of rational numbers, the sum a + b and the
product ab also represent rational numbers. Moreover, the following
properties hold.
A1.
A2.
A3.
A4.
M1.
M2.
a + (b + c) = (a + b) + c for all a, b, c.
a + b = b + a for all a, b.
a + 0 = a for all a.
For each a, there is an element −a such that a + (−a) = 0.
a(bc) = (ab)c for all a, b, c.
ab = ba for all a, b.
14
1. Introduction
M3. a · 1 = a for all a.
M4. For each a = 0, there is an element a−1 such that aa−1 = 1.
DL a(b + c) = ab + ac for all a, b, c.
Properties A1 and M1 are called the associative laws, and properties A2 and M2 are the commutative laws. Property DL is the
distributive law; this is the least obvious law and is the one that
justiﬁes “factorization” and “multiplying out” in algebra. A system
that has more than one element and satisﬁes these nine properties is
called a ﬁeld. The basic algebraic properties of Q can proved solely
on the basis of these ﬁeld properties. We will not pursue this topic
in any depth, but we illustrate our claim by proving some familiar
properties in Theorem 3.1 below.
The set Q also has an order structure ≤ satisfying
O1.
O2.
O3.
O4.
O5.
Given a and b, either a ≤ b or b ≤ a.
If a ≤ b and b ≤ a, then a = b.
If a ≤ b and b ≤ c, then a ≤ c.
If a ≤ b, then a + c ≤ b + c.
If a ≤ b and 0 ≤ c, then ac ≤ bc.
Property O3 is called the transitive law. This is the characteristic
property of an ordering. A ﬁeld with an ordering satisfying properties
O1 through O5 is called an ordered ﬁeld. Most of the algebraic and
order properties of Q can be established for an ordered ﬁeld. We will
prove a few of them in Theorem 3.2 below.
The mathematical system on which we will do our analysis will
be the set R of all real numbers. The set R will include all rational
numbers, all algebraic numbers, π, e, and more. It will be a set that
can be drawn as the real number line; see Fig. 3.1. That is, every
real number will correspond to a point on the number line, and
every point on the number line will correspond to a real number.
In particular, unlike Q, R will not have any “gaps.” We will also
see that real numbers have decimal expansions; see 10.3 on page 58
and §16. These remarks help describe R, but we certainly have not
deﬁned R as a precise mathematical object. It turns out that R can
be deﬁned entirely in terms of the set Q of rational numbers; we
indicate in the enrichment §6 one way this can be done. But then
it is a long and tedious task to show how to add and multiply the
§3. The Set R of Real Numbers
15
FIGURE 3.1
objects deﬁned in this way and to show that the set R, with these
operations, satisﬁes all the familiar algebraic and order properties
we expect to hold for R. To develop R properly from Q in this way
and to develop Q properly from N would take us several chapters.
This would defeat the purpose of this book, which is to accept R as
a mathematical system and to study some important properties of
R and functions on R. Nevertheless, it is desirable to specify exactly
what properties of R we are assuming.
Real numbers, i.e., elements of R, can be added together and
multiplied together. That is, given real numbers a and b, the sum
a+b and the product ab also represent real numbers. Moreover, these
operations satisfy the ﬁeld properties A1 through A4, M1 through
M4, and DL. The set R also has an order structure ≤ that satisﬁes
properties O1 through O5. Thus, like Q, R is an ordered ﬁeld.
In the remainder of this section, we will obtain some results for
R that are valid in any ordered ﬁeld. In particular, these results
would be equally valid if we restricted our attention to Q. These
remarks emphasize the similarities between R and Q. We have not
yet indicated how R can be distinguished from Q as a mathematical
object, although we have asserted that R has no “gaps.” We will
make this observation much more precise in the next section, and
then we will give a “gap ﬁlling” axiom that ﬁnally will distinguish R
from Q.
3.1 Theorem.
The following are consequences of the ﬁeld properties:
(i) a + c = b + c implies a = b;
(ii) a · 0 = 0 for all a;
(iii) (−a)b = −ab for all a, b;
(iv) (−a)(−b) = ab for all a, b;
(v) ac = bc and c = 0 imply a = b;
(vi) ab = 0 implies either a = 0 or b = 0;
for a, b, c ∈ R.
16
1. Introduction
Proof
(i) a + c = b + c implies (a + c) + (−c) = (b + c) + (−c), so by A1,
we have a + [c + (−c)] = b + [c + (−c)]. By A4, this reduces to
a + 0 = b + 0, so a = b by A3.
(ii) We use A3 and DL to obtain a · 0 = a · (0 + 0) = a · 0 + a · 0,
so 0 + a · 0 = a · 0 + a · 0. By (i) we conclude 0 = a · 0.
(iii) Since a + (−a) = 0, we have ab + (−a)b = [a + (−a)] · b =
0 · b = 0 = ab + (−(ab)). From (i) we obtain (−a)b = −(ab).
(iv) and (v) are left to Exercise 3.3.
(vi) If ab = 0 and b = 0, then 0 = b−1 · 0 = 0 · b−1 = (ab) · b−1 =
a(bb−1 ) = a · 1 = a.
3.2 Theorem.
The following are consequences of the properties of an ordered ﬁeld:
(i) If a ≤ b, then −b ≤ −a;
(ii) If a ≤ b and c ≤ 0, then bc ≤ ac;
(iii) If 0 ≤ a and 0 ≤ b, then 0 ≤ ab;
(iv) 0 ≤ a2 for all a;
(v) 0 = b.
Proof
(i) Suppose a ≤ b. By O4 applied to c = (−a) + (−b), we have
a + [(−a) + (−b)] ≤ b + [(−a) + (−b)]. It follows that −b ≤ −a.
(ii) If a ≤ b and c ≤ 0, then 0 ≤ −c by (i). Now by O5 we have
a(−c) ≤ b(−c), i.e., −ac ≤ −bc. From (i) again, we see bc ≤ ac.
(iii) If we put a = 0 in property O5, we obtain: 0 ≤ b and 0 ≤ c
imply 0 ≤ bc. Except for notation, this is exactly assertion
(iii).
(iv) For any a, either a ≥ 0 or a ≤ 0 by O1. If a ≥ 0, then a2 ≥ 0
by (iii). If a ≤ 0, then we have −a ≥ 0 by (i), so (−a)2 ≥ 0,
i.e., a2 ≥ 0.
(v) Is left to Exercise 3.4.
§3. The Set R of Real Numbers
17
(vi) Suppose 0 0. We will use expressions like “obviously”
and “clearly” in place of very simple arguments, but we will
not use these terms to obscure subtle points.]
18
1. Introduction
(ii) There are four easy cases here. If a ≥ 0 and b ≥ 0, then ab ≥ 0,
so |a| · |b| = ab = |ab|. If a ≤ 0 and b ≤ 0, then −a ≥ 0, −b ≥ 0
and (−a)(−b) ≥ 0 so that |a| · |b| = (−a)(−b) = ab = |ab|.
If a ≥ 0 and b ≤ 0, then −b ≥ 0 and a(−b) ≥ 0 so that
|a| · |b| = a(−b) = −(ab) = |ab|. If a ≤ 0 and b ≥ 0, then
−a ≥ 0 and (−a)b ≥ 0 so that |a| · |b| = (−a)b = −ab = |ab|.
(iii) The inequalities −|a| ≤ a ≤ |a| are obvious, since either
a = |a| or else a = −|a|. Similarly −|b| ≤ b ≤ |b|. Now four
applications of O4 yield
−|a| + (−|b|) ≤ a + b ≤ |a| + b ≤ |a| + |b|
so that
−(|a| + |b|) ≤ a + b ≤ |a| + |b|.
This tells us a + b ≤ |a| + |b| and also −(a + b) ≤ |a| + |b|.
Since |a + b| is equal to either a + b or −(a + b), we conclude
|a + b| ≤ |a| + |b|.
3.6 Corollary.
dist(a, c) ≤ dist(a, b) + dist(b, c) for all a, b, c ∈ R.
Proof
We can apply inequality (iii) of Theorem 3.5 to a − b and b − c to
obtain |(a − b) + (b − c)| ≤ |a − b| + |b − c| or dist(a, c) = |a − c| ≤
|a − b| + |b − c| ≤ dist(a, b) + dist(b, c).
The inequality in Corollary 3.6 is very closely related to an
inequality concerning points a, b, c in the plane, and the latter inequality can be interpreted as a statement about triangles: the length
of a side of a triangle is less than or equal to the sum of the lengths
of the other two sides. See Fig. 3.2. For this reason, the inequality
in Corollary 3.6 and its close relative (iii) in Theorem 3.5 are often
called the Triangle Inequality.
3.7 Triangle Inequality.
|a + b| ≤ |a| + |b| for all a, b.
A useful variant of the triangle inequality is given in Exercise 3.5(b).
Exercises
19
FIGURE 3.2
Exercises
3.1 (a) Which of the properties A1–A4, M1–M4, DL, O1–O5 fail for N?
(b) Which of these properties fail for Z?
3.2 (a) The commutative law A2 was used in the proof of (ii) in
Theorem 3.1. Where?
(b) The commutative law A2 was also used in the proof of (iii) in
Theorem 3.1. Where?
3.3 Prove (iv) and (v) of Theorem 3.1.
3.4 Prove (v) and (vii) of Theorem 3.2.
3.5 (a) Show |b| ≤ a if and only if −a ≤ b ≤ a.
(b) Prove ||a| − |b|| ≤ |a − b| for all a, b ∈ R.
3.6 (a) Prove |a + b + c| ≤ |a| + |b| + |c| for all a, b, c ∈ R. Hint : Apply the
triangle inequality twice. Do not consider eight cases.
(b) Use induction to prove
|a1 + a2 + · · · + an | ≤ |a1 | + |a2 | + · · · + |an |
for n numbers a1 , a2 , . . . , an .
3.7 (a) Show |b| b, then a ≤ b.
20
§4
1. Introduction
The Completeness Axiom
In this section we give the completeness axiom for R. This is the
axiom that will assure us R has no “gaps.” It has far-reaching consequences and almost every signiﬁcant result in this book relies on it.
Most theorems in this book would be false if we restricted our world
of numbers to the set Q of rational numbers.
4.1 Definition.
Let S be a nonempty subset of R.
(a) If S contains a largest element s0 [that is, s0 belongs to S and
s ≤ s0 for all s ∈ S], then we call s0 the maximum of S and
write s0 = max S.
(b) If S contains a smallest element, then we call the smallest
element the minimum of S and write it as min S.
Example 1
(a) Every ﬁnite nonempty subset of R has a maximum and a
minimum. Thus
max{1, 2, 3, 4, 5} = 5 and
max{0, π, −7, e, 3, 4/3} = π
and
min{1, 2, 3, 4, 5} = 1,
min{0, π, −7, e, 3, 4/3} = −7,
max{n ∈ Z : −4 M1 .
Example 3
(a) If a set S has a maximum, then max S = sup S. A similar
remark applies to sets that have inﬁmums.
(b) If a, b ∈ R and a = 7 for all
r ∈ Q, by the Rational Zeros Theorem 2.2. However, sup(B) =

3
7. The set B is not bounded below; if this isn’t obvious, think
about the graph of y = x3 . Clearly B has no minimum. Starting
with the next section,√we would write inf(B) = −∞.
(c) The set C = {m + n 2 : m, n ∈ Z} isn’t bounded above or
below, so it has no maximum or minimum. We could write
sup(C) = +∞ and inf(C) = −∞.
2
(d) The
√ set√D = {x ∈ R : x 0,
then
1
0,
then
b 0
and b > 0 such that a 0 and b > 0, then for some positive integer n, we have na > b.
This tells us that, even if a is quite small and b is quite large,
some integer multiple of a will exceed b. Or, to quote , given enough
time, one can empty a large bathtub with a small spoon. [Note that
if we set b = 1, we obtain assertion (*), and if we set a = 1, we
obtain assertion (**).]
Proof
Assume the Archimedean property fails. Then there exist a > 0 and
b > 0 such that na ≤ b for all n ∈ N. In particular, b is an upper
bound for the set S = {na : n ∈ N}. Let s0 = sup S; this is where we
are using the completeness axiom. Since a > 0, we have s0 0,
an 0, the Archimedean property shows there exists an
n ∈ N such that
n(b − a) > 1,
and hence bn − an > 1.
(2)
From this, it is fairly evident that there is an integer m between an
and bn, so that (1) holds. However, the proof that such an m exists is
a little delicate. We argue as follows. By the Archimedean property
again, there exists an integer k > max{|an|, |bn|}, so that
−k −k, we
have m − 1 in K, so the inequality an 0, then there exists n ∈ N such that
1
n
> 0. Then there exists a ∈ A and
b ∈ B, so that a > sup A − 2 and b > sup B − 2 . Then a + b ∈ A + B
and so sup(A + B) ≥ a + b > sup A + sup B − . Since > 0 is
arbitrary, we conclude sup(A + B) ≥ sup A + sup B.
The exercises for this section clear up some loose ends. Most of
them extend results in §4 to sets that are not necessarily bounded.
30
1. Introduction
Exercises
5.1 Write the following sets in interval notation:
(a) {x ∈ R : x = Q and α is not empty,
(ii) If r ∈ α, s ∈ Q and s > 0 there exists a number N
n>N
such that
implies |sn − s| .
(1)
If (sn ) converges to s, we will write limn→∞ sn = s, or sn → s. The
number s is called the limit of the sequence (sn ). A sequence that
does not converge to some real number is said to diverge.
Several comments are in order. First, in view of the Archimedean
property, the number N in Deﬁnition 7.1 can be taken to be a positive
integer if we wish. Second, the symbol [lower case Greek epsilon]
in this deﬁnition represents a positive number, not some new exotic
number. However, it is traditional in mathematics to use and δ
[lower case Greek delta] in situations where the interesting or challenging values are the small positive values. Third, condition (1) is
an inﬁnite number of statements, one for each positive value of .
The condition states that to each > 0 there corresponds a number
N with a certain property, namely n > N implies |sn − s| . The
value N depends on the value , and normally N will have to be
large if is small. We illustrate these remarks in the next example.
Example 2
Consider the sequence sn =
1
n
3n+1
7n−4 .
If we write sn as
1
3+ n
4
7− n
and note
and n4 are very small for large n, it seems reasonable to conclude
lim sn = 37 . In fact, this reasoning will be completely valid after we
36
2. Sequences
have the limit theorems in §9:

3 + n1
lim 3 + lim( n1 )
3+0
3
lim sn = lim
=
4
1 = 7 − 4 · 0 = 7.
7− n
lim 7 − 4 lim( n )
However, for now we are interested in analyzing exactly what we
mean by lim sn = 37 . By Deﬁnition 7.1, lim sn = 37 means
for each > 0 there exists a number N such that

3
n > N implies 3n+1
7n−4 − 7 .
(1)
As varies, N varies. In Example 2 of the next section we will show
19
that, for this particular sequence, N can be taken to be 49
+ 47 .
Using this observation, we ﬁnd that for equal to 1, 0.1, 0.01, 0.001,
and 0.000001, respectively, N can be taken to be approximately 0.96,
4.45, 39.35, 388.33, and 387,755.67, respectively. Since we are interested only in integer values of n, we may as well drop the fractional
part of N . Then we see ﬁve of the inﬁnitely many statements given
by (1) are:

3n + 1 3

n > 0 implies

3n + 1 3
− 4 implies
7n − 4 7

3n + 1 3

− 39 implies
7n − 4 7

3n + 1 3
− 388 implies
7n − 4 7

3n + 1 3

− 387,755 implies
7n − 4 7
Table 7.1 partially conﬁrms assertions (2) through (6). We could go
on and on with these numerical illustrations, but it should be clear
we need a more theoretical approach if we are going to prove results
Example 3
§7. Limits of Sequences
37
TABLE 7.1
n
sn = 3n+1
7n−4
is approximately
|sn − 37 |
is approximately
2
3
4
5
6
40
400
0.7000
0.5882
0.5417
0.5161
0.5000
0.4384
0.4295
0.2714
0.1597
0.1131
0.0876
0.0714
0.0098
0.0010
(a) lim n12 = 0. This will be proved in Example 1 of the next
section.
(b) The sequence (an ) where an = (−1)n does not converge. Thus
the expression “lim an ” is meaningless in this case. We will
discuss this example again in Example 4 of the next section.
(c) The sequence cos( nπ
3 ) does not converge. See Exercise 8.7.
1/n
appears to converge to 1. We will prove
(d) The sequence n
lim n1/n = 1 in Theorem 9.7(c) on page 48.
(e) The sequence (bn ) where bn = (1+ n1 )n converges to the number
e that should be familiar from calculus. The limit lim bn and
the number e will be discussed further in Example 6 in §16 and
in §37. Recall e is approximately 2.7182818.
We conclude this section by showing that limits are unique. That
is, if lim sn = s and lim sn = t, then we must have s = t. In short,
the values sn cannot be getting arbitrarily close to diﬀerent values
for large n. To prove this, consider > 0. By the deﬁnition of limit
there exists N1 so that

n > N1 implies |sn − s|
n > N2 implies |sn − t| max{N1 , N2 }, the Triangle Inequality 3.7 shows

|s − t| = |(s − sn ) + (sn − t)| ≤ |s − sn | + |sn − t| ≤ + = .
2 2
38
2. Sequences
This shows |s − t| for all > 0. It follows that |s − t| = 0; hence
s = t.
Exercises
7.1 Write out the ﬁrst ﬁve terms of the following sequences.
1
3n+1
(a) sn = 3n+1
(b) bn = 4n−1
n

(c) cn = 3n
(d) sin( 4 )
7.2 For each sequence in Exercise 7.1, determine whether it converges. If
it converges, give its limit. No proofs are required.
7.3 For each sequence below, determine whether it converges and, if it
converges, give its limit. No proofs are required.
2
n
+3
(a) an = n+1
(b) bn = nn2 −3
(c) cn = 2−n
(d) tn = 1 + n2
n
(e) xn = 73 + (−1)
(f ) sn = (2)1/n
(g) yn = n!
(h) dn = (−1)n n
3
(−1)n
+8n
(i) n
(j) 7n
2n3 −3
2
−18
(k) 9n
6n+18
(m) sin(nπ)
(o) n1 sin n
n
(q) 3n!
2
+3
(s) 4n
3n2 −2
(l) sin( nπ
2 )
2nπ
(n) sin( 3 )
n+1
+5
(p) 22n −7
(r) (1 + n1 )2
6n+4
(t) 9n
2 +7
7.4 Give examples of
(a) A sequence (xn ) of irrational numbers having a limit lim xn
that is a rational number.
(b) A sequence (rn ) of rational numbers having a limit lim rn that
is an irrational number.
7.5 Determine the following limits. No proofs are required, but show any
relevant algebra.

(a) lim sn where sn = n2 + 1 − n,

(b) lim( n2 + n − n),

(c) lim( 4n2 + n − 2n).
1
Hint for (a): First show sn = √n2 +1+n
.
39
In this section we give several examples of proofs using the deﬁnition
of the limit of a sequence. With a little study and practice, students
should be able to do proofs of this sort themselves. We will sometimes refer to a proof as a formal proof to emphasize it is a rigorous
mathematical proof.
Example 1
Prove lim n12 = 0.
Discussion. Our task is to consider an arbitrary > 0 and show
there exists a number N [which will depend on ] such that n > N
implies | n12 − 0| . So we expect our formal proof to begin with
“Let > 0” and to end with something like “Hence n > N implies
| n12 − 0| .” In between the proof should specify an N and then
verify N has the desired property, namely n > N does indeed imply
| n12 − 0| .
As is often the case with trigonometric identities, we will initially
work backward from our desired conclusion, but in the formal proof
we will have to be sure our steps are reversible. In the present example, we want | n12 − 0| and we want to know how big n must be.
So we will operate on this inequality algebraically and try to “solve”
for n. Thus we want n12 . By multiplying both sides by n2 and
dividing both sides by , we ﬁnd we want 1
we put N =
√1 .

Formal Proof
Let > 0. Let N =
n2
1

√1 .

> and hence >
proves lim n12 = 0.
Example 2
3
Prove lim 3n+1
7n−4 = 7 .
√1

implies | n12 −0| . This suggests
√1 which

| n12 − 0| N implies n >
1
.
n2
Thus n > N implies
implies
. This
40
2. Sequences
Discussion. For each > 0, we need to to decide how big n must
3
be to guarantee | 3n+1
7n−4 − 7 | . Thus we want

21n + 7 − 21n + 12
19
or

7(7n − 4) .

7(4n − 4)
Since 7n − 4 > 0, we can drop the absolute value and manipulate
the inequality further to “solve” for n:
19

or
19
+ 4
or
19
4
+ 7
19
+ 47 . Incidentally, we
Our steps are reversible, so we will put N = 49
19
could have chosen N to be any number larger than 49
+ 47 .
Formal Proof
19
19
Let > 0 and let N = 49
+ 47 . Then n > N implies n > 49
+ 47 ,
19
19
19
hence 7n > 7 + 4, hence 7n − 4 > 7 , hence 7(7n−4) , and hence
3
3n+1
3
| 3n+1
7n−4 − 7 | . This proves lim 7n−4 = 7 .
Example 3
3
Prove lim 4nn3+3n
= 4.
−6
Discussion. For each > 0, we need to determine how large n
has to be to imply
3

4n + 3n

3n + 24

n3 − 6 − 4 or n3 − 6 .
By considering n > 1, we may drop the absolute values; thus we need
. This time it would be
to ﬁnd how big n must be to give 3n+24
n3 −6
very diﬃcult to “solve” for or isolate n. Recall we need to ﬁnd some
N such that n > N implies 3n+24
, but we do not need to ﬁnd
n3 −6
the least such N . So we will simplify matters by making estimates.
The idea is that 3n+24
is bounded by some constant times nn3 = n12
n3 −6
for suﬃciently large n. To ﬁnd such a bound we will ﬁnd an upper
bound for the numerator and a lower bound for the denominator.
For example, since 3n + 24 ≤ 27n, it suﬃces for us to get n27n
3 −6 .
To make the denominator smaller and yet a constant multiple of n3 ,
3
we note n3 − 6 ≥ n2 provided n is suﬃciently large; in fact, all we
54
n2
n3
2
≥ 6 or n3 ≥ 12 or n > 2. So it suﬃces to get

or n > 54
, provided n > 2.
need is
27n
n3 /2
41
or
Formal Proof

54
Let > 0 and let N = max{2, }. Then n > N implies n > 54
,
hence n542 , hence n27n
3 /2 . Since n > 2, we have
also 27n ≥ 3n + 24. Thus n > N implies
n3
2
≤ n3 − 6 and
3n + 24
54
27n
≤ 1 3 = 2 ,
3
n −6
n
2n
and hence

3

4n + 3n

n3 − 6 − 4 ,
as desired.
Example 3 illustrates direct proofs of even rather simple limits
can get complicated. With the limit theorems of §9 we would just
write

3

4 + n32
lim 4 + 3 · lim( n12 )
4n + 3n
lim
= 4.
=
= lim
n3 − 6
1 − n63
lim 1 − 6 · lim( n13 )
Example 4
Show that the sequence an = (−1)n does not converge.
Discussion. We will assume lim(−1)n = a and obtain a contradiction. No matter what a is, either 1 or −1 will have distance at
least 1 from a. Thus the inequality |(−1)n − a| = 1 in the deﬁnition
of the limit, we see that there exists N such that
n>N
implies
|(−1)n − a| N , we see that
|1 − a| > 0 and show there exists N
such that

n > N implies | sn − s| .
This time we cannot expect to obtain N explicitly in terms of
because of the general nature of the problem. But we can hope to
show such N exists. The trick here is to violate our training in algebra
and “irrationalize the denominator”:
√ √

( sn − s)( sn + s)

sn − s

√ .
sn − s =
=√

sn + s
sn + s
Since sn → s we will be able to make the numerator small [for large
n]. Unfortunately, if s = 0 the denominator will also be small. So we
consider two cases. If s > 0, the denominator is bounded below by

s and our trick will work:

|sn − s|
| sn − s| ≤ √
,
s

so we will select N so that |sn − s| for n > N . Note that N

exists, since we can apply the deﬁnition of limit to s just as well
as to . For s = 0, it can be shown directly that lim sn = 0 implies

lim sn = 0; the trick of “irrationalizing the denominator” is not
needed in this case.
Formal Proof
Case I: s > 0. Let > 0. Since lim sn = s, there exists N such that

n > N implies |sn − s| .
43
Now n > N implies

s
|sn − s|
|sn − s|
√ ≤ √
| sn − s| = √
.
sn + s
s
s
Case II: s = 0. This case is left to Exercise 8.3.
Example 6
Let (sn ) be a convergent sequence of real numbers such that sn = 0
for all n ∈ N and lim sn = s = 0. Prove inf{|sn | : n ∈ N} > 0.
Discussion. The idea is that “most” of the terms sn are close to s
and hence not close to 0. More explicitly, “most” of the terms sn are
within 12 |s| of s, hence most sn satisfy |sn | ≥ 12 |s|. This seems clear
from Fig. 8.1, but a formal proof will use the triangle inequality.
Formal Proof
Let = 12 |s| > 0. Since lim sn = s, there exists N in N so that
n>N
implies
|sn − s| N
|s| = |s − sn + sn | ≤ |s − sn | + |sn |
|s| |s|
+
= |s|
2
2
|s|
, |s1 |, |s2 |, . . . , |sN | ,
2
FIGURE 8.1
(1)
44
2. Sequences
then we clearly have m > 0 and |sn | ≥ m for all n ∈ N in view of (1).
Thus inf{|sn | : n ∈ N} ≥ m > 0, as desired.
Formal proofs are required in the following exercises.
Exercises
8.1 Prove the following:
n
(a) lim (−1)
=0
n
2n−1
(c) lim 3n+2 = 23
1
(b) lim n1/3
=0
n+6
(d) lim n2 −6 = 0
8.2 Determine the limits of the following sequences, and then prove your
claims.
(a) an = n2n+1
(b) bn = 7n−19
3n+7
4n+3
(c) cn = 7n−5
(d) dn = 2n+4
5n+2
(e) sn = n1 sin n
8.3 Let (sn ) be a sequence of nonnegative real numbers, and suppose

lim sn = 0. Prove lim sn = 0. This will complete the proof for
Example 5.
8.4 Let (tn ) be a bounded sequence, i.e., there exists M such that |tn | ≤ M
for all n, and let (sn ) be a sequence such that lim sn = 0. Prove
lim(sn tn ) = 0.
8.5 1
(a) Consider three sequences (an ), (bn ) and (sn ) such that an ≤
sn ≤ bn for all n and lim an = lim bn = s. Prove lim sn = s.
This is called the “squeeze lemma.”
(b) Suppose (sn ) and (tn ) are sequences such that |sn | ≤ tn for all
n and lim tn = 0. Prove lim sn = 0.
8.6 Let (sn ) be a sequence in R.
(a) Prove lim sn = 0 if and only if lim |sn | = 0.
(b) Observe that if sn = (−1)n , then lim |sn | exists, but lim sn does
not exist.
8.7 Show the following sequences do not converge.
(a) cos( nπ
(b) sn = (−1)n n
3 )

(c) sin( 3 )
1
This exercise is referred to in several places.
§9. Limit Theorems for Sequences
8.8 Prove the
√ following [see Exercise 7.5]:

(a) lim[ n2 + 1 − n] = 0
(b) lim[ n2 + n − n] =

(c) lim[ 4n2 + n − 2n] = 14
45
1
2
8.9 2 Let (sn ) be a sequence that converges.
(a) Show that if sn ≥ a for all but ﬁnitely many n, then lim sn ≥ a.
(b) Show that if sn ≤ b for all but ﬁnitely many n, then lim sn ≤ b.
(c) Conclude that if all but ﬁnitely many sn belong to [a, b], then
lim sn belongs to [a, b].
8.10 Let (sn ) be a convergent sequence, and suppose lim sn > a. Prove
there exists a number N such that n > N implies sn > a.
§9
Limit Theorems for Sequences
In this section we prove some basic results that are probably already familiar to the reader. First we prove convergent sequences
are bounded. A sequence (sn ) of real numbers is said to be bounded
if the set {sn : n ∈ N} is a bounded set, i.e., if there exists a constant
M such that |sn | ≤ M for all n.
9.1 Theorem.
Convergent sequences are bounded.
Proof
Let (sn ) be a convergent sequence, and let s = lim sn . Applying
Deﬁnition 7.1 with = 1 we obtain N in N so that
n>N
implies
|sn − s| N implies |sn | . Our choice of = 1 was quite arbitrary.
2
This exercise is referred to in several places.
46
2. Sequences
9.2 Theorem.
If the sequence (sn ) converges to s and k is in R, then the sequence
(ksn ) converges to ks. That is, lim(ksn ) = k · lim sn .
Proof
We assume k = 0, since this result is trivial for k = 0. Let > 0 and
note we need to show |ksn − ks| for large n. Since lim sn = s,
there exists N such that

n > N implies |sn − s| N
implies
|ksn − ks| .
9.3 Theorem.
If (sn ) converges to s and (tn ) converges to t, then (sn +tn ) converges
to s + t. That is,
lim(sn + tn ) = lim sn + lim tn .
Proof
Let > 0; we need to show
|sn + tn − (s + t)| for large
n.
We note |sn + tn − (s + t)| ≤ |sn − s| + |tn − t|. Since lim sn = s, there
exists N1 such that

n > N1 implies |sn − s|
n > N2 implies |tn − t|
n>N implies |sn +tn −(s+t)| ≤ |sn −s|+|tn −t|.
2 2
9.4 Theorem.
If (sn ) converges to s and (tn ) converges to t, then (sn tn ) converges
to st. That is,
lim(sn tn ) = (lim sn )(lim tn ).
§9. Limit Theorems for Sequences
47
Discussion. The trick here is to look at the inequality
|sn tn − st| = |sn tn − sn t + sn t − st|
≤ |sn tn − sn t| + |sn t − st| = |sn | · |tn − t| + |t| · |sn − s|.
For large n, |tn −t| and |sn −s| are small and |t| is, of course, constant.
Fortunately, Theorem 9.1 shows |sn | is bounded, so we will be able
to show |sn tn − sn t| is small.
Proof
Let > 0. By Theorem 9.1 there is a constant M > 0 such that
|sn | ≤ M for all n. Since lim tn = t there exists N1 such that

n > N1 implies |tn − t|
n > N2 implies |sn − s|

, because t could be 0.] Now if N =
[We used 2(|t|+1)
max{N1 , N2 }, then n > N implies
|sn tn − st| ≤ |sn | · |tn − t| + |t| · |sn − s|

+ |t| ·
.
≤M·
2M
2(|t| + 1)
2 2
To handle quotients of sequences, we ﬁrst deal with reciprocals.
9.5 Lemma.
If (sn ) converges to s, if sn = 0 for all n, and if s = 0, then (1/sn )
converges to 1/s.
Discussion. We begin by considering the equality

1

− 1 = s − sn .
sn s sn s
For large n, the numerator is small. The only possible diﬃculty would
be if the denominator were also small for large n. This diﬃculty was
solved in Example 6 of §8 where we proved m= inf{|sn | : n∈N}>0.
Thus

1

− 1 ≤ |s − sn | ,
sn
s
m|s|
48
2. Sequences
and it is clear how our proof should proceed.
Proof
Let > 0. By Example 6 of §8, there exists m > 0 such that |sn | ≥ m
for all n. Since lim sn = s there exists N such that
n>N
implies |s − sn | · m|s|.
Then n > N implies

1

sn

1 |s − sn |
|s − sn |
=

.

s
|sn s|
m|s|
9.6 Theorem.
Suppose (sn ) converges to s and (tn ) converges to t. If s = 0 and
sn = 0 for all n, then (tn /sn ) converges to t/s.
Proof
By Lemma 9.5, the sequence (1/sn ) converges to 1/s, so
lim
tn
1
1
t
= lim
· tn = · t =
sn
sn
s
s
by Theorem 9.4.
The preceding limit theorems and a few standard examples allow
one to easily calculate many limits.
9.7 Theorem (Basic Examples).
(a) limn→∞ ( n1p ) = 0 for p > 0.
(b) limn→∞ an = 0 if |a| 0.
Proof
(a) Let > 0 and let N = ( 1 )1/p . Then n > N implies np > 1
and hence > n1p . Since n1p > 0, this shows n > N implies
| n1p − 0| . [The meaning of np when p is not an integer will
be discussed in §37.]
(b) We may suppose a = 0, because limn→∞ an = 0 is obvious for
1
a = 0. Since |a| 0. By
§9. Limit Theorems for Sequences
49
the binomial theorem [Exercise 1.12], (1 + b)n ≥ 1 + nb > nb,
so
1
1
|an − 0| = |an | =
N implies n > b
Now consider > 0 and let N = b
1
and hence |an − 0| .
1/n
(c) Let sn = (n ) − 1 and note sn ≥ 0 for all n. By Theorem 9.3
it suﬃces to show lim sn = 0. Since 1 + sn = (n1/n ), we have
n = (1 + sn )n . For n ≥ 2 we use the binomial expansion of
(1 + sn )n to conclude
1
1
n = (1 + sn )n ≥ 1 + nsn + n(n − 1)s2n > n(n − 1)s2n .
2
2
2
Thus n > 12 n(n − 1)s2n , so s2n
2
sn 1, so lim( a1 )1/n =
1 from above. Lemma 9.5 now shows lim(a1/n ) = 1.
Example 1
Prove lim sn = 14 , where
sn =
n3 + 6n2 + 7
.
4n3 + 3n − 4
Solution
We have
sn =
1+
4+
6
7
n + n3
3
− n43
n2
.
By Theorem 9.7(a) we have lim n1 = 0 and lim n13 = 0. Hence by
Theorems 9.3 and 9.2 we have
lim 1 +
6
7
+
n n3
= lim(1) + 6 · lim
1
n
+ 7 · lim
1
n3
= 1.
50
2. Sequences
Similarly, we have
lim 4 +
4
3

n2 n3
= 4.
Hence Theorem 9.6 implies lim sn = 14 .
Example 2
Find lim nn−5
2 +7 .
Solution
Let sn = nn−5
2 +7 . We can write sn as
does not converge. So we write
sn =
1
n
5
1− n
7
n+ n

1+
5
n2
7
n2
, but then the denominator
.
Now lim( n1 − n52 ) = 0 by Theorems 9.7(a), 9.3 and 9.2. Likewise
lim(1 + n72 ) = 1, so Theorem 9.6 implies lim sn = 01 = 0.
Example 3
2 +3
Find lim nn+1
.
Solution
We can write
n2 +3
n+1
as
n + n3
1 + n1
1+
or
1
n
+
3
n2
1
n2
.
Both fractions lead to problems: either the numerator does not con2 +3
verge or else the denominator converges to 0. It turns out nn+1
does
2
+3
not converge and the symbol lim nn+1
is undeﬁned, at least for the
present; see Example 6. The reader may have the urge to use the symbol +∞ here. Our next task is to make such use of the symbol +∞
legitimate. For a sequence (sn ), lim sn = +∞ will signify that the
terms sn are eventually all large. Here is the precise deﬁnition.
9.8 Definition.
For a sequence (sn ), we write lim sn = +∞ provided
for each M > 0 there is a number N such that
n > N implies sn > M .
§9. Limit Theorems for Sequences
51
In this case we say the sequence diverges to +∞.
Similarly, we write lim sn = −∞ provided
for each M N implies sn 0 and show
there exists N [which will depend on M ] such that

n > N implies
n + 7 > M.

To see how big N must be we “solve” for n in the inequality n+7 >

M. This inequality holds provided n > M − 7 or n > (M − 7)2 .
Thus we will take N = (M − 7)2 .
Formal Proof
Let M > 0 and let N = (M −7)2 . Then n > N implies n > (M −7)2 ,

hence n > M − 7, hence n + 7 > M . This shows lim( n + 7) =
+∞.
52
2. Sequences
Example 6
2 +3
Give a formal proof that lim nn+1
= +∞; see Example 3.
Discussion. Consider M > 0. We need to determine how large n
2 +3
must be to guarantee nn+1
> M . The idea is to bound the fraction
n2 +3
n+1
below by some multiple of
n2
n2
n2
n
= n; compare Example 3 of §8.
Since
+3>
and n + 1 ≤ 2n, we have
suﬃces to arrange for 12 n > M .
n2 +3
n+1
>
n2
2n
= 12 n, and it
Formal Proof
Let M > 0 and let N = 2M . Then n > N implies 12 n > M , which
implies
n2 + 3
n2
1
>
= n > M.
n+1
2n
2
2
+3
= +∞.
Hence lim nn+1
The limit in Example 6 would be easier to handle if we could
apply a limit theorem. But the limit Theorems 9.2–9.6 do not apply.
WARNING. Do not attempt to apply the limit Theorems 9.2–9.6
to infinite limits. Use Theorem 9.9 or 9.10 below or Exercises 9.9–
9.12.
9.9 Theorem.
Let (sn ) and (tn ) be sequences such that lim sn = +∞ and lim tn > 0
[lim tn can be finite or +∞]. Then lim sn tn = +∞.
Discussion. Let M > 0. We need to show sn tn > M for large n.
We have lim sn = +∞, and we need to be sure the tn ’s are bounded
away from 0 for large n. We will choose a real number m so that
0 m for large n. Then all we need is
sn > M
m for large n.
Proof
Let M > 0. Select a real number m so that 0 N1
implies
tn > m;
§9. Limit Theorems for Sequences
53
see Exercise 8.10. Since lim sn = +∞, there exists N2 so that
M
n > N2 implies sn >
.
m
Put N = max{N1 , N2 }. Then n > N implies sn tn >
M
m ·m
= M.
Example 7
2 +3
Use Theorem 9.9 to prove lim nn+1
= +∞; see Example 6.
Solution
We observe
n2 +3
n+1
=
3
n+ n
1
1+ n
= sn tn where sn = n +
3
n
and tn =
1
1 .
1+ n
It
is easy to show lim sn = +∞ and lim tn = 1. So by Theorem 9.9, we
have lim sn tn = +∞.
Here is another useful theorem.
9.10 Theorem.
For a sequence (sn ) of positive real numbers, we have lim sn = +∞
if and only if lim( s1n ) = 0.
Proof
Let (sn ) be a sequence of positive real numbers. We have to show
lim sn = +∞ implies
lim
1
sn
=0
(1)
lim sn = +∞.
(2)
and
lim
1
sn
=0
implies
In this case the proofs will appear very similar, but the thought
processes will be quite diﬀerent.
To prove (1), suppose lim sn = +∞. Let > 0 and let M = 1 .
Since lim sn = +∞, there exists N such that n > N implies sn >
M = 1 . Therefore n > N implies > s1n > 0, so

1

n > N implies − 0 .
sn
That is, lim( s1n ) = 0. This proves (1).
To prove (2), we abandon the notation of the last paragraph and
1
begin anew. Suppose lim( s1n ) = 0. Let M > 0 and let = M
. Then
54
2. Sequences
> 0, so there exists N such that n > N implies | s1n − 0| =
Since sn > 0, we can write
n>N
implies
0N
M N0 .
(a) Prove that if lim sn = +∞, then lim tn = +∞.
(b) Prove that if lim tn = −∞, then lim sn = −∞.
(c) Prove that if lim sn and lim tn exist, then lim sn ≤ lim tn .
9.10 (a) Show that if lim sn = +∞ and k > 0, then lim(ksn ) = +∞.
(b) Show lim sn = +∞ if and only if lim(−sn ) = −∞.
(c) Show that if lim sn = +∞ and k −∞, then
lim(sn + tn ) = +∞.
(b) Show that if lim sn = +∞ and lim tn > −∞, then lim(sn + tn ) =
+∞.
(c) Show that if lim sn = +∞ and if (tn ) is a bounded sequence, then
lim(sn + tn ) = +∞.
9.12 3 Assume all sn = 0 and that the limit L = lim | sn+1
sn | exists.
(a) Show that if L N .
(b) Show that if L > 1, then lim |sn | = +∞. Hint : Apply (a) to
the sequence tn = |s1n | ; see Theorem 9.10.
9.13 Show

0

1
n
lim a =
+∞
n→∞

does not exist
if
if
if
if
|a| 1
a ≤ −1.
9.14 Let p > 0. Use Exercise 9.12 to show

0
an ⎨
+∞
lim p =
n→∞ n

does not exist
if
if
if
|a| ≤ 1
a>1
a 1 case, use Exercise 9.12(b).
3
This exercise is referred to in several places.
56
2. Sequences
9.15 Show limn→∞
an
n!
= 0 for all a ∈ R.
9.16 Use Theorems 9.9 and 9.10 or Exercises 9.9–9.15 to prove the following:
4
(a) lim nn2+8n
+9 = +∞
n
(b) lim[ 2n2 + (−1)n ] = +∞
n
(c) lim[ 3n3 −
3n
n! ]
= +∞
9.17 Give a formal proof that lim n2 = +∞ using only Deﬁnition 9.8.
9.18 (a) Verify 1 + a + a2 + · · · + an =
1−an+1
1−a
for a = 1.
(…