Math 4023
Lecture 1
Orientation
(See handouts.)
1 The Set of Natural Numbers
The underlying space for the analysis in this course is the set R of real numbers. This
space should be familiar to to you all - it can be viewed as all the points on line stretching
from ,1 to +1. But one does not always realize how articial this representation of the
real numbers is. For even though we have a certain intuition for the real numbers coming
from the physical notion for what a continuous line should be - there is really no object
in nature upon which to model this line at the microscopic level (since the properties of
matter and space are not very well understood in this realm). Indeed, the only way we
can get a sure understanding of set of real numbers is by rst writing down certain axioms
that eectively dene the real numbers and then proving the (perhaps familiar) properties
of real numbers from these axioms. Then and only then can we begin a rigorous discussion
of analysis over R.
We shall begin, however, with N, the set of natural numbers . This is just the set f1; 2; 3; : : : g
of all positive integers. Historically, this was the rst set of numbers discovered by man;
it is the set of numbers by which we count objects. In order to put the set N on rmer
mathematical ground, we will simply declare as axioms its most basic properties (thereby
leaving nothing to our imagination).
The Peano Axioms for N:
N1. 1 2 N.
N2. If n 2 N, then its \successor", n + 1, belongs to N.
N3. 1 is not the successor of any element in N
N4. If n; m 2 N, and n and m 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 .
Some of you may be familiar with the Well-Ordering Axiom for the natural numbers.
The Well-Ordering Axiom
Every non-empty subset of N has a least element.
In the context of this course, this is not a sixth axiom for N; but rather a property of N
which we shall prove using the Peano Axioms.
Proof.
1
2
Let E be a non-empty subset of N. We have two possibilities: either 1 2 E or 1 2= E .
If 1 2 E , then since 1 is the least element of N (by Axiom N3), 1 is certainly the least
element of the subset E N.
If 1 2= E , then set
F = fp j p < n ; 8 n 2 E g :
Certainly, F 6= N since E and F are disjoint and E N is nonempty. So in order to avoid
a contradiction with Axiom N5, there must exist a p 2 F such p + 1 2= F . Therefore, there
exists a m 2 E such that p + 1 m. We claim that m is the least element of E . To see
this, pick n 2 E and note that p < n. So
q n,p2N :
But then
m p+1 p+q =n :
Hence, m n. So m is the smallest element of E . The Principle of Mathematical Induction. Assume that for each natural number n,
a statement P (n) is given. If
(i) P (1) is a true statement;
(ii) whenever P (k) is a true statement, then P (k + 1) is also a true statement.
then P (n) is a true statement for every n 2 N .
This principle can be extended as follows: a
The Extended Principle of Mathematical Induction. Assume that for each natural
number n, a statement P (n) is given. If
(i) P (m) is a true statement;
(ii) whenever P (k) is a true statement, then P (k + 1) is also a true statement and
k m.
then P (n) is a true statement for every n m.
Example. Prove
using mathematical induction.
Let P (n) be the statement
Then
n > n + 1 if n 2
2
n > n+1 :
2
P (2) , 4 > 3
3
which is certainly true. Now suppose k 2 and that P (k) is true. Then
(k + 1) = k + 2k + 1 > (k + 1) + 2k + 1 :
2
But
So
2
(k + 1) + 2k + 1 = 3k + 2 < k + 2
;
if k > 1 :
(k + 1) > (k + 1) + 1
is true. So P (k) implies P (k +1) is true. The extended principle of mathematical induction
then implies that the statement is true for all n 2. 2
4
Math 4023
Lecture 2
2. Integers, Rational Numbers, and Algebraic Numbers
In the set N of natural numbers only the operations of addition and multiplication can be
dened. For allowing the operations of subtraction and division quickly take us out of the
set N;
2 2 N and 3 2 N but 2 , 3 = ,1 2= N
1 2 N and 2 2 N but 1 2 = 21 6= N
The set Z of integers is formed by expanding N to obtain a set that is closed under
subtraction as well as addition.
Z= f0; ,1; +1; ,2; +2; ,3; +3; : : : g :
The new set Z is not closed under division, however. One therefore expands Z to include
fractions as well and arrives at the number eld Q, the rational numbers . More formally,
the set Q of rational numbers is the set of all ratios of integers:
p
Q=
q j p; q 2 Z; q 6= 0
The rational numbers appear to be a very satisfactory algebraic system until one begins
to tries to solve equations like
x =2 :
It turns out that there is no rational number that satises this equation. To see this,
suppose there exists integers p; q such that
:
2 = pq
We can without loss of generality assume that p and q have no common divisors (i.e., that
the fraction pq is reduced as far as possible). We have
2q = p
so p is even. Hence p is even. Therefore, p is of the form p = 2k for some k 2 Z. But then
2q = 4k
or
q = 2k
so q is even, so p and q have a common divisor - a contradiction since p and q are can be
assumed to be relatively prime. Thus, no such p and q exist. 2
2
2
2
2
2
2
p
2
2
p
Yet 2 certainly exists; at the very least one can represent the number 2 geometrically
as the diagonal length of a square whose sides have unit length. To handle such numbers
we must widen our number eld even further.
5
Denition 2.1. A number is called an algebraic number if if satises a polynomial
equation
anxn + an, xn, + + a x + a = 0
(2.1)
1
1
1
0
where the coecients an; an, ; : : : ; a are all integers, an 6= 0, and n > 0.
1
0
We note that rational numbers are always algebraic numbers since a rational number pq is
a solution of
qx + (,p) = 0 :
On the other hand, not every real number is algebraic. Certain numbers like or e
are known not to satisfy any algebraic equation like (2.1). Numbers which can not be
represented as the solutions of equations like (2.1) are called transcendental. It is in
general a very dicult problem
p to decide whether a given number is transcendental or
p
algebraic. e; , e , and 2 are all transcendental. The last case, however, was not
proved until 1934. It is unknown whether , e, or e + are algebraic or transcendental.
2
However, we do have a test to decide whether an algebraic number is rational or not.
Theorem 2.2. (Rational Zeros Theorem) Suppose that an; an, ; : : : ; a are integers and
1
that r is a rational number satisfying the polynomial equation
0
anxn + an, xn, + + a x + a = 0
1
1
1
0
where n 1 , an 6= 0 and a 6= 0. Write r = pq where p and q are integers such that q 6= 0
and having no common factors except 1. Then q divides an and p divides a .
0
0
Proof. We rst recall the Fundamental Theorem of Arithmetic which states that every
integer n has a factorization
n = (p )a1 (p )a2 (pk )a
1
k
2
where the fpi g are distinct prime numbers and that this factorization is unique up to the
ordering of factors and the sign of pairs of factors.
Suppose that r = pq has the stated properties. So
an
n
p
q
n,1
+ an, pq
1
+ + a
1
p +a =0 :
q
0
6
Multiplying both sides by qn yields
an pn + an, qpn, + a pqn, + a q = 0 ;
(2.2)
1
1
or
,
1
1
0
a qn = p ,anpn, , an, qpn, , , a qn, :
Hence, p divides aoqn. Since p and q are assumed to have no common factors, p must
divide ao .
Alternatively, we can also rewrite (2.2) as
1
0
1
2
1
1
,
anp = q ,an, pn, , , a pqn, , a qn,
1
1
2
1
0
1
and so q divides an p. Since p and q are assumed to have no common factors, q must divide
an . Example: Show that the solution of x , 6 is irrational.
According to the theorem above, if pq is a solution of x , 6 = 0 then p must divide 6 and
2
2
q must divide 1. Thus, the only possible rational solutions are
x = 1; 2; 3; 6 :
But then
So no rational solution exists.
x = 1; 4; 9; 36 6= 6 :
2
3. Fields and Ordered Fields
The rational numbers Q and the real numbers R are examples of what is called an ordered
eld . More generally, a eld is a set F upon which operations of \addition" and \multiplication" are dened and for which the following axioms are satised:
A1. a + (b + c) = (a + b) + c for all a; b; c 2 F .
A2. a + b = b + a for all a; b 2 F .
A3. There exists 0 2 F such that a + 0 = a for all a 2 F .
A4. For each a 2 F there is an element ,a 2 F such that a + (,a) = 0.
M1. a(bc) = (ab)c for all a; b; c 2 F .
M2. ab = ba for all a; b 2 F .
M3. There exists 1 2 F such that a 1 = a for all a 2 F .
M4. For each a 2 F , there exists a, 2 F such that aa, = 1.
DL. a(b + c) = ab + bc for all a; b; c 2 F .
1
1
7
Theorem 3.1. Suppose F is a eld. Then if a; b; c 2 F
(i)
(ii)
(iii)
(iv)
(v)
(vi)
a + c = b + c implies a = b.
a 0 = 0 for all a 2 F .
(,a)b = ,(ab) for all a; b 2 F .
(,a)(,b) = ab for all a; b 2 F .
ac = bc and c 6= 0 implies a = b.
ab = 0 implies either a = 0 or b = 0.
An ordered eld is a eld F with the an order relation satisfying the following axioms:
O1. Given a; b 2 F , then either a b or b a.
O2. If a b and b a, then a = b.
O3. If a b and b c, then a c.
O4. If a b, then a + c b + c.
O5. If a b and 0 c, then ac bc.
In the theorem below we use the notation a b to mean a b and a 6= b.
Theorem 3.2. If F is an ordered eld is an ordered eld and a; b; c 2 F , then:
(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 a for all a 2 F .
(v) 0 1.
(vi) if 0 a, then 0 a, .
(vii) if 0 a b, then 0 b, a, .
2
1
1
1
Taking F = Q and to coincide with the the usual numerical inequality , the above
properties should seem quite elementary. However, in our context (the axiomatic development of number elds) these statements are properties which must rst be proved before
they can be employed. This we shall do next time.
8
Math 4023
Lecture 3
3. Fields and Ordered Fields, Cont'd
Recall that an ordered eld is a F together with a binary relation satisfying the following
axioms:
O1. Given a; b 2 F , then either a b or b a.
O2. If a b and b a, then a = b.
O3. If a b and b c, then a c.
O4. If a b, then a + c b + c.
O5. If a b and 0 c, then ac bc.
In the theorem below we use the notation a b to mean a b and a 6= b.
Theorem 3.2. If F is an ordered eld is an ordered eld and a; b; c 2 F , then:
(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 a for all a 2 F .
(v) 0 1.
(vi) if 0 a, then 0 a, .
(vii) if 0 a b, then 0 b, a, .
2
1
1
1
Taking F = Q and to coincide with the the usual numerical inequality , the above
properties should seem quite elementary. However, in our context (the axiomatic development of number elds) these statements are properties which must rst be proved before
they can be employed.
Proof.
(i) Suppose that a b. Then by O4 we have
a + ((,a) + (,b)) b + ((,a) + (,b))
or
(ii) If c 0, then
,b ,a :
0 = c + (,c) 0 , c = ,c
by (i). So 0 ,c. If a b, then by O5 we have
a(,c) b(,c) ;
9
or
,ac ,bc :
Again by (i) we have
bc ac :
(iii) If we put a = 0 and apply Axiom O5 we obtain
0 b and 0 c ) 0 bc :
Except for the notation, this precisely statement (iii).
(iv) For any a 2 F , either a 0 or 0 a by O1. If 0 a, then 0 a by (iii).
2
If a 0; then 0 ,a by (i) and so 0 (,a) by (iii). But by statement (iv) of
Theorem 3.1, (,a)(,a) = a , hence 0 a .
(v) Well, 0 1 means that 0 1 and 0 6= 1. Now 0 6= 1 is certainly true; since
otherwise Axiom M3 and statement (ii) of Theorem 3.1 would contradict each
other. (We assume the eld F does consists of more than one element.) To see
that 0 1, we simply apply Axiom M3 and statement (iv) above to the case when
a = 1.
01 =11=1 :
(vi) Suppose that 0 a; i.e 0 a and 0 6= a. If a, 0 then by (i) we have
0 (,a, ), and so by (iii)
2
2
2
2
1
1
0 a(,a, ) = ,1 :
1
But then (i) implies
1 ,0 = 0
which contradicts (v). Hence we cannot have a, 0; consequently 0 a , .
(vii) Assume 0 a b. In view of (vi) we need only show that b, a, . But
since a 6= b, we know a, 6= b, ; for if a, = b, , then 1 = a, a = b, a, so
b = bb, a = a. So we really only need to show b, a, .
1
1
1
1
1
1
1
1
1
1
1
1
1
Since 0 a b, we have, in particular, 0 a and 0 b, hence 0 ab by (iii). In
fact 0 ab by Statement (vi) of Theorem 3.1. But then 0 (ab), by (vi). We
now apply Axiom 05 with c = (ab), .
1
1
a(ab), b(ab),
1
1
The statement now follows from the identity
(ab), = a, b,
1
1
1
:
10
Math 4023
Lecture 4
3. Fields and Ordered Fields, Cont'd
Denition 3.3. Let F be an ordered eld. If a 2 F , then the absolute value jaj of a is
the element of F dened by
a ; if 0 a
jaj = ,a ; if a 0 :
Denition 3.4. Let a; b be arbitrary elements of an ordered eld F . The distance
between a and b is the element dist(a; b) 2 F dened by
dist(a; b) = ja , bj :
Theorem 3.5. Let F be an ordered eld. Then:
(i) 0 jaj for all a 2 F .
(ii) jabj = jaj jbj for all a; b 2 F .
(iii) ja + bj jaj + jbj for all a; b 2 F .
Proof.
(i) This obvious from the denition of jaj.
(ii) There are four easy cases here.
If 0 a and 0 b, then 0 ab by Theorem 3.2 (iii). So
jabj = ab = jaj jbj
by Denition 3.3.
If a 0 and b 0, then 0 ab by Theorem 3.2 (ii). So
jabj = ab = (,jaj) (,jbj) = jaj jbj
by Denition 3.3.
If a 0 and 0 b, then ab 0 by Theorem 3.2 (ii). So
jabj = ,ab = , (,jaj) jbj = jaj jbj
11
by Denition 3.3.
If 0 a and b 0, then ab 0 by Theorem 3.2 (ii). So
jabj = ,ab = ,jaj (jbj) j = jaj jbj
by Denition 3.3.
(iii) The inequalities
,jaj a jaj
follow from the facts that either a = jaj or a = ,jaj and ,jaj jaj. Similarly,
,jbj b jbj :
Applying Axiom O4 four times we get
,jaj + (,jbj) ,jaj + b a + b a + jbj jaj + jbj
so that
(3.1)
, (jaj + jbj) a + b
and
(3.2)
a + b jaj + jbj
Now (3.1) implies
(3.3)
,(a + b) jaj + jbj :
Since ja + bj is either equal to a + b or ,(a + b), (3.2) and (3.3) imply
ja + bj jaj + jbj :
Corollary 3.6. If F is an ordered eld, then
dist(a; c) dist(a; b) + dist(a; c) ; 8 a; b; c 2 F :
Homework:
1. Problem 2.1.
2. Prove Theorem 3.1.
3. Problem 3.1.
12
4. The Completeness Axiom
Thus far, we have been fairly careful up in constructing the number elds N, Z, Q from
the Peano Axioms. A rigorous construction of the real numbers from the rational numbers
takes quite a bit of work, however. Much more work than we have time for in this course.
However, it will be important for us to understand the property of the real numbers that
distinguishes the reals from the rationals; for it is precisely this property of the reals that
permits the rigorous development of calculus. So, henceforth we shall not worry about
the problem of constructing the set of real numbers, we shall simply assume that the real
numbers exist and lay down as axioms the properties of R that we need for analysis.
Denition 4.1. Let S be a nonempty subset of R.
(a) If so is an element of S with the property that
s 2 S ) s so
then so is called the maximum of S .
(b) If so is an element of S with the property that
s 2 S ) so s
then so is called the minimum of S .
Examples:
(a) Every nite subset of R has a maximum. For example,
max f1; 3; 6; ,2g = 6 :
(b) Not every subset of R has a maximum. The subsets N; Zand Q have no maximum.
The subset N, however, does have a minimal element; viz., 1.
(c) The subset
n
po
x2Qj1x 2
has a minimum, 1, but it does not have a maximal element.
(d) The subset
1
1
1
1; 2 ; 3 ; 4 ; : : :
does not have a minimal element.
Below we present some denitions that will give us a handle on the counterexamples
mentioned above.
13
Denition 4.2. Let S be a nonempty subset of R.
(a) If a real number M has the property that
s M ; 8s 2 S ;
then M is called an upper bound of S and the set S is said to be bounded
from above.
(b) If a real number m has the property that
m s ; 8s 2 S ;
then m is called a lower bound of S and the set S is said to be bounded from
below.
(c) A subset S R is said to be bounded if it is bounded from above and bounded
from below. Thus, S is bounded if there exist real numbers m; M such that
S [m; M ] :
Denition 4.3. Let S be a nonempty subset of R.
(a) Suppose S is bounded from above. The suprenum, or least upper bound, of
S is the number sup S 2 R dened by
sup S = min fx 2 R j x is an upper bound of S g :
(b) Suppose S is bounded form below. The inmum or greatest lower upper
bound, of S is the number inf S 2 R dened by
sup S = max fx 2 R j x is a lower bound of Sg :
There is a problem with this denition, however. We have already seen examples of subsets
of R which do not have a maximum or a minimum; how, for example, do we know that
the set
fx 2 R j x is an upper bound of S g
has a minimal element?
This problem we shall resolve by regarding as an axiom the following of R.
4.4 The Completeness Axiom. Every non-empty subset S of R that is bounded from
above has a least upper bound.
14
Math 4023
Lecture 5
4. The Completeness Axiom, Cont'd
Corollary 4.5. Every non-empty subset S of R that is bounded from below has a greatest
lower bound inf S .
Proof. Let T be the set f,s j s 2 S g. Since S is bounded from below there is an m 2 R
such that m s for all s 2 S . This implies ,s ,m for all s 2 S and so t ,m for all
t 2 T . So T is bounded from above, hence by the Completeness Axiom, sup T exists. Let
u = sup T . We shall show that ,u = inf S .
More precisely, we shall show that
,u s ; 8 s 2 S
(4.1)
and that
(4.2)
t s ; 8s 2 S ) t ,u :
Now by denition, since u is the least upper bound of T ,
,s u ; 8 s 2 S
(4.3)
and
,s q ; 8 s 2 S ) u q
(4.4)
Now from Theorem 3.2 (i) we know (4.3) is equivalent to (4.1). Setting q = ,t, (4.4) reads
,s ,t ; 8 s 2 S ) u ,t
or, using Theorem 3.4 (i) again,
t s ; 8 s 2 S ) t ,u
which is precisely (4.2). 15
Theorem 4.6. (The Archimedean Property of R) If a > 0 and b > 0, then for some
positive integer n, we have
na > b :
Proof. Assume the Archimedian Property fails. Then there exists a > 0 and b > 0 such
that
na b ; 8 n 2 N :
In particular, b is an upper bound for the set
S = fna j n 2 Ng :
Let so = sup S ; this is where we are using the Completeness Axiom. Since a > 0, we have
so < so + a
and so
so , a < so
(where we have used Axiom O4 of ordered elds and the property a + (,a) = 0). Since so
is the least upper bound for S , so , a cannot be an upper bound for S . If follows that
so , a < noa
for some no 2 N. But then
so < (no + 1)a :
But (no + 1)a belongs to S , so so can not be an upper bound for S ; hence we have reached
a contradiction. Therefore, our assumption that the Archimedean Property does not hold
must be false. Theorem 4.7. (The Denseness of Q) If a; b 2 R and a < b, then there is a rational
number r such that a < r < b.
Proof. It suces to show that there exist integers m and n > 0 such that
na < m < nb :
Since 0 < b , a, the Archimedean Property shows that there exists an n 2 N such that
1 < n(b , a)
or
an + 1 < bn :
16
At this point it seems obvious that there is an integer lying between an and bn. Rather
than make a plausibility argument, we shall provide an explicit construction of such an
integer.
By the Archimedean Property again, there also exists positive integers k0; k00 such that
janj < k0
Set
;
jbnj < k00 :
k = maxfk0 ; k00 g :
Then
,k < an < bn < k :
The set
fj 2 Zj ,k < j < k and an < j g
is nite and nonempty. Set
m = min fj 2 Zj ,k < j < k and an < j g
so that
an < m ; but m , 1 an :
Then we have
m = (m , 1) + 1 an + 1 < bn :
Now we have found an m 2 Z such that
an < m < bn :
Homework:
pg. 14; 3.7, 3.8
pg. 20-21; 4.3, 4.4, 4.5, 4.7, 4.10, 4.11, 4.13, 4.16
17
Math 4023
Lecture 6
Solutions to Homework Problems
3.7
(a) Show that jbj < a if and only if ,a < b < a.
)
First of all, since 0 jbj and jbj < a, we have 0 < a, by O5. Hence, ,a < 0, by
Theorem 3.2 (i).
If 0 b, we have jbj = b and so b < a. Since ,a < 0, by Theorem 3.2 (i), we have
,a < b by O3. Hence ,a < b < a.
If b 0, we have jbj = ,b. So ,b < a. So ,a < b by Theorem 3.2 (i). Also, since
0 jbj < a, and b 0, we have b < a. Hence ,a < b < a.
(
Assume, ,a < b < a.
If 0 b, then jbj = b and so jbj = b < a.
If b 0, then jbj = ,b. But ,a < b, implies by Theorem 3.2, that ,b = jbj < a.
So in either case, jbj < a.
(b) Show that ja , bj < c if and only if b , c < a < b + c.
)
From Part (a) we have,
,c < a , b < c :
Adding b to each term and applying Axiom O4, we get
b,c<a<b+c :
(
Adding ,b to each term in the inequalities b , c < a < b + c and applying O4, we
get ,c < a , b < c. From part (a) we can deduce
ja , bj < c
18
(c) Show that ja , bj c if and only if b , c a b + c.
)
In view of Part (b), we need only show that if ja , bj = c then b , c a b + c.
If 0 a , b, we have c = ja , bj = a , b. So adding b to both sides produces
b + c = a. So certainly a b + c. But we also have, 2c = a , (b , c). Now the fact
that c = ja , bj implies 0 2c. Hence, 0 a , (b , c), hence b , c a.
If a , b 0, then c = ,(a , b) = b , a. Then we have c , b = ,a or b , c = a, hence
b , c a. As above we must also have 0 c. so a a + 2c = b , c + 2c = b + c.
Consequently, b , c a b + c.
( First of all, b , c b + c implies 0 c. We have already seen from Part (b)
that if b , c < a < b + c, then ja , bj < c. It therefore suces to show that if
a = b , c or a = b + c, then ja , bj c.
If a = b , c, then a , b = ,c 0. Hence ja , bj = b , a = c. So ja , bj c.
If a = b + c, then a , b = c 0. Hence ja , bj = a , b = c. So ja , bj c.
3:8 Let a; b 2 R. Show that that a b for every b > b, then a b.
1
1
Let's change notation a bit. We want to show that if a c for every c > b, then a b.
Suppose the contrary; i.e., that a c for every c > b, but b < a.
Since b < a, a , b > 0, hence a,b > 0, hence b < b + a,b = b a . But we also have
b + a < a + a = 2a, hence b a < a. We have thus found a number c = b a , such that b < c
and c < a. But by hypothesis, if b < c, then c a. We thus arrived at a contradiction.
Hence, the statement b < a must be false.
+
2
2
+
2
2
4.3 - 4.4
(a) S = [0; 1]
(b) S = (0; 1)
(c) S = f2; 7g
2
+
inf S = 0
;
sup S = 1 :
inf S = 0
;
sup S = 1 :
inf S = 2
;
sup S = 7 :
19
(d) S = f; eg
1
(e) S = n j n 2 N
inf S = e
;
sup S = :
inf S = 0
;
sup S = 1 :
inf S = 0
;
sup S = 0 :
inf S = 0
;
sup S = 3 :
(f) S = f0g
(g) S = [0; 1] [ [2; 3]
S
(h) S = 1
n [2n; 2n + 1]
=1
inf S = 1
;
T
(i) S = 1
n ,n; 1 + n
inf S = 0
sup S = D:N:E: :
1
1
=1
(j) S = 1 ,
1
3n
;
sup S = 1 :
;
sup S = 1 :
jn2N
inf S = 32
n
o
(k) S = n + ,n j n 2 N
(
n
1)
inf S = 0
;
sup S = D:N:E: :
(l) S = fr 2 Q j r < 2g
inf S = D:N:E:
;
sup S = 2 :
(m) S = r 2 Q j r < 4
2
inf S = ,2
sup S = 2 :
;
sup S = 2 :
(n) S = r 2 Q j r < 2
2
;
p
inf S = , 2
p
(o) S = fx 2 R j x < 0g
inf S = D:N:E:
;
sup S = 0 :
20
(p) S = 1; ; ; 10
2
3
inf S = 1
;
sup S = 10 :
inf S = 0
;
sup S = 16 :
,
T
(r) S = 1
n 1 , n; 1 + n
inf S = 1
;
sup S = 1 :
;
sup S = 21 :
(q) S = f0; 1; 2; 4; 8; 16g
1
1
=1
1
(s) S = n j n 2 N and n is prime
inf S = 0
N.B.1 is not prime.
(t) S = x 2 R j x < 8
3
inf S = D:N:E
;
p
sup S = 3 8 :
(u) S = x j x 2 R
2
inf S = 0
;
sup S = D:N:E: :
(v) S = fcos(n=3) j n 2 Ng
inf S = ,1
(w) S = fsin(n=3) j n 2 Ng
p
inf S = , 23
;
;
sup S = 1 :
p
sup S = 23 :
4.5 Let S be a nonempty subset of R that is bounded from above. Prove that if sup S
belongs to S then, sup S = max S .
This is just denitions. The maximum of S is the element m of S such that s m for all
s 2 S . sup S is the element of R such that (i) s sup S for all s 2 S and (ii) if s t for
all s 2 S , then s t. If sup S belongs to S , then condition (i) on sup S says that it is the
maximum of S .
21
4.7 Let S and T be nonempty bounded subsets of R.
(a) Prove that if S T , then inf T inf S sup S sup T .
To show that inf S sup S , one just needs the denitions of inf S and sup S ,
and the transitivity of . Explicitly, if s 2 S , then by denition inf S s and
s sup S ; hence by the Ordering Axiom O3, inf S sup S .
Since inf T is less than or equal to every element of T , it must in particular be
less than or equal to every element of S , since S is a subset of T . Thus, inf T is
a lower bound of S . Hence,
inf T inf S :
Similarly, since sup T is greater than or equal to every element of T , it must in
particular be greater than or equal to every element of S , since S is a subset of T .
Thus, sup T is an upper bound of S . Hence,
sup S sup T :
Putting these inequalities together we get
inf T inf S sup S sup T :
(b) Prove that sup (S [ T ) = max fsup S; sup T g.
Certainly maxfsup S; sup T g is greater than or equal to any element of S or T .
So max fsup S; sup T g is an upper bound for S [ T .
It is also clear that every upper bound for S [ T must be greater than or equal to
every element of S and every element of T . This implies
q is upper bound for S [ T ) sup S q and sup T q
which in turn implies
q is an upper bound for S [ T
) max fsup S; sup T g q :
Since max fsup S; sup T g is an upper bound for S [ T and every other upper
bound for SUT is greater than of equal to max fsup S; sup T g, we must have
sup (S [ T ) = max fsup S; sup T g :
22
4.10 Prove that if a > 0, then there exists n 2 N such that
1 <a<n :
n
We shall break this problem up into three mutually exclusive cases.
(i) a < 1
By the Density of Q Theorem, we now that there exists a rational number pq , with
p; q 2 N, such that
0 < pq < a :
Since p 2 N, we know that
1p
dividing both sides of this inequality by the positive integer q gives us
1p :
q q
Therefore,
1 p <a
1 <a :
q q
q
On the other hand, since q is a natural number,
1q :
Since by hypothesis, a < 1, we then have a < q. Setting n = q we thus have
1 <a<n :
n
(ii) a = 1
Since 0 < 1, we have
or
0+1<1+1
1<2 :
Dividing this inequality by the positive number 2 yields
1 <1 :
2
23
So setting n = 2, we have
(iii) 1 < a
1 <a<n :
n
Recall the Archimedean Property of R: if b > 0 and c > 0, then for some n 2 N,
we have c < nb. Taking b = 1 and c = a, we have the statement that there exists
n 2 N such that
a<n :
But since 1 n, we have by Theorem 3.2 (vii),
1 1 :
n
Since by hypothesis, 1 < a, we have
1 <a<n :
n
4.11 Consider a; b 2 R where a < b: Use the Density of Q Theorem to show that there are
innitely many rationals between a and b.
This is an easy argument. According to the Density of Q Theorem there exists a rational
number r such that
a<r <b :
Applying the Density of Q Theorem again, we know there exists a rational number r
between r and b;
a<r <r <b :
Applying the Density of Q Theorem a third time, we conclude that there exists a rational
number r between r and b; hence
a<r <r <r <b :
It is evident that we can continue in this manner, to construct an innite number of
rationals lying between a and b. (Alternatively, one could devise a proof by contradiction
to show that this process of creating a new rational numbers between a and b can not
terminate.) 1
1
2
1
1
3
2
2
1
2
3
4.13 Prove that the following are equivalent for real numbers a; b; c.
(i) ja , bj < c.
(ii) b , c < a < b + c.
(iii) a 2 (b , c; b + c).
24
Since
(b , c; b + c) fx 2 R j b , c < x < b + cg
(ii) and (iii) are equivalent by denition. (i) and (ii) are equivalent by Problem 3.7(b). 4.16 Show that sup fr 2 Q j r < ag = a for each a 2 R.
Let S = fr 2 Q j r < ag. Any element of S is certainly less than or equal to a; so a is an
upper bound of S . It suces to show that if t is any other upper bound of S , then a t.
Suppose not; i.e. suppose there is an element t 2 R, such that
s t for all s 2 S
and
t<a :
Then by the Density of Q Theorem, there is a rational number r such that
t<r<a :
But then r 2 S , since r < a. But then t is not an upper bound for S since t < r. Hence
we have a contradiction, unless a t. f
if a b for all
1
25
Math 4023
Lecture 7
5. The Symbols +1 and ,1
The symbols +1 and ,1 are very useful even though they do not represent real numbers.
We stress, however, that these symbols are used strictly for ease of notation.
(,1; a) = fx 2 R j x < ag
(,1; a] = fx 2 R j x ag
(a; +1) = fx 2 R j a < xg
[a; +1) = fx 2 R j a xg
(,1; +1) = R
If S is a subset of S we also write
sup S = +1 if S is not bounded from above;
inf S = ,1 if S is not bounded from below:
7. Limits of Sequences
Denition. A sequence in R is a function s whose domain is a set of the form
fn 2 Zj n mg
and whose range is a subset of C .
Thus, a sequence prescribes, for each n m, a real number s(n); although it is more
common to denote the value of a sequence s at n by sn. Sometimes we denote a sequence
s by the set of its values; viz.,
(sm ; sm ; sm ; : : : )
+1
or
or if m = 1
+2
( sn ) 1
m
(sn )n2N :
26
Examples.
The following are sequences in R:
(a)
sn = n1 ; 1 n
(b)
sn = (,1)n ; ,4 n
(c)
sn = cos n
; 2n
3
(d)
sn = n 1 ; 1 n
2
n
The limit of a sequence (if it exists) is the value that sn approaches as n gets very large.
More precisely,
Denition. A sequence (sn) of real numbers is said to converge to the real number S
provided that for each > 0, there exists a number N 2 Z such that
n > N ) jsn , S j < :
If a sequence (sn ) converges to S we will write
or
lim s = S
n!1 n
sn ! S :
The number S is called the limit of the sequence (sn). If a sequence does not converge to
any real number, then the sequence is said to diverge.
Example.
Consider the sequence
sn = (,1)n ; 1 n :
This sequence does not converge to any real number. For although jsn j = 1 for all n 2 N,
the sign of sn alternates as n alternates between positive and negative numbers and there
is no way to stay within of -1 and +1 simultaneously without rst restricting 1 . But
27
the denition requires the choice of to be arbitrary, hence the limit cannot exist. Thus,
this sequence diverges.
More formally, suppose (sn = (,1)n ) does converge. Then chosing = 1, we must be able
to nd an a and a N such that
jsn , aj < 1 for all n > N:
Now is n is even, sn = a, and if n is odd, sn = ,1. Thus we need both
j1 , aj < 1
and
The rst inequality implies
while the second implies
j , 1 , aj < 1 :
0<a<2
,2 < a < 0 :
Noting that there is no overlap between these two intervals we conclude that the conditions
for the existence of a limit cannot be satised. Thus, (sn) does not converge.
28
Math 4023
Lecture 8
7. Limits of Sequences, Cont'd
Denition. A sequence (sn) of real numbers is said to converge to the real number S
provided that for each > 0, there exists a number N 2 Z such that
n > N ) jsn , S j < :
If a sequence (sn ) converges to S we will write
nlim
!1 sn = S
or
sn ! S :
The number S is called the limit of the sequence (sn). If a sequence does not converge to
any real number, then the sequence is said to diverge.
Proposition. If a sequence (sn ) has a limit S , then that limit is unique.
Proof.
Suppose
lim s = S
n!1 n
and also
lim s = T:
n!1 n
That is to say, suppose for every > 0, there is an N 2 N such that
N < n ) jsn , S j < 2
and for every > 0, there is an N 2 N such that
N < n ) jsn , T j < 2 :
For any n > max fN ; N g we must have
jT , S j = j(sn , S ) , (sn , T )j jsn , S j + jsn , T j < 2 + 2 < :
Thus,
0 jT , S j < for all > 0 :
Thus, jT , S j = 0, hence T = S . 1
1
2
2
1
2
8. Limit Theorems for Sequences
29
Denition. A sequence (sn )nm in R is said to be bounded if there exists an M 2 R
jsnj M for all n m .
such that
Theorem 9.1. Convergent sequences are bounded.
Proof. Let (sn)nm be a convergent sequence and let s = lim sn . According to the
denition of the limit of a sequence, for any > 0 there is an N 2 N such that
n > N ) jsn , sj < :
Adding jsj to both sides of the inequality on the right we get
(1)
jsn , sj + jsj < + jsj :
By the Triangle Inequality, we have
(2)
jsnj = jsn , s + sj jsn , sj + jsj ;
and so combining the inequalities (1) and (2) we get
(3)
jsnj < + jsj :
We can in particular choose = 1 and nd an N 2 N such that
(4)
jsn j < jsj + 1 for all n > N .
Thus, the set of terms fsn j n > N g is bounded from above. Set
M = max fjsj + 1; sm ; sm ; : : : ; sN g
+1
then we have
sn M for all n m ;
and so the sequence is bounded. Theorem 9.2. If the sequence (sn ) converges to s and k 2 R, then the sequence (ksn)
converges to ks.
Proof. We assume k 6= 0 since the result is trivial for k = 0. We need to show that for any
> 0 there is an N 2 N such that
jksn , ksj < for all n > N .
30
Since (sn ) converges, we know that for any 0 > 0 there is an N 0 2 N such that
jsn , sj 0 for all n > N 0 :
In particular, we can choose 0 = jkj and nd an N 0 such that
jsn , sj jk j for all n > N 0 :
Multiplying both sides of the inequality on the left by the positive number jkj we get
jkj jsn , sj = jksn , ksj < :
So setting N = N 0 we get the statement we need
jksn , ksj < for all n > N .
Homework:
7.1, 7.3, 8.1(a), 8.2(a), 8.5, 8.9, 8.10
31
Math 4023
Lecture 9
9. Limit Theorems for Sequences, Cont'd
Theorem 9.3. If (sn ) converges to s and (tn) converges to t, then (sn + tn) converges to
s + t.
Proof. Let > 0. We need to show that there is an N such that
jsn + tn , s , tj < for all n > N :
Since lim sn = s and lim tn = t, there exists numbers N and N such that
jsn , sj < 2 for all N > N ;
jtn , tj < 2 for all N > N :
1
2
1
2
Using the Triangle Inequality we have
jsn + tn , s , tj = j(sn , s) + (tn , t)j jsn , sj + jtn , tj :
So setting N = max fN ; N g we have
1
2
jsn + tn , s , tj < 2 + 2 = for all n > N :
Theorem 9.4. If (sn) converges to s and (tn) converges to t, then (sn tn) converges to st.
Proof.
Let > 0. We need to construct a number N such that
jsn tn , stj < for all n > N :
By Theorem 9.1, since (sn) converges it is bounded and so there exists a constant M > 0
such that jsn j M for all n. Since (tn) converges we can nd a number N such that
for all n > N :
jtn , tj < 2M
1
1
32
We can also nd a number N such that
jsn , sj < 2 (jtj+ 1) for all n > N
2
2
:
If N = max fN ; N g, then for all n > N we have
1
2
jsn tn , stj = jsntn , sn t + snt , stj jsn (tn , t) + t (sn , s)j
jsnj jtn , tj + jtj jsn , sj
+ jtj < M 2M
2 (jtj + 1)
<
:
33
Math 4023
Lecture 10
9. Limit Theorems for Sequences, Cont'd
Last time we proved
Theorem 9.3. If (sn ) converges to s and (tn) converges to t, then (sn + tn) converges to
s + t.
Theorem 9.4. If (sn) converges to s and (tn) converges to t, then (sn tn) converges to st.
We need two more basic results in our repetoire.
Theorem 9.5. If (sn ) converges to s 6= 0, and if sn 6= 0 for all n, then
s.
1
s
n
converges to
1
Proof. Let > 0. We need to construct a number N such that
1
s
, 1s < for all n > N
n
Now
1
s
:
, sn 1 1 = js , sn j :
, 1s = s ss
n
n s sn The troublesome point here is that we need to show that
1
s
n
is bounded, overwise the right hand side might diverge. Since (sn) converges to s, we know
that there exists an N such that
jsn , sj < j2sj ; for all n > N :
We claim that
jsj js j ; for all n > N
n
2
1
1
1
as well. For suppose
jsnj < j2sj ; for all n > N
1
:
34
The Triangle Inequality then would imply
jsj = jsn , s , sn j jsn , sj + jsnj < js2j + j2sj = jsj ;
which is a contradiction since jsj can not be greater than itself. Now set
j
s
j
m = min 2 ; js j ; js j ; : : : ; jsN1 j ;
1
2
then
m jsn j for all n :
Note that m exists and is positive because it is the minimum of a nite set of positive
numbers. This implies
1 1 for all n :
jsnj m
Now since (sn) converges, we also know that there is a number N such that
1
jsn , sj < m jsj for all n > N .
1
So setting N = N , we have
1
1
s
, 1s = 1s s1 js , snj < j1sj
n
n
1 (m jsj) = m
for all n > N :
Theorem 9.6. Suppose that (sn) converges to s, (tn) converges to t. If s 6= 0 and sn 6= 0
for all n, then
t
n
lim s = st :
n
Proof. By Lemma 9.5, s
1
n
converges to s . By Theorem 9.4 then
1
lim stn = lim (tn) lim s1 = st :
n
n
Homework:
35
Prove
(a)
(b)
1 = 0 ; for p > 0 :
lim
n!1 np
lim an = 0 ; if jaj < 1 :
n!1
lim n =n = 1 :
(c)
n!1
(d)
n!1
1
lim a =n = 1 :
1
36
Math 4023
Lecture 11
9. Limit Theorems for Sequences, Cont'd
Denition. Let (sn) be a sequence in R. If for each M > 0 there is a number N 2 N
such that
sn > M for all n > N ;
we say that the sequence (sn) diverges to +1 and write
lim (sn ) = +1 :
If for each M < 0 there is a number N 2 N such that
sn < M for all n > N ;
we say that the sequence (sn) diverges to ,1 and write
lim (sn ) = ,1 :
Theorem 9.9. Let (sn) and (tn) be sequences such that lim (sn) = +1 and lim (tn) > 0.
Then
lim (sntn) = +1 :
Proof.
Fix M > 0. Select m 2 R such that 0 < m < lim (tn). Whether lim (tn ) = +1 or not, it
is clear that there exists N such that
tn > m for all n > N :
(See Homework Problem 8.10.) Since
lim (sn) = +1
there exists N such that
M < s for all n > N :
m n
Put N = max fN ; N g. Then we have
sntn > M
m M = M for all n > N :
1
1
2
2
1
2
37
Theorem 9.10. Let (sn ) be a sequence in R such that sn > 0 for all n. Then
lim (sn) = +1 if and only if lim s1 = 0 :
n
Proof.
)
Suppose lim (sn ) = +1. Choose > 0 and let M = . Since lim (sn) = +1,
there exists an N 2 N such that n > N implies sn > M = . Therefore, n > N
implies > s . Hence
1
1
1
n
1
s
n
(
so s
1
n
,
0 =
1 < for all n > N ;
sn
converges to 0. Suppose lim s = 0. Let M > 0 and let = M . Then > 0 and so there exists
an N such that
1
= 1 <= 1
,
0
sn
sn
M for all n > N :
1
1
n
Since sn > 0, we can write
M < sn for all n > N ;
hence lim (sn) = +1. 10. Monotonic Sequences and Cauchy Sequences
In this section we obtain two theorems that will allow us to conclude the convergence of a
sequence without knowing its limit in advance.
Denition 10.1. A sequence (sn) of real numbers is called non-decreasing if
sn sn for all n
and (sn ) is called non-increasing if
sn sn for all n :
+1
+1
38
A sequence that is either non-decreasing or non-increasing is called monotonic.
Examples:
The sequence
1
= 1; 1 ; 1 ; : : :
n n2N
2 3
is non-increasing, hence monotonic.
The sequence
1 , 21n
= 12 ; 43 ; 78 ; : : :
n2N
is non-decreasing, hence monotonic.
Theorem 10.2. All bounded monotonic sequences converge.
Proof. Let (sn) be a bounded non-decreasing sequence. Let
S = fsn j n 2 Ng :
Since S is bounded, the Completeness Axiom implies that S has a least upper bound, call
it u. We shall show that
lim (sn ) = u :
Choose > 0. We need to show that there exists N 2 N such that
jsn , uj < for all n > N :
Since u is the least upper bound, and because is positive, u , is not an upper bound
for S . Therefore, there exist some N 2 N such that
u , < sN :
But then since the sequence (sn ) is non-decreasing, we must have
u , < sn for all n > N ;
or
, < sn , u for all n > N ;
or
u , sn < for all n > N :
Again since u is an upper bound for S , we must have 0 u,sn. Therefore, jsn , uj = u,sn
and so we ahve
jsn , uj < for all n > N :
The case when (sn ) is a non-increasing sequence is similar. 39
Math 4023
Lecture 12
10. Monotonic Sequences and Cauchy Sequences
Theorem 10.4.
(i) If (sn) is an unbounded non-decreasing sequence, then
lim (sn ) = +1 :
(ii) If (sn) is an unbounded non-increasing sequence, then
lim (sn ) = ,1 :
Proof.
(i) Let (sn) be an unbounded nondecresing sequence. Let M > 0. Since the set
fsn j n 2 Ng
is unbounded, but is bounded from below by s (since (sn) is non-decreasing), it
must be unbounded from above. Hence, for some N 2 N, we must have
1
sN > M :
Since (sn) is non-decreasing, this implies
M < sn for all n > N ;
so
lim (sn ) = +1 :
(ii) The proof is similar to that of part (i).
Corollary 10.5. If (sn ) is a monotonic sequence then either (sn ) converges, or (sn ) diverges to +1, or (sn ) diverges to ,1.
40
Lemma. Suppose (sn ) is a non-decreasing sequence that is bounded from above, then
(sn ) converges to some s 2 R and
sn s for all n
Let M be an upper bound for fsn j n 2 Ng. Since (sn) is non-decreasing we have
s sn M for all n 2 N :
1
Thus,
jsnj max fjs j ; jM jg
1
and so (sn ) is bounded. Hence it converges by Theorem 10.2. Let
s = lim (sn ) :
In HW Problem 8.9 it is shown that if sn a for all but nitely many n then lim (sn) a.
Suppose now that one element of the sequence, say sN > s. Then because (sn) is nondecreasing we must have
s < sN sn for all n > N ;
or even
s < s +2sN < sN sn for all n > N :
Hence, there must be an innite number the sn greater than s s and so sn must converge
to a number greater than s s which is strictly greater that s lim (sn). Thus we have
a contradiction unless every sn is less than or equal to s. + N
2
+ N
2
Proposition. Suppose that for all n 2 N, we are given a closed interval In such that
In In for all n 2 N :
+1
Then
\
n2N
In 6= ; :
T
If in addition, the corresponding sequence of lengths converges to 0, then n2N In consists
of exactly one point.
Proof.
Write In = [an; bn ] with an bn. Since In In, we must have
+1
an an
+1
;
bn bn
+1
; for all n 2 N :
41
We then have
a an bn b for all n
so both (an ) and (bn) are bounded montonic sequence, hence they converge. Set
1
1
a = lim (an) ;
b = lim (bn) :
In view of the Lemma above, we must have
an a for all n :
Also, since for any nite n, we have am bn for all m, we have
a = lim (an) bn for any n
(See HW Problem 8.9). Therefore
an a bn for all n 2 N ;
hence a lies in every interval In, n 2 N. Hence,
a2
Now suppose
\
n2N
In :
lim (jbn , anj) = 0
and that
a0 2
\
n2N
In :
Since both a and a0 belong to In, we must have
0 ja , a0 j jbn , anj for all n :
But then by HW Problem 8.9 again, we must have
0 ja , a0 j lim (jbn , anj) = 0 :
Hence, ja , a0 j = 0, or a = a0 . Thus far, we've seen that monotonic sequences either have limits or they diverge in a very
special way.
42
There is another way to get a handle on a sequence and show that its limit exists, without
actually knowing the limit precisely.
Suppose (sn) is a bounded sequence in R, not necessarily monotonic. It may or may not
converge; but whatever the case the cruxt of the matter is the behavior of sets of the form
SN = fsn j n > N g :
For example, if lim (sn) does exists, it must clearly lie in the interval [uN ; vN ] where
uN = inf SN ;
vN = sup SN :
Since
SN 0 SN
it is clear that
if N 0 < N ;
;
u u u
1
2
3
and
v v v Since (un) and (vn ) are bounded monotonic sequences, their limits must exist. Set
1
2
3
lim sup (sn) = lim (vn) = Nlim
sup fsn j n > N g
!1
lim inf (sn) = lim (un) = Nlim
inf fsn j n > N g
!1
We emphasize that in general
lim sup (sn ) 6= sup fsn j n 2 Ng :
For example, if
then
but
sn = 1 + (,1)n
n
+1
lim sup (sn) = 1
sup fsn j n 2 Ng = 2 :
One can think of lim sup(sn ) as the largest number that innitely many sn can get close
to and lim inf (sn) as the smallest number that innitely many sn can approach.
43
Theorem 10.7. Let (sn) be a sequence in R.
(i) If lim (sn) is dened (as a real number or 1), then
lim inf (sn ) = lim (sn ) = lim sup (sn) :
(ii) If lim inf (sn) = lim sup(sn), then lim (sn ) is dened and
lim (sn ) = lim inf (sn ) = lim sup (sn) :
Homework:
1. Prove
(a)
(b)
1 = 0 ; for p > 0 :
lim
n!1 np
lim an = 0 ; if jaj < 1 :
n!1
(c)
nlim
!1 n
(d)
n!1
1
=n = 1
:
lim a =n = 1 :
1
2. Prove that if (sn ) is a non-decreasing sequence that is bounded from above, then (sn )
converges to some s 2 R and sn s for all n.
3. Problems 9.1, 9.9, 10.2
44
Math 4023
Lecture 13
Let me begin by recalling the denitions of lim sup and lim inf
Suppose (sn) is a bounded sequence in R, not necessarily monotonic. It may or may not
converge; but whatever the case the cruxt of the matter is the behavior of sets of the form
SN = fsn j n > N g :
For example, if lim (sn) does exists, it must clearly lie in the interval [uN ; vN ] where
uN = inf SN ;
vN = sup SN :
Since
SN 0 SN ; if N 0 < N ;
it is clear that
u u u and
v v v Since (un) and (vn ) are bounded monotonic sequences, their limits must exist. Set
lim sup (sn) = lim (vn) = Nlim
sup fsn j n > N g
!1
lim inf (sn) = lim (un) = Nlim
inf fsn j n > N g
!1
1
2
1
2
3
3
We emphasize that in general
lim sup (sn ) 6= sup fsn j n 2 Ng :
One can think of lim sup(sn ) as the largest number that innitely many sn can get close
to and lim inf (sn) as the smallest number that innitely many sn can approach.
Theorem 10.7. Let (sn ) be a sequence in R.
(i) If lim (sn) is dened (as a real number or 1), then
lim inf (sn ) = lim (sn ) = lim sup (sn) :
(ii) If lim inf (sn ) = lim sup (sn ), then lim (sn) is dened and
lim (sn ) = lim inf (sn ) = lim sup (sn) :
45
Proof.
(i) Suppose that lim sn = s 2 R and let > 0. Since (sn ) converges there exists an
N 2 N such that
So
Set
jsn , sj < for all n > N :
sn < s + for all n > N :
vn sup fsn j n > N g :
Clearly,
vn s + for all n > N :
Since vn s + for all but nitely many n, we know (from Homework Problem
8.9(b) that
lim sup (sn) lim (vn) s + :
Since lim sup (sn) s + for all > 0. We conclude that
lim sup (sn) s lim (sn) :
A similar argument shows that
lim inf (sn) s = lim (sn ) :
Since
we conclude
lim inf (sn) lim sup (sn )
lim inf (sn ) = lim (sn ) = lim sup (sn) :
Now suppose that
lim (sn ) = +1 :
Since (sn) diverges to +1, we know that for any real number M , there exists an
N 2 N such that
M < sn for all n > N :
Set
um = inf fsn j n > mg :
It follows that for every m > N , um > M . In other words,
lim (um ) lim inf (sn ) = +1 :
46
One similarly shows that
lim sup (sn) = +1 :
(ii)
Suppose that
lim inf (sn) = s = lim sup (sn )
where s is a real number. Let > 0. Since s = lim sup (sn ), there exists a number
N such that
js , sup fsn j n > N gj < :
Thus,
sup fsn j n > N g < s + and so
sn < s + for all n > N :
Similarly, since s = lim inf (sn ), there exists a number N such that
1
1
1
1
2
js , inf fsn j n > N gj < :
2
Thus,
s , < inf fsn j n > N g
2
and so
s , < sn for all n > N :
Set N = max fN ; N g. Then we have
2
1
2
s , < sn < s + or
jsn , sj < Thus,
The cases when
for all n > N ;
for all n > N :
lim (sn) = s :
lim inf (sn ) = lim sup (sn) = 1
are easy, and are left as homework.
47
Math 4023
Lecture 14
Recall
Theorem 10.7. Let (sn ) be a sequence in R.
(i) If lim (sn) is dened (as a real number or 1), then
lim inf (sn ) = lim (sn ) = lim sup (sn) :
(ii) If lim inf (sn ) = lim sup (sn ), then lim (sn) is dened and
lim (sn ) = lim inf (sn ) = lim sup (sn) :
This tells us that if a sequence (sn) converges, then for large N the numbers lim inf (sn) and
lim sup (sn ) must get closer and closer together. But this implies that all the numbers in
the set fsn j n > N g must get closer together as N gets large. This idea will be formalized
as follows:
Denition 10.8. A sequence (sn) of real numbers is called a Cauchy sequence if for
each > 0 there exists a natural number N such that
jsn , sm j < for all n; m > N :
Here is an easy lemma.
Lemma 10.9. Convergent sequences are Cauchy sequences.
Proof.
Suppose that lim (sn ) = s. Choose > 0. Since (sn ) converges we can always nd a
number N such that
jsn , sj < 2 and jsm , sj < 2 for all m; n > N :
By the Triangle Inequality then
jsn , sm j = jsn , s , (sm , s)j jsn , sj + jsm , sj < 2 + 2 = 48
for all n; m > N . Hence, (sn ) is a Cauchy sequence. Homework: 10.5, 10.7, 10.10
Also: Prove that
lim (sn ) = +1
,
lim inf (sn ) = lim sup (sn) = +1
49
Math 4023
Lecture 15
Recall
Denition. A sequence (sn ) of real numbers is called a Cauchy sequence if for each
> 0 there exists a natural number N such that
jsn , sm j < for all n; m > N :
and
Lemma 10.9. Convergent sequences are Cauchy sequences.
It is also easy to prove the following.
Lemma 10.10. Cauchy sequences are bounded.
Proof.
Let (sn ) be a Cauchy sequence. Then by denition, for any > 0 there is an N such that
jsn , sm j < for all n; m > N :
Choose = 1, then
jsn , sm j < 1 for all n; m > N :
In particular,
jsn , sN j < 1 for all n > N :
+1
Thus,
jsnj < jsN j + 1 for all n > N :
+1
Set
M = max fjsN j + 1; js j ; js j ; : : : ; jsN jg :
+1
Then
hence (sn) is bounded. 1
2
jsn j M for all n ;
50
Lemma 10.10. Cauchy sequences are bounded.
Theorem 10.11. A sequence (sn ) converges if and only if it is a Cauchy sequence.
Proof.
)
This is proved in Lemma 10.9.
(
Let (sn) be a Cauchy sequence. By Lemma 10.10 we know that (sn) is bounded. So we
cannot have lim (sn) = 1. By Theorem 10.7 we need only show that
lim inf (sn) = lim sup (sn ) :
Choose > 0. Since (sn ) is Cauchy, there exists N such that
jsn , sm j < for all n; m > N ;
or equivalently
sm , < sn < sm + for all n; m > N :
This shows that sm + is an upper bound for the set
SN = fsn j n > N g
so
vN sup SN sm + for all m > N :
or
vN , sm for all m > N
But the latter relation implies that vN , is a lower bound for the set SN . Thus
vN , inf SN uN :
or
vN uN + :
(1)
Since SN SN SN , the sequence (vN ) is non-increasing and the sequence
+1
+2
uN = inf SN
51
is non-decreasing,
lim sup (sn ) = Nlim
v vN
!1 N
(2)
and
lim inf (sn) Nlim
u uN
!1 N
(3)
But then (1), (2) and (3) imply
lim sup (sn) vN < uN + lim inf (sn) + (4)
for every > 0. Since
and
for every > 0, we must have
lim inf (sn) lim sup (sn )
lim sup (sn) lim inf (sn) + lim sup (sn ) = lim inf (sn ) :
Hence the sequence converges. Homework: 10.5, 10.7, 10.10
Also: Prove that
lim (sn ) = +1
,
lim inf (sn ) = lim sup (sn) = +1
52
Math 4023
Lecture 16
REVIEW SESSION
11. Subsequences
11.1Denition. Suppose that (sn )n2N is a sequence in R. A subsequence of (sn) is a
sequence (tk )k2N such that for each k there is a positive integer n such that
n < n < n < < nk < nk < and
tk = sn :
1
2
3
+1
k
Thus, a subsequence is just a selection of some (possibly all) of the sn arranged in order.
Note, however, that a subsequence must be innitely long, since we must dene a tk for
each k 2 N.
There are alternative ways of approaching this concept. One could choose rst an innite
subset fn ; n ; : : : g elements of which are arranged in increasing order, and then set
tk = sn :
1
2
k
Alternatively, given a sequence (sn )n2N , any monotonically increasing function : N ! N
denes a subsequence (tk )k2N of (sn)n2N ; viz.,
tk = s k ; 8 k 2 N :
(
)
11.2Theorem. If a sequence (sn) converges, then every subsequence of (sn) converges
to the same limit.
Proof.
Let (sn ) denote a subsequence of (sn). It is easy to prove by induction that that nk k
for all k. Let s = lim (sn). Then for any > 0, there exists N such that
jsn , sj < for all n > N :
Now k > N implies nk > N ; hence
jsn , sj < for all nk > N :
Consequently,
lim (sn ) = s :
k
k
k
53
Math 4023
Lecture 19
11. Subsequences
11.1 Denition. Suppose that (sn )n2N is a sequence in R. A subsequence of (sn) is a
sequence (tk )k2N such that for each k there is a positive integer n such that
n < n < n < < nk < nk < 1
2
3
and
+1
tk = sn
k
:
Thus, a subsequence is just a selection of some (possibly all) of the sn arranged in order.
Note, however, that a subsequence must be innitely long, since we must dene a tk for
each k 2 N.
There are alternative ways of approaching this concept. One could choose rst an innite
subset fn ; n ; : : : g elements of which are arranged in increasing order, and then set
1
2
tk = sn
k
:
Alternatively, given a sequence (sn )n2N , any monotonically increasing function : N ! N
denes a subsequence (tk )k2N of (sn)n2N ; viz.,
tk = s k ;
(
)
8k2N :
11.2 Theorem. If a sequence (sn) converges, then every subsequence of (sn) converges
to the same limit.
Proof.
Let (sn ) denote a subsequence of (sn). It is easy to prove by induction that that nk k
for all k. Let s = lim (sn). Then for any > 0, there exists N such that
k
jsn , sj < for all n > N :
Now k > N implies nk > N ; hence
jsn , sj < for all k > N :
k
54
Consequently,
lim (sn ) = s :
k
Example
,
Let sn = (,1)n 1 , n . The corresponding sequence is
2
3
4
5
6
1
0; 2 ; , 3 ; 4 ; , 5 ; 6 ; , 7 ; 1
This sequence does not converge. In fact, it is relatively easy to show that
lim inf (sn ) = ,1
:
lim sup (sn ) = +1
This sequence however, has two convergent subsequences,
tn = s n
rn = s n
2
2
+1
which respectively converge to -1 and 1.
The goal of this section is to prove the Bolzano-Weierstrass Theorem which states that
every bounded sequence has a convergent subsequence.
55
Math 4023
Lecture 20
11. Subsequences, Cont'd
11.3 Theorem. Every sequence has a monotonic subsequence.
Proof.
Let's say that the nth term sn of a sequence is dominant if it is greater than every other
term that follows it;
sn is dominant , sm < sn 8 m > n :
(1)
Case 1. Suppose there are innitely many dominant terms. Let (sn ) be any subsequence
consisting solely of the dominant terms. Then
k
sn +1 < sn
k
k
for all k by (1), and so (sn ) is monotonically decreasing.
k
Case 2. Suppose there are only nitely many dominant terms. Select n such that sn1 is
beyond all the dominant terms in the sequence. Since sn1 is not dominant, there exists
n > n such that sn2 sn1 . Since n is not dominant, there exists n > n such that
sn3 sn2 sn1 . Continuing in this manner, we construct a nondecreasing sequence
(sn ). 1
2
1
k
2
3
2
56
Math 4023
Lecture 21
11. Subsequences, Cont'd
Recall,
11.3 Theorem. Every sequence has a monotonic subsequence.
We'll now prove:
11.4 Corollary. Let (sn ) be any sequence. There exists a monotonic subsequence whose
limit is lim sup (sn ) and a monotonic subsequence whose limit is lim inf (sn ).
Proof.
For any N 2 N, set
and
Case 1. v = ,1.
vN = sup fsn j n > N g
v = Nlim
(v ) = lim sup (sn ) :
!1 N
In this case, (sn ) diverges to ,1, since
sN vN , ;
1
8N 2N :
Thus, the sequence (sn) itself converges to lim sup (sn ).
Case 2. ,1 < v < +1.
Select a sequence (tN ) that is monotonically increasing to v. If v = +1, then
tN = N will do; if ,1 < v < +1, then tN = v , N will do. We then have
1
tN < v vN = sup fsm j m > N g :
Since tN is less than the least upper bound of fsm j m > N g, we know there must
exist an element sm , with m > N such that
tN < sm :
In this way we construct a subsequence (sn ) of (sn) such that
N
tN < sn vN :
N
57
By the squeeze principle this subsequence converges to v. By Theorem 11.3, this
subsequence itself has a monotonic subsequence
sn
Nk
which, by Theorem 11.2, must converge to the same limit v.
Case 3. v = +1.
As just as in the preceding case, we can construct a monotonic sequence
sn
Nk
however, we can not apply Theorem 11.2 since (sn ) does not converge. However,
Theorem 11.2 is easily extended by the following fact.
N
Lemma. If a sequence (sn) diverges to 1, then so does every subsequence of
(sn).
Proof. (deferred to Homework)
This Lemma allows us to conclude that in the present case the subsequence
sn
Nk
diverges to lim sup (sn) = v = +1.
Similarly, one proves that there exists a subsequence of (sn ) that converges to lim inf (sn ).
11.5 The Bolzano-Weierstrass Theorem. Every bounded sequence has a convergent
subsequence.
Proof. If (sn) is a bounded sequence, then lim inf (sn) and lim sup (sn) are real numbers, and so the monotonic subsequences constructed in Corollary 11.4 converge to real
numbers. Homework:
1. Prove that if a sequence (sn ) diverges to +1, then so does every subsequence of (sn ).
2. Problem 11.3, pg. 55
3. Problem 11.6, pg. 56
58
Math 4023
Lecture 22
11. Subsequences, Cont'd
11.6 Denition. Let (sn) be a sequence in R. a subsequential limit is any real number
or symbol 1 that is the limit of some subsequence of (sn).
11.7 Theorem. Let (sn) be any sequence in R and let S denote the set of subsequential
limits of (sn).
(i) S is nonempty.
(ii) sup S = lim sup (sn) and inf S = lim inf (sn ).
(iii) lim (sn ) exists if and only if S has exactly one element.
Proof.
(i) This follows immediately from Corollary 11.4.
(ii) Let t be the limit of a subsequence (sn ) of (sn ). By Theorem 10.7 we have
k
lim inf (sn ) = t = lim sup (sn ) :
k
k
Since nk > k, we have
fsn j k > N g fsn j n > N g
k
so
inf fsn j n > N g inf fsn j k > N g sup fsn j k > N g sup fsn j n > N g
k
k
which implies
lim inf (sn) lim inf (sn ) = t = lim sup (sn ) lim sup (sn )
k
k
This chain of inequalities implies that
lim inf (sn) inf S sup S lim sup (sn ) :
Since, by Corollary 11.4, both lim inf (sn ) and lim sup (sn) belong to S , we have
inf S = lim inf (sn)
sup S = lim sup (sn ) :
(iii) This follows from (ii) and Theorem 10.7.
59
11.8 Theorem. Let S denote the set of subsequential limits of a sequence (sn). Suppose
(tn) is a sequence in S \ R and that t = lim (tn). Then t belongs to S .
,Proof.
sn
k;i
Suppose (tk ) is a sequence of limits of subsequences of (sn) and let us denote by
i2N the corresponding subsequence of (sn ) whose limit is tk ; i.e.,
lim s
i!1 n
k;i
= tk :
Noting that each tk is by hypothesis nite, we can use the denition of convergence to
conclude that for each positive number k , we can nd a Nk such that
sn , tk < 1
k ; 8 i > Nk :
Set
m =N
m = min fn ;i j n ;i > N ; n ;i > m g
m = min fn ;i j n ;i > N ; n ;i > m g
..
.
mk = min fnk;i j nk;i > Nk ; nk;i > mk, g :
Then (sm ) is a subsequence of (sn) and
(1)
jsm , tk j < k1 ; 8 k 2 N :
1
k;i
1
1
2
2
2
2
2
1
3
3
3
3
3
2
1
k
k
Case (i). Suppose t = lim (tk ) 2 R. Then
jsm , tj jsm , tk j + jtk , tj < k1 + jtk , tj ; 8 k 2 N :
k
It follows easily that
k
lim (sm ) = t
k
and so t 2 S .
Case (ii). Suppose t = +1. We have from (1)
tk , k1 < sm :
Since (tk ) , k diverges to innity, we must have (sm ) diverge to innity as well. But
+1 2 S . So the statement of the theorem is true in this case as well.
k
1
k
Case (iii). Suppose t = ,1. This is similar to Case (ii) above.
60
Math 4023
Lecture 22
11. Subsequences, Cont'd
11.8 Theorem. Let S denote the set of subsequential limits of a sequence (sn). Suppose
(tn) is a sequence in S \ R and that t = lim (tn). Then t belongs to S .
Suppose (tk ) is a sequence of limits of subsequences of (sn) and let us denote by
i2N the corresponding subsequence of (sn ) whose limit is tk ; i.e.,
,Proof.
sn
k;i
lim s
i!1 n
k;i
= tk :
Noting that each tk is by hypothesis nite, we can use the denition of convergence to
conclude that for each positive number k , we can nd a Nk such that
sn , tk < 1
k ; 8 i > Nk :
Set
m = n ;N1
m = min fn ;i j i > N and n ;i > m g
m = min fn ;i j i > N and n ;i > m g
..
.
mk = min fnk;i j i > Nk and nk;i > mk, g :
Then (sm ) is a subsequence of (sn) and
(1)
jsm , tk j < k1 ; 8 k 2 N :
1
k;i
1
1
2
2
2
2
1
3
3
3
3
2
1
k
k
Case (i). Suppose t = lim (tk ) 2 R. Then
jsm , tj jsm , tk j + jtk , tj < k1 + jtk , tj ; 8 k 2 N :
k
It follows easily that
and so t 2 S .
k
lim (sm ) = t
k
Case (ii). Suppose t = +1. We have from (1)
tk , k1 < sm
k
:
61
Since (tk ) , k diverges to innity, we must have (sm ) diverge to innity as well. But
+1 2 S . So the statement of the theorem is true in this case as well.
1
k
Case (iii). Suppose t = ,1. This is similar to Case (ii) above.
12. lim sup's and lim inf's
Recall that if (sn ) is any sequence of real numbers and if S is the set of subsequential limits
of (sn ) then
lim inf (sn) Nlim
inf fsn j n > N g = inf S ;
!1
lim sup (sn) Nlim
sup fsn j n > N g = sup S :
!1
Theorem 12.1. If (sn) converges to a positive real number s and (tn ) is any sequence,
then
lim sup (sn tn) = s lim sup (tn) :
Here we allow the conventions
s (+1) = +1
s (,1) = ,1
for all s > 0.
Proof.
Let = lim sup (tn). We have three cases.
Case 1. 2 R.
By Corollary 11.4, there exists a subsequence (tn ) of (tn) such that
k
lim (tn ) = :
k
Since (sn) converges to s, so does any subsequence of (sn ). Hence,
lim (sn ) = s :
k
Applying Theorem 9.4, we have
lim (sn tn ) = s :
k
k
62
But, according to Theorem 11.7, the limit of a subsequence of (sn tn) must be
bounded from above by lim sup (sntn ). Thus we have
s = s lim sup (tn) lim sup (sn tn) :
(1)
Now since the limit s of (sn ) is assumed to be positive, we know that except for
nitely many terms the terms sn must all be positive. By Lemma 9.5, then we
know that
lim s1 = 1s :
n
Working through
the argument leading to (1), except replacing the series (sn ) by
the series s and (tn) by (sntn), we obtain
1
n
or
1 lim sup (s t ) lim sup (t )
nn
n
s
lim sup(sn tn) s lim sup (tn) :
(2)
Comparing (1) and (2), we conclude
lim sup(sn tn) = s lim sup (tn) :
Case 2. = +1.
By Corollary 11.2 there exists a subsequence (tn ) of (tn) such that
k
lim (tn ) = +1 :
k
Since (sn) converges to s > 0, we have
lim (sn ) = s > 0 :
k
But then Theorem 9.9 tells us that
lim (sn tn ) = +1 :
k
k
But by Theorem 11.7, lim sup (sn tn) must be greater than or equal to the limit of
any subsequence of (sntn ). Hence
lim sup (sn tn) = +1 :
Case 3. = ,1.
63
In this case, since lim inf (tn) is always less than lim sup (tn), we must have
lim inf (tn) = lim sup (tn) = ,1 ;
hence, by Theorem 10.7,
lim (tn) = ,1 :
Consequently,
lim (sn(,tn )) = +1 ;
by Theorem 9.9, and so
lim (sntn) = ,1 :
Hence,
Homework.
12.3, 12.4, 12.8,
lim sup (sn tn) = ,1 :
64
Math 4023
Lecture 25
12. lim sup's and lim inf's
Recall that if (sn ) is any sequence of real numbers and if S is the set of subsequential limits
of (sn ) then
lim inf (sn) Nlim
inf fsn j n > N g = inf S ;
!1
lim sup (sn) Nlim
sup fsn j n > N g = sup S :
!1
Theorem 12.1. If (sn) converges to a positive real number s and (tn ) is any sequence,
then
lim sup (sn tn) = s lim sup (tn) :
Here we allow the conventions
s (+1) = +1
s (,1) = ,1
for all s > 0.
Proof.
Let = lim sup (tn). We have three cases.
Case 1. 2 R.
By Corollary 11.4, there exists a subsequence (tn ) of (tn) such that
k
lim (tn ) = :
k
Since (sn) converges to s, so does any subsequence of (sn ). Hence,
lim (sn ) = s :
k
Applying Theorem 9.4, we have
lim (sn tn ) = s :
k
k
But, according to Theorem 11.7, the limit of a subsequence of (sn tn) must be
bounded from above by lim sup (sntn ). Thus we have
(1)
s = s lim sup (tn) lim sup (sn tn) :
65
Now since the limit s of (sn ) is assumed to be positive, we know that except for
nitely many terms the terms sn must all be positive. By Lemma 9.5, then we
know that
lim s1 = 1s :
n
Working through
the argument leading to (1), except replacing the series (sn ) by
the series s and (tn) by (sntn), we obtain
1
n
or
1 lim sup (s t ) lim sup (t )
nn
n
s
lim sup(sn tn) s lim sup (tn) :
(2)
Comparing (1) and (2), we conclude
lim sup(sn tn) = s lim sup (tn) :
Case 2. = +1.
By Corollary 11.2 there exists a subsequence (tn ) of (tn) such that
k
lim (tn ) = +1 :
k
Since (sn) converges to s > 0, we have
lim (sn ) = s > 0 :
k
But then Theorem 9.9 tells us that
lim (sn tn ) = +1 :
k
k
But by Theorem 11.7, lim sup (sn tn) must be greater than or equal to the limit of
any subsequence of (sntn ). Hence
lim sup (sn tn) = +1 :
Case 3. = ,1.
In this case, since lim inf (tn) is always less than lim sup (tn), we must have
lim inf (tn) = lim sup (tn) = ,1 ;
hence, by Theorem 10.7,
lim (tn) = ,1 :
66
Consequently,
lim (sn(,tn )) = +1 ;
by Theorem 9.9, and so
lim (sntn) = ,1 :
Hence,
Homework.
12.3, 12.4, 12.8,
lim sup (sn tn) = ,1 :
67
Math 4023
Lecture 26
12. lim sup's and lim inf's, Cont'd
Theorem 12.2. Let (sn ) be any sequence of nonzero real numbers. Then we have
sn+1 lim inf s n
lim inf jsnj
=n lim sup js
1
=n lim sup sn
nj
1
+1
sn
Proof. The middle inequality is obvious.
To prove the rst inequality, it suces to show that
K=
= :
In order to show that K , it suces to prove that
(1)
M
sn+1 < lim inf s ) M lim inf jsn j =n
1
n
(This is just an application of generalization of the proposition
a for all < b
)
ba
or even more specically
a for all < 0
)
0a :
See Homework Problem 3.8.)
So suppose
M<
sn+1 lim inf s =
n
There there then exists an N such that
M<
lim
N !1
sn+1 inf s sn+1 inf s n
n
jnN
jn>N
:
Thus,
(2)
M
sn+1 < s n
for all n > N :
:
:
68
For n > N , we can write
jsnj
= ssn
n,
1
sn,1 s
n,
2
sN +1 s
N
jsN j :
Applying (2) we see that
jsN j (M )n,N < jsnj for all n > N :
Since M and N are xed in this argument, a = M ,N jsN j is a positive constant. We may
thus write
M n a < j sn j
or, equivalently,
for all n > N ;
Ma =n < jsn j =n for all n > N ;
Now recall that if a is a positive number, then
(3)
1
1
lim a
n!1
=n = 1
1
;
Hence, hence any subsequence of the right hand side of (3) converges
to M . On the
=n
other hand, by Corollary 11.4, we can nd a subsequence of jsnj
that converges to
lim inf jsn j =n. But then (3) implies
1
1
Ma =n < jsn j =n
1
which implies
1
k
for all k > N ;
k
k
M lim inf jsn j =n :
1
Thus (1) is true, so
sn+1 lim inf s n
lim inf jsnj =n
1
The last inequality is proved similarly (see text). Homework: 12.3, 12.4, 12.8,
14. Series
Suppose (ai )i2N is any sequence of elements of R. For any nite n we can compute
n
X
i
=1
ai a + a + a + + an :
1
2
3
69
Set
sn =
n
X
i
ai :
=1
We can thus associate with the original sequence (ai ) a sequence (sn) of partial sums. The
sequence (sn) may or may not converge. The formal symbol
1
X
i
an
=0
is called the innite series corresponding to (ai ). It dened by
n
X
lim s = lim
n!1 n n!1 i
ai
=1
whenever this limit is exists; if the sequence (sn) diverges to +1, we write
1
X
i
ai = +1
=1
and say that the innite series diverges to +1; if the sequence (sn ) diverges to ,1, we
write
1
X
ai = ,1
i
=1
and say that the innite series diverges to ,1; otherwise, we simply say that the innite
series diverges.
Denition 14.3. We say that a series P an satises the Cauchy criterion if its sequence
(sn ) of partial sums is a Cauchy sequence:
for each > 0 there exists an N 2 N such that
m; n > N ) jsn , sm j < .
One can equally well phrase the Cauchy criterion as follows
for each > 0 there exists an N 2 N such that
(1)
nm>N )
n X
a
<
i
i m
:
=
We will usually employ this version of the Cauchy criterion.
Theorem 10.11 and the above denition imply the following.
Theorem 14.4. A series converges if and only if it satises that Cauchy Criterion.
70
Math 4023
Lecture 27
14. Series
Denition 14.3. We say that a series PPan satises the Cauchy criterion if its sequence
of partial sums is Cauchy. Equivalently,
there exists an N such that
nm>N
an satises the Cauchy criterion if for each > 0
)
n
X
a
n i m
< :
=
Theorem 14.4. A series is convergent if and only if it satises the Cauchy criterion.
Corollary 14.5. If a series P an converges then
lim ja j = 0 :
n!1 n
Proof. According to Theorem 14.4 and Denition 14.3, a series
if for every > 0 there is an N such that
nm>N
)
n
X
an
i m
P
an converges if and only
< :
=
Choosing n = m, we have
) jan j < ;
which implies precisely that the sequence (janj) converges to zero. n>N
Theorem 14.6. (The Comparison Test). Let P an be a series where an 0 for all n.
P
P
(i) If an converges and jbn j an for all n, then bn converges.
P
P
(ii) If an = +1 and bn an for all n, then bn = +1.
P
P
(iii) If an = ,1 and bn an for all n, then bn = ,1.
Proof.
71
(i) For n m, we have
n X
b
i
(1)
i m
n
X
i m
jbi j =
=
n
X
i m
ai =
n X
a
i
i m
:
=
=
The rst inequality is just the generalized triangle inequality, the second
inequality
P and the equality on the right hand side hold by hypothesis.
Since an converges, by Theorem 14.4, it satises the Cauchy criterion;
for any > 0, there exists N such that
nm>N
n X
a
<
i
)
i m
:
=
In view of (1) then, for this same N then
nm>N
)
n X
bi i m
< :
=
Hence,
14.4.
P
bn satises the Cauchy criterion and so converges by Theorem
P
(ii) Let (sn) and (tn) be the sequences of partial sums for an and
Since bn an for all n, we have tn sn for all n. By hypothesis,
1
X
n
=1
P
bn .
an nlim
!1 sn = +1 :
Hence, by Homework Problem 9.9,
+1 = nlim
!1 tn 1
X
n
bn :
=1
(iii) This is similar to the proof of (ii).
Examples of Series.
The following two series provide convenient reference points for Comparison Tests.
1. The p-Series
1
X
1
p
n n
=1
72
The integral test shows that a p-series converges if p > 1 and diverges if
p 1.
2. The Geometric Series
1
X
i
=0
n
X
ari = nlim
!1 i
=1
!
ari = nlim
!1
1 , rn
+1
a 1,r
8
<
a
if jrj < 1
=:1,r
undened otherwise
(The third equality is proved by induction on n.) Thus,
1
X
i
=1
converges if jrj < 1.
ari
73
Math 4023
Lecture 28
14. Series, Cont'd
Denition. A series P an is said to be absolutely convergent if the series P janj
converges.
Corollary 14.7. Absolutely convergent series are convergent.
Proof. This follows from the Comparison Test (Theorem 14.6) and the obvious fact
an jan j :
Theorem 14.9. (The Root Test). Let P an be a series and let = lim sup janj =n.
1
P
an converges if < 1.
P
(ii) The series an diverges if > 1.
(iii) Otherwise, = 1 and the Root Test gives no information.
(i) The series
Proof.
(i) Suppose < 1 and select > 0 so that + < 1. Then by denition,
there is a natural number N such that
n
o
, < sup janj =n j n > N < + :
1
In particular, we have
jan j =n < + for all n > N:
1
Therefore,
janj < ( + )n for all n > N:
Since 0 < + < 1, the geometric series
1
X
n
=1
( + )n
74
converges. The comparison test then shows that the series
1
X
n N
=
also converges. Hence
an
+1
1
X
n
an
=1
converges.
(ii) If > 1, then by Corollary 11.4, there is a subsequence
jan j =n
1
k
k
converging to . It follows that janj > 1 for innitely many choices of n.
Hence,
lim ja j 6= 0 :
n!1 n
P
Thus, by Corollary 14.5. The series
(iii) Consider the series
an cannot converge.
1
X
1
n n
1 1
X
:
n
n
One can use the Integral Test to show that the rst series converges, and
that the second diverges. But in both cases
2
=1
=1
=n = 1 :
lim
j
a
j
n
n!1
1
Hence, the root test is inconclusive when = 1.
75
Math 4023
Lecture 29
14. Series, Cont'd
Theorem 14.8. (The Ratio Test). Let P an be a series such that an 6= 0 for all n.
an+1 lim sup an <
(i)
P
an converges absolutely if
(ii)
P
an diverges if lim inf aa+1 > 1.
1.
n
n
(iii) Otherwise,
an+1 lim inf a n
1
and the test gives no information.
an+1 lim sup a n
Proof. According to Theorem 12.2, we have
lim sup jan j =n lim sup aan n
lim inf aan lim inf janj =n
n
+1
1
+1
(i) If
then the series
(ii) If
P
1
an+1 lim sup a <
1
n
an converges by the Root Test.
an+1 lim inf a n
>1
then the series diverges by the Root Test.
(iii) The series
1
X
1
n n
;
=1
both satisfy
1
X
1
n n
2
=1
an+1 nlim
!1 a =
n
1 :
:
76
However, the rst series diverges and the second converges. The Ratio
Test must therefore be inconclusive when
an =1 :
lim
n!1 an +1
Remark. Note that since the Root Test was used to prove the Ratio Test,
the Root Test
P is a bit more rened than the Ratio Test. Indeed, there
exists series an for which the Ratio Test is inconclusive, but for which
the Root Test is conclusive.
Example 1.
Consider the series
1
X
n
=0
1 +1+ 1
2 , ,n = 2 + 41 + 12 + 16
8 64
(
1)
n
If we apply the ratio test nd
8
<1
an+1 an = : 28
Therefore
if n is even
if n is odd
1 = lim inf an < 1 < lim sup an = 2
a a 8
n
n
and so the ratio test is inconclusive.
+1
+1
However,
janj
=n =
1
Since
1
(,1)n ,n n
2
(
=
2
2
1,n
n
,1n,n
= 2 1 2, if n is even
= 2, 1 2, if n is odd
n
n
1
,1 = 1
lim
2
=
lim
2
n!1
n
n
1
lim sup jan j = 12 < 1
n
1
1
77
and so the series converges.
Example 2.
Determine if the series
1
X
2
n
=1
converges.
n,1
n
Well, at rst nothing seems to work. The Ratio and Root tests are quickly seen to be
inconclusive, and a comparison test with the harmonic series does not work because
n,1 = 1 , 1 < 1 ;
n
n n n
but we need the opposite inequality to conclude that
2
2
1
X
2
n
=1
from the fact that
n , 1 = +1
n
1
X
1 = +1 :
n n
=1
We'll argue as follows. Suppose
1
X
2
n
=1
Set
sN =
N
X
N 1 X
N 1
n,1 = X
,
= rN + tN
n
n n n n
2
n
=1
We know that
n,1 = S 2 R :
n
2
=1
lim (t ) =
N !1 N
converges. Theorem 9.3 then implies
=1
1
X
1
n n
2
=1
lim (s + t ) = Nlim
(r )
N !1 N N
!1 N
78
converges. But
1
X
1 = +1 :
n n
P
Hence, we have a contradiction if nn,2 converges. Therefore
lim (r ) =
N !1 N
=1
1
1
X
n , 1 = +1 :
n
2
n
=1
Homework: 14.2, 14.4, 14.7, 14.12, 15.1, 15.4, 15.6
79
Math 4023
Lecture 30
15. Other Convergence Tests
I will simply cite without proof the Integral Test and the Alternating Series Test.
Theorem 15.2. (The Integral Test). Suppose P an is a series for which each element
an of the underlying series is P
the value at n of a continuous positive decreasing function
f : [1; +1] ! R. The series an is convergent if and only if the improper integral
Z
1
1
f (x) dx
is convergent.
Theorem 15.3. (The Alternating Series Test). If
a a a an 0
1
and
then the alternating series
converges.
2
3
lim a = 0
n!1 n
1
X
n
;
(,1)n an
=0
Homework: 14.2, 14.4, 14.7, 14.12, 15.1, 15.4, 15.6
Examples:
1.
1
X
n
=1
n
n!
2
When factorials appear in the expressions for an, the ration test is usually decisive.
80
(n+1)2 (n+1)! lim n!1 nn2! n!
(n + 1)
= nlim
!1
n
(n + 1)!
n
+
2
n
+
1
1
= nlim
lim
!1
n!1 n + 1
n
=10
=0
2
2
2
2
This implies
an+1 lim sup a <
n
and so the series converges.
2.
1
X
n
1
(,1)n
=1
This series cannot converge since
lim ja j = 1 6= 0
n!1 n
:
(See Corollary 14.5).
3. (ProblemP14.7) pShow that if
p > 1, then (an) converges.
Well, by Corollary 14.5, since
P
P
an is a convergent series of nonnegative numbers and
an converges,
lim a = 0
n!1 n
:
This implies that we can nd an N such that
an < 1
for all n > N :
This and the fact that p is assumed to be greater than 1 implies
0 (an )p < an < 1
Hence
1
X
n N
=
+1
for all n > N :
(an )p
81
converges since
1
X
n N
=
an
+1
P
converges (the comparison test). Thus the series (an)p converges (As far as the question of convergence goes, the number n at which we begin the \innite summation" is
immaterial.)
4. (Problem 15.4(a))
1
X
n
=2
1
pn log(
n) :
One could try the integral test directly, but an antiderivative of
1
f (x) = px log(
x)
is not so easy to nd. We'll make an estimate rst.
I rst claim that
p
log(n) < n
for all suciently large n. To see this, we simply dierentiate the dierence d(x) p
x , log(x),
px , 2
1
1
0
d (x) = 2px , x = 2x
> 0 for all x > 4 :
and so d(x) is monotonically increasing for x > 4 and moreover, d(5) = :627:: and so d(x)
is positive for all x > 4. Thus,
p1n < log(1 n) for all n > 4 ;
hence
1 = p1 p1 < p 1
n
n n
n log(n) for all n > 4 :
Now since
1 1
X
n n
diverges by the integral test,
1
X
1
pn log(
n)
n
diverges by the comparison test.
=5
=2
82
Math 4023
Lecture 34
17. Continuous Functions
Recall that a function f is a rule that assigns to each element of a set A an element of a
set B. The set A is called the domain of the function f . We will use the notation
dom(f ) = A :
We will be primarily concerned with functions f such that dom(f ) R and such that
f (x) 2 R for all x 2 dom(f ). Properly speaking, the symbol f represents the function
while the notation f (x) is reserved for the value of the function at x. However, a function
is often given by specifying its values without mentioning its domain; e.g.,
f (x) = x :
2
In such a case, it is generally assumed that domain of f is its so-called natural domain ;
the largest subset of R upon which f (x) is a well-dened real number. Thus, the function
f (x) = x1
is shorthand for \the function f whose domain is D = fx 2 R j x 6= 0g and whose value at
x 2 D is given by x ",
1
Denition 17.1. Let f be a real-valued function whose domain is a subset D of R. The
function f is said to be continuous at xo 2 D if, for every sequence (xn ) of points in D
converging to xo, we have
lim (f (xn )) = f (xo ) :
If f is continuous at every point of a set S D, then f is said to be continuous on S .
The function f is said to be continuous if it is continuous on D.
Theorem 17. 2. Let f be a real-valued function with domain D R. Then f is
continuous at xo 2 D if and only if
(i) for each > 0 there exists a > 0 such that
x 2 D and jx , xo j < Proof.
)
jf (x) , f (xo )j < :
83
=) Assume that f is continuous at zo but that condition (i) fails. Then there
exists an > 0 such that there is no such that
x 2 D and jx , xo j < )
jf (x) , f (xo )j < :
)
jf (x) , f (xo )j < In particular, the statement
x 2 D and jx , xo j < n1
must be false for all n. This implies that for each n 2 N, there is an
xn 2 D such that
jxn , xo j < n1 and jf (xn ) , f (xo )j :
Note that the sequence (xn) converges to xo but the sequence (f (xn ))
cannot converge to f (xo ) since
jf (xn ) , f (xo )j > 0
for all n. Thus, if f is continuous, then condition (i) cannot hold without
contradiction. Hence, continuity implies condition (i).
(= Assume condition (i) holds and consider an arbitrary sequence (xn) in
D such that
We need to prove that
lim (xn) = xo :
lim (f (xn )) = f (xo ) :
Choose > 0. By (i), there exists such that
x 2 D and jx , xo j < ) jf (x) , f (xo )j < :
Since (xn ) converges to xo , there exists an N such that
(3)
jxn , xo j < for all n > N :
But then (2) and (3) together imply
jf (xn ) , f (xo )j < for all n > N :
Hence
lim (f (xn )) = f (xo ) :
(2)
84
Example 1.
Show that the function
8
< x2 sin 1
f (x) = :
x ; x 6= 0
0 ; x=0
is continuous at x = 0.
Let (xn) be a sequence in R converging to 0. We need to show that
lim (f (xn )) = 0 :
Now for any x 6= 0, we have
so for x 6= 0, we have
,1 sin x1 1
x sin 1 x x 2 sin x1 jf (x) , 0j =
j j
x
2
and certainly
jf (0) , 0j = 0 :
Hence,
(f (x) , 0) x
(4)
2
for all x 2 R:
Now choose > 0. Since (xn ) converges to 0, we can nd an N such that
jxn , 0j = jxnj p
for all n > N :
In view of (4), this implies
jf (xn ) , 0j (xn ) for all n > N :
2
Hence, (f (xn )) converges to 0, and so the function f is continuous at 0.
85
Theorem 17.3. Let f be a real-valued function with domain D R. If f is continuous
at xo 2 D, then jf j and kf , k 2 R, are also continuous at xo.
Proof. Since f is continuous at xo, we know that if (xn ) is any sequence converging to xo,
then
lim (f (xn )) = f (xo ) :
By Theorem 9.2
lim (an ) = L
) lim (kan) = kL :
Hence,
lim (kf (xn )) = kf (xo ) [kf ] (xo ) ;
and so kf is continuous.
On the other hand, the inequalities
0 jjf (xn )j , jf (xo )jj jf (xn ) , f (xo )j :
and the squeeze principle imply that
lim jf (xn )j = jf (xo )j jf j (xo ) ;
and so jf j is continuous at xo .
86
Math 4023
Lecture 35
17. Continuous Functions, Cont'd
Theorem 17.3. Let f be a real-valued function with domain D R. If f is continuous
at xo 2 D, then jf j and kf , k 2 R, are also continuous at xo.
Proof. Since f is continuous at xo, we know that if (xn ) is any sequence converging to xo,
then
lim (f (xn )) = f (xo ) :
By Theorem 9.2
lim (an ) = L
) lim (kan) = kL :
Hence,
lim (kf (xn )) = kf (xo ) [kf ] (xo ) ;
and so kf is continuous.
On the other hand, the inequalities
0 jjf (xn )j , jf (xo )jj jf (xn ) , f (xo )j :
and the squeeze principle imply that
lim jf (xn )j = jf (xo )j jf j (xo ) ;
and so jf j is continuous at xo .
Theorem 17.4. Let f and g be real-valued functions that are continuous at xo 2 R.
Then
(i) f + g is continuous at xo ;
(ii) fg is continuous at xo ;
(iii) f=g is continuous at xo if g(xo) 6= 0.
Proof. As with the preceding theorem, the proofs of these statements boils down
to properties of limits of sequences.
87
By Theorem 9.4
)
lim (an) = A and lim (bn ) = B
Hence
(
lim (an + bn ) = A + B
lim (anbn ) = AB
lim (f (xn )) = f (xo ) and lim (g(xn)) = g(xo )
)
(
lim (f (xn ) + g(xn)) = f (xo ) + g(xo ) [f + g] (xo )
lim (f (xn )g(xn )) = f (xo )g(xo ) [fg] (xo )
Similarly, statement (iii) is proved by refering to Theorem 9.5:
an 6= 0 and lim (an) = A 6= 0
) lim a1 = a1
n
to show that
1
lim g(x ) = g(1x )
n
and then applying (ii) to show that
o
lim ([f=g] (xn )) = lim f (xn ) g(x1 ) = fg((xxo)) [f=g] (xo ) :
n
o
Theorem 17.5. If f is continuous at xo and g is continuous at f (xo ), then g f
is continuous at xo.
Proof. Let
D = fx 2 dom(f ) j f (x) 2 dom(g)g :
(D is the natural domain of g f .) Since f is continuous at xo , given any sequence
(xn) in D, (f (xn )) converges to f (xo ) and moreover, by construction,
f (xn ) 2 dom(g) for all n :
Hence, (f (xn )) is a sequence in dom(g) converging to f (xo ). Since g is continuous
at f (xo ), (g (f (xn ))) = (g f (xn )) converges to g f (xo ). Hence, g f is continuous
at xo . Homework: 17.2, 17.5, 17.6, 17.12
88
Math 4023
Lecture 35
17. Continuous Functions, Cont'd
Theorem 17.3. Let f be a real-valued function with domain D R. If f is
continuous at xo 2 D, then jf j and kf , k 2 R, are also continuous at xo .
Proof. Since f is continuous at xo , we know that if (xn) is any sequence converging
to xo , then
lim (f (xn )) = f (xo ) :
By Theorem 9.2
lim (an ) = L
)
lim (kan) = kL :
Hence,
lim (kf (xn )) = kf (xo ) [kf ] (xo ) ;
and so kf is continuous.
On the other hand, the inequalities
0 jjf (xn )j , jf (xo )jj jf (xn ) , f (xo )j :
and the squeeze principle imply that
lim jf (xn )j = jf (xo )j jf j (xo ) ;
and so jf j is continuous at xo.
Theorem 17.4. Let f and g be real-valued functions that are continuous at
xo 2 R. Then
(i) f + g is continuous at xo ;
(ii) fg is continuous at xo ;
(iii) f=g is continuous at xo if g(xo) 6= 0.
Proof. As with the preceding theorem, the proofs of these statements boils down
to properties of limits of sequences.
89
By Theorem 9.4
)
lim (an) = A and lim (bn ) = B
Hence
(
lim (an + bn ) = A + B
lim (anbn ) = AB
lim (f (xn )) = f (xo ) and lim (g(xn)) = g(xo )
)
(
lim (f (xn ) + g(xn)) = f (xo ) + g(xo ) [f + g] (xo )
lim (f (xn )g(xn )) = f (xo )g(xo ) [fg] (xo )
Similarly, statement (iii) is proved by refering to Theorem 9.5:
an 6= 0 and lim (an) = A 6= 0
) lim a1 = a1
n
to show that
1
lim g(x ) = g(1x )
n
and then applying (ii) to show that
o
lim ([f=g] (xn )) = lim f (xn ) g(x1 ) = fg((xxo)) [f=g] (xo ) :
n
o
Theorem 17.5. If f is continuous at xo and g is continuous at f (xo ), then g f
is continuous at xo.
Proof. Let
D = fx 2 dom(f ) j f (x) 2 dom(g)g :
(D is the natural domain of g f .) Since f is continuous at xo , given any sequence
(xn) in D, (f (xn )) converges to f (xo ) and moreover, by construction,
f (xn ) 2 dom(g) for all n :
Hence, (f (xn )) is a sequence in dom(g) converging to f (xo ). Since g is continuous
at f (xo ), (g (f (xn ))) = (g f (xn )) converges to g f (xo ). Hence, g f is continuous
at xo . Homework: 17.2, 17.5, 17.6, 17.12
90
Math 4023
Lecture 36
18. Properties of Continuous Functions
Denition. A real-valued function f is said to be bounded if there exists M 2 R
such that
jf (x)j M for all x 2 dom(f ) :
Theorem 18.1. Let f be a real-valued function that is continuous on a closed
interval [a; b]. Then f is bounded on [a; b]. Moreover, there exist xo ; yo 2 [a; b]
such that
f (xo ) f (x) f (yo ) for all x 2 [a; b] :
Proof. Assume that f is not bounded on [a; b]. Then to each n 2 N, there is a
point xn 2 [a; b], such that jf (xn )j > n. By the Bolzano-Weierstrass Theorem,
there exists a subsequence (xn ) converging to some point xo 2 [a; b]. Since f is
continuous on [a; b], we must have
lim (f (xn )) = f (xo) 2 R
But we also have
lim (f (xn )) lim (n) = +1 :
Thus, we have a contradiction. Thus, f must be bounded if it is continuous.
k
k
k
Now let
M = sup ff (x) j x 2 [a; b]g :
By the preceding paragraph, M is nite. For each n 2 N there exists yn 2 [a; b],
such that
M , n1 < f (yn ) M :
(Conceivably, we could have yn = y for all n and f (y ) = M . But since the sup
is the least upper bound, if f (yi ) < M , then there has to be another point yj
such that f (yi ) < f (yj ) M; otherwise f (yi ) would be the least upper bound of
ff (x) j x 2 [a; b]g.) Hence we have
lim (f (yn )) = M :
By the Bolzano-Weierstrass Theorem, there is a convergent subsequence (yn ) of
(yn) that converges to a point yo 2 [a; b]. Since f is continuous, we must have
lim (f (yn )) = f (yo ) :
1
1
k
k
91
But since (f (yn )) is a subsequence of (f (yn )) we must also have
k
lim (f (yn )) = M :
k
Hence, there is a yo 2 [a; b] such that
f (yo ) = sup ff (x) j x 2 [a; b]g ;
i.e., f attains a maximal value on [a; b].
One argues similarly to show that f attains a minimal value on [a; b].
Theorem 18.2. (Intermediate Value Theorem). If f is a continuous real-valued
function on a nite interval I , then whenever a; b 2 I , a < b, and f (a) < y < f (b)
or f (b) < y < f (a); there exists an least one x 2 I such that y = f (x).
Proof. We assume f (a) < y < f (b), the other case being similar. Set
S = fx 2 [a; b] j f (x) < yg :
Since a belongs to S , S is non-empty, hence by the completeness axiom
xo = supS
exists. It is also clear that xo 2 [a; b]. For each n > N,
xo , n1
is not an upper bound for S and so there exists an element sn 2 S such that
xo , n1 < sn
It is clear that
and so by the continuity of f
(1)
and
f (sn ) < y :
lim (sn) = xo
f (xo ) = lim (f (sn )) y :
92
Now set
Since
we have
tn = min b + xo + n1
xo tn xo + n1 ;
:
8n2N ;
lim (tn) = xo:
But since tn xo = sup S , we know that
f (tn ) y
8n :
Again continuity implies
(2)
f (xo ) = lim (f (tn )) y :
Comparing (1) and (2), we conclude f (xo ) = y. 93
Math 4023
Lecture 37
18. Properties of Continuous Functions, Cont'd
Last time we proved
Theorem 18.2. (Intermediate Value Theorem). If f is a continuous real-valued
function on a nite interval I , then whenever a; b 2 I , a < b, and f (a) < y < f (b)
or f (b) < y < f (a); there exists an least one x 2 I such that y = f (x).
The Intermediate Value Theorem is extremely important to the theory of calculus.
One of its simplest consequences is given below
Corollary 18.3. If f is a continuous real-valued function on an interval I , then
the set
f (I ) = ff (x) j x 2 I g
is also an interval or a single point.
Proof. A nonempty subset J R is an interval if
)
inf J < y < sup J
(3)
Set
= inf (f (I ))
;
y2J :
= sup (f (I ))
and suppose
<y< :
By the denition of innum and suprenum then, there must exist elements f (x ); f (x ) 2
f (I ) such that
< f (x ) < y < f (x ) < :
By the preceding theorem, then there must exist x 2 I such that y = f (x). Hence
y 2 f (I ), so the condition (3) is satised, so f (I ) is an interval. 1
1
2
Here is another application of the Intermediate Value Theorem.
2
94
Let f be a continuous function mapping the interval [0; 1] into itself. I claim there
is at least one point xo 2 [0; 1] such that
f (xo ) = xo :
Consider the function
g(x) = f (x) , x :
It is also continuous on [0; 1]. We have
g(0) = f (0) , 0 0
and
g(1) = f (1) , 1 0 :
Therefore,
g(1) 0 g(0) :
Now if g(1) or g(0) equals zero, then we can, respectively, take xo = 1 or xo = 0.
Otherwise,
g(1) < 0 < g(0)
and so by the Intermediate Value Theorem we must have an xo 2 [0; 1] such that
g(xo ) = 0 ;
consequently,
f (xo ) = xo :
Denition. We say that a function f is strictly increasing if for all point x ; x 2
dom(f ),
x <x
) f (x ) < f (x ) :
1
1
2
1
2
2
Theorem 18.5. Let f be a strictly increasing function on an interval I such that
f (I ) is an open interval. Then f is a continuous function.
Proof. Consider a point x 2 I . We can assume that x is not an endpoint of I .
Then f (x ) is not an endpoint of I either, and so there exists an > 0 such that
0
0
0
(f (x ) , ; f (x ) + ) f (I ) :
0
0
But then there exist x ; x 2 I such that
f (x ) = f (x ) , f (x ) = f (x ) + 1
2
1
0
2
0
95
and
x <x < :
1
0
2
Set
= min fx , x ; x , x g :
But now we have constructed a > 0 such that
1
jx , x j < 0
)
0
2
0
jf (x) , f (x )j < :
0
Hence, f (x) is continuous at x . Since x is arbitrary, we conclude that f is
continuous at every interior point of I . 0
0
Remark. The condition that the interval I be open can be relaxed
without any change in the statement. (Of course, there's then
something left to be proved.)
Theorem 18.4. Let f be a continuous strictly increasing function on some interval I . Then f , is a well-dened function with domain f (I ). Moreover, f , is
continuous on f (I ).
1
1
Theorem 18.6. Let f be a one-to-one continuous function on an interval I . Then
f is either strictly increasing or strictly decreasing.
Homework: 18.4, 18.6, 18.9
96
Math 4023
Lecture 38
18. Properties of Continuous Functions, Cont'd
Theorem 18.4. Let f be a continuous strictly increasing function on some interval I . Then f , is a well-dened function with domain f (I ). Moreover, f , is
continuous on f (I ).
1
1
Theorem 18.6. Let f be a one-to-one continuous function on an interval I . Then
f is either strictly increasing or strictly decreasing.
Homework: 18.4, 18.6, 18.9
19. Uniform Continuity
Recall the , denition of continuity (Theorem 17.2).
A function f is continuous at xo 2 dom(f ) if for each > 0
there exists a > 0 such that
(1)
jx , xo j < )
jf (x) , f (xo )j < :
Notice that this only denes continuity at xo . If we wish to wish to show that
f is continuous at another point x , we have to show that given > 0, we can
construct a number x1; > 0 such that
1
jx , x j < 1
)
jf (x) , f (x )j < :
1
1
If we wish to show that f is continuous on an interval I dom(f ), we will have
to verify (1) for each point xo 2 I . Clearly, this would be an insurmountable task
unless the number x ; did not really depend of the point xo . This observation
motivates the following dention.
o
97
Denition 19.1. Let f be a real-valued function dened on a set S R. Then
f is uniformly continuous on S if
for each > 0 there exists a > 0 such that
x; y 2 S ; jx , yj < ) jf (x) , f (y)j < :
We will say that f is uniformly continuous if f is uniformly continuous on dom(f ).
Theorem 19.2. If f is a continuous function of a closed interval [a; b], f is
uniformly continuous on [a; b].
Proof. Assume that f is not uniformly continuous on [a; b]. Then there exists an
> 0 such that for each > 0, the statement
\ jx , yj < )
jf (x) , f (y)j < "
is false. In other words, for each > 0, there exists a pair x; y 2 [a; b], such that
jx , yj < and jf (x) , f (y)j :
In particular, for each n 2 N, there exists a pair xn ; yn 2 [a; b] such that
jxn , ynj < n1 and jf (xn ) , f (yn )j :
(2)
By the Bolzano-Weierstrass Theorem, there must exist a subsequence (xn ) of
(xn) that converges to some number xo 2 [a; b]. (We know that the limit point
belongs to the interval since the interval is closed.) Let (yn ) be the corresponding
sequence for (yn ). We have
k
k
jxn , yn j < n1 k1 for all k ;
k
k
k
and so cleary (yn ) also converges to xo . Since f is continuous on [a; b] it is
continuous at xo, hence
k
lim (f (xn )) = f (xo) = lim (f (yn )) :
k
Consequently,
k
lim jf (xn ) , f (yn )j = 0 :
k
But by (2)
k
jf (xn ) , f (yn )j > 0
k
k
98
or all k. Hence, we have a contradiction. We conclude that f must be uniformly
continuous on [a; b]. Remark. The preceding proof relied on the following two properties of the interval
[a; b]. First, we needed the interval to be bounded, so that we could apply the
Bolzano-Weierstrass Theorem to construct the convergent subsequences (xn ) and
(yn ). Second, we need the fact that [a; b] is closed to ensure that the limits of
the (xn ) and (yn ) lie in [a; b]. Theorem 19.2 is thus easily generalized to the
statement
k
k
k
k
If f is continuous on a closed and bounded set S of R, then
f is uniformly continuous on S .
Example 1. Show that the function
f (x) = x1
2
is uniformly continuous on [; 1], > 0.
By Problem 17.6 in the homework, we know that f is continuous everywhere on
the real line except at x = 0. Therefore, it is continuous on [; 1]. By the preceding
theorem then, f is uniformly continuous on [; 1].
Theorem 19.4. If f is uniformly continuous on a set S and (sn) is a Cauchy
sequence in S , then (f (sn)) is a Cauchy sequence.
Proof. Let (sn ) be a Cauchy sequence in S and let > 0. Since f is uniformly
continuous on S , there exists a > 0 such that
(3)
x; y 2 S ; jx , yj < )
jf (x) , f (y)j < :
Since (sn) is Cauchy, there exists an N such that
jsn , sm j < for all n; m > N :
But then (3) implies
jf (sn) , f (sm )j < Hence, (f (sn)) is Cauchy. for all n; m > N
:
99
Example 2. Show that
f (x) = x1
2
is not uniformly continuous on (0; 1). Let
sn = n1 for n 2 N :
Then (sn) is obviously a Cauchy sequence in (0; 1). However, since
f (sn) = n ;
2
(f (sn )) can not be a Cauchy sequence (it diverges). Therefore, f can not be
uniformly continuous on (0; 1).
100
Math 4023
Lecture 39
19. Uniform Continuity, Cont'd
Denition. We say that a function f~ is an extension of a function f if
(i) dom(f ) dom(f~)
(ii) f~(x) = f (x) for all x 2 dom(f ).
Theorem 19.5. A real-valued function f on an open interval (a; b)
is uniformly continuous on (a; b) if and only if it can be extended to
a continuous function f~ on [a,b].
Proof.
(=
Suppose f can be extended to a function f~ that is continuous on [a; b]. Then, by
Theorem 19.2, f~ is uniformly continuous on [a; b]. Hence,
f = f~ a;b
(
)
is uniformly continuous on (a; b).
=)
Now suppose f is uniformly continuous on (a; b). In order to construct a continuous
extension f~ of f to [a; b], we need to dene numbers f~(a) and f~(b) such that if
(sn) and (tn) are sequences in (a; b) such that lim (sn ) = a and lim (tn) = b, then
lim (f (sn )) = f~(a)
and
lim (f (tn )) = f~(b) :
However, we must rst show
(i) lim (f (sn )) and lim (f (tn )) exist;
(ii) these limits do not depend on the choice of the sequences (sn )
and (tn) (so long as they, respectively, converge to a and b).
To prove (i), we note that by Theorem 19.4, the sequences (f (sn )) and (f (tn )) are
both Cauchy (because f is assumed to be uniformly convergent, and because the
101
fact that (sn ) and (tn) are converge implies they are Cauchy). Thus, both (f (sn ))
and (f (tn )) converge.
To prove (ii) we suppose (rn ) and (sn) are two sequences in (a; b), both converging
to a. Let (qn) be the sequence
(
qn = s
r2
if n is even
if n is odd ;
n
+1
2
n
so that
(qn ) = (s ; r ; s ; r ; s ; r ; : : : ) :
Its obvious that (qn ) also converges to a. Hence by (i),
1
1
2
2
3
3
(f (qn ))
converges; say, to y 2 R. But then, since (f (rn )) and (f (sn )) can be regarded as
subsequences of (f (qn )), they must also converge to y. Hence
lim (f (sn )) = lim (f (rn )) :
We can thus unambiguously dene f~(a) as the limit
lim (f (sn ))
where (sn) is any sequence in (a; b) with limit a. We similarly dene f~(b). 102
Math 4023
Lecture 40
20. Limits of Functions
Denition 20.1. Let S be a subset of R, let a be a real number or
symbol 1 that is the limit of some sequence in S and let L be a
real number or a symbol 1. We write
lim f (x) = L
x!a
S
if
(i) f is a function dened on S ;
(ii) for every sequence (sn) in S with limit a we have
lim (f (sn )) = L :
n!1
Denition . Let a 2 R. By a deleted neighborhood of a we a
shall mean an open interval (a , ; a + ) about a with the point a
removed; i.e., a set of the form
fx 2 (a , ; a + j x 6= a)g :
Denition 20.3. Let a 2 R and f be a real-valued function dened
on some domain D R.
(a) We write
lim f (x) = L
x!a
provided that there exists some deleted neighborhood S of a, such
that if (xn ) is any sequence in S with limit a, then
lim f (xn ) = L :
n!1
(b) We write
lim f (x) = L
x!a+
103
provided there exists an open interval S = (a; a + ), such that if
(xn) is any sequence in S with limit a, then
nlim
!1 f (xn ) = L
(c) We write
:
lim f (x) = L
x!a,
provided there exists an open interval S = (a , ; a), such that if
(xn) is any sequence in S with limit a, then
lim f (xn ) = L :
n!1
(d) We write
lim f (x) = L
x! 1
+
provided there exists an open interval S = (M; +1), such that if
(xn) is any sequence in S with limit +1, then
nlim
!1 f (xn ) = L
(e) We write
:
lim f (x) = L
x!,1
provided there exists an open interval S = (,1; M ), such that
if (xn ) is any sequence in S with limit ,1, then
lim f (xn ) = L :
n!1
Note that a function f is continuous at a point a 2 dom(f ) if and only if
lim f (x) = lim+ f (x) = lim, f (x) = f (a) :
x!a
x!a
x!a
Theorem 20.4. Let f and g be functions for which the limits L =
limx!a f (x) and L = limx!a g(x) exist are nite. Then
s
2
s
(i) limx!a (f + g) exists and equals L + L ;
(ii) limx!a (fg) exists and equals L L
s
s
1
1
2
2
1
104
(iii) limx!a (f=g) exists and equals L =L , provided L 6= 0 and
g 6= 0 for x 2 S .
s
Proof. Suppose
1
2
2
lim (f (x)) = L
lim (g(x)) = L
x!a
x!a
1
2
Then there exist deleted neighborhoods S and T of a such that if if (sn ) is any
sequence in S converging to a and (tn) is any sequence in T converging to a, we
have
lim (f (sn )) = L
n!1
lim (f (tn )) = L :
n!1
1
2
Set
U =S\T :
Then for any sequence (un) in U we have
lim (f (un) + g(un)) = nlim
n!1
!1 (f (un )) + lim (g (un )) = L + L :
1
2
The proofs of the remaining statements are similar.
Theorem 20.5. Let f be a function for which the limit
L = xlim
f (x)
!a
S
exists and is nite. If g is a function dened on
T = ff (x) j x 2 S g [ fLg
that is continuous at L, then
lim g f (x)
x!a
S
exists and equals g(L).
Proof. Note that g f is dened on S by our assumptions. Consider a sequence
(xn) in S with limit a. Then
lim f (xn )
n!1
is a sequence in T with limit L. Since g is continuous at L it follows that
lim (g f (xn )) = nlim
n!1
!1 (g (f (xn ))) = g nlim
!1 (f (xn )) = g (L) :
105
Note how similar these proofs of properties of limits are to the proofs in Section
17. Indeed, a closer inspection reveals that all we have done is let the number L
representing the limit of the values of the function near the point x = a play the
role of the value f (a) of the function at a.
Thus, it might seem that having two separate sections, one to deal with continuity,
the other to deal with limits is almost a tautology. However, it really underscores
a key point. The value of a function does not have to be dened at a point a in
order to compute its limit.
For example, the function
f (x) = sin(x x)
is not continuous at x = 0 since it is not even dened at that point. However,
xlim
!a f (x) = 1
:
We can therefore dene an extension f~ of f
8
<
sin(x) ; if x 6= 0
~
f (x) = : x
1 ; if x = 0
which is continuous on the whole real line.
If this still seems like a mote point, recall that functions of the form
g(xo) ;
f (x) = g(xx) ,
,x
o
x 6= xo
commonly arise in the denition of derivatives and that it must be possible to
extend f to a continuous function at xo if the derivative of g is to exist at xo .
Theorem 20.6. Let f be a function dened on a subset S of R, let
a be a real number that is the limit of some sequence in S , and let
L be a real number. Then
lim f (x) = L
x!a
S
106
if and only if
for each > 0, there exists a > 0 such that x 2 S and
jx , aj < imply jf (x) , Lj < .
Proof. To prove this theorem one simple substitutes L for f (a) in the proof of
Theorem 17.2.
Theorem 20.10. Let f be a function dened on some deleted neigh-
borhood of a. Then limx!a f (x) exists if and only if the limits
limx!a, f (x) and limx!a+ f (x) both exist and are equal, in which
case all three limits are equal.
Homework: 19.1, 19.4(a), 20.1, 20.14, 20.16, 20.17
107
Math 4023
Lecture 41
23. Power Series
From any sequence (an ) of real numbers one has an associated power series
1
X
(1)
n
an xn :
=0
As it stands, the expression (1) is a purely formal object. However, by plugging
specic values for x, say x = xo 2 R, we can obtain a numerical series
1
X
(2)
n
an (xo)n
=0
which may or may not converge. If the series (2) does converge, it has a unique
limit and so we can dene a function
f (xo ) = Nlim
!1
(3)
N
X
n
an (xo )n
=0
whose domain consists of all xo 2 R such that (2) converges. It turns out that the
domain of such a function is either the single point x = 0, or an interval containing
0; it is never a disconnected subset of the real line.
Theorem 23.1. Let P an xn be a power series, and let
8
8
1
>
0
>
>
>
>
<
<
= lim sup janj =n = > s 2 R ; R = > 1 2 R :
s
>
>
>
:1
:
0
1
Then the power series converges for jxj < R and diverges for jxj > R.
P
Proof. Fix x 2 P
R and regard
anxn as a numerical series. By the root test
(Theorem 14.9), an xn converges absolutely if
lim sup janxn j =n < 1
1
and it diverges if
lim inf janxn j =n > 1 :
1
108
But if
1 > lim sup janxn j =n = jxj lim sup janj =n = jxj 1
then
Also
1
jxj < 1 = R :
1 < lim inf jan xnj =n lim sup jan xnj =n = jxj 1
1
and so
jxj > R :
The statements thus follow. Recall that it is often easier to calculate
an+1 lim a n
()
than
lim sup janj =n ;
and that, due to Theorem 12.2, if the limit of the ratio of successive coecients exists it necessarily coincides with lim sup jan j =n . Therefore, whenever () converges
we have
an 1
R = a +1 = lim a :
n
lim a 1
1
n
+1
n
Example 1. Calculate the radius of convergence of
1
X
n
=0
xn
2nn
2
We have
lim
Therefore,
+1 (n+1)2
2n
1
n2
1
2n
= 21
!
n
= lim 2(n + 2n + 1) = 21 :
2
2
)
R= 1 =2 :
1
2
109
Example 2. Calculate the radius of convergence of
1
X
n
=0
We have
lim
1
n
n
(
+1)!
1
!
Therefore,
!
xn
n!
n
!
= lim (n + 1)! = lim n +1 1 = 0 :
)
=0
R=1 :
In the examples above we have dealt with power series centered about the origin.
One can also consider generalized power series , which are formal sums of the form
1
X
n
an (x , xo )n :
=0
Convergence of these series can be tested in the same manner as above, but now
R=
1
lim sup jan j =n
1
is interpreted as the greatest number for which
jx , xo j < R
)
1
X
n
=0
an (x , xo)n converges:
110
Math 4023
Lecture 42
24. Uniform Convergence
Suppose
1
X
(1)
n
an xn
=0
is a power series with radius covergence R. Then for each x 2 (,R; R), we have a
well dened number
N
X
lim
anxn
N !1
n
=0
and so (1) denes a real-valued function with domain (,R; R). We now intend to
study the properties of such functions. The rst question we would like to address
is this: are such functions continuous? To answer this question we must make a
brief digression.
Denition 24.1. Let (fn ) be a sequence of real-valued functions
dened on a set S R. The sequence (fn ) converges pointwise to a
function f if
lim f (x) = f (x)
n!1 n
8x2X :
The condition of pointwise convergence can be rephrased as follows:
for each > 0 and x 2 S , there exists an N;x 2 N such
that
jfn (x) , f (x)j < 8 n > N;x :
I have appended the subscripts and x to N to emphasize the fact that the chosen
N may depend on the choice of and the choice of x. To isolate the situation for
which N can be chosen independently of x, we have the following denition.
Denition 24.2. Let (fn ) be a sequence of real-valued functions
dened on a set S R. The sequence (fn) is said to converge
uniformly to a function f dened on S if
for each > 0 there exists a number N such that
jfn (x) , f (x)j < 8 x 2 S and 8 n > N
:
111
Example 1.
Let
for x 2 (,1; 1), and let
fn(x) = (1 , jxj)n
0 if x 6= 0
f (x) = 1 if x = 0
Because
(2)
0 if 0 < a < 1
1 if a = 0
it is clear that (fn) converges pointwise to f . However, this sequence of functions
does not converge uniformly to f . To see this let us try to nd an N such that
jfn (x) , f (x)j < 12 8 x 2 (,1; 1) ; 8 n > N :
In particular, we must have
j(1 , x)n , 0j < 12 8 x 2 (0; 1) ; 8 n > N :
But for any xed N we can always
nd an x 2 (0; 1) so that the above inequality
1
,
+1
is false; e.g., take x = 1 , 2
and n = N + 1, then
j(1 , x)n j = 2, +1 = 21
and so (2) is false. We conclude that the sequence (fn ) does not converge uniformly
to f
lim an =
n!1
N
n
N
The following theorem justies our interest in uniform convergence.
Theorem 24.3. The uniform limit of a sequence of continuous functions is continuous. More precisely, let (fn ) be a sequence of continuous functions dened on a set S , and suppose that
lim (fn ) = f uniformly on S :
Then f is continuous.
Proof. Let > 0. Since f is the uniform limit of (fn ), there exists an N 2 N, such
that
n>N
) jfn (x) , f (x)j < 3 ; 8 x 2 S :
112
In particular,
jfN (x) , f (x)j < 3 ; 8 x 2 S :
+1
Since fN
+1
is continuous at xo there is a > 0 such that
x 2 S and jx , xoj < jfN (x) , fN (xo )j < 3 :
)
+1
+1
Now it follows from the triangle inequality that
jf (x) , f (xo )j jf (x) , fN (x)j + jfN (x) , fN (xo )j + fN (xo ) , f (xo 3 + 3 + 3 = +1
+1
+1
+1
if x 2 S and jx , xo j < . This proves that f is continuous at xo.
)
113
Math 4023
Lecture 43
24. Uniform Convergence, Cont'd
The following theorem justies our interest in uniform convergence.
Theorem 24.3. The uniform limit of a sequence of continuous functions is continuous. More precisely, let (fn ) be a sequence of continuous functions dened on a set S , and suppose that
lim (fn ) = f uniformly on S :
Then f is continuous.
Proof. Let > 0. Since f is the uniform limit of (fn ), there exists an N 2 N, such
that
n>N
) jfn (x) , f (x)j < 3 ; 8 x 2 S :
In particular,
jfN (x) , f (x)j < 3 ; 8 x 2 S :
+1
Since fN
+1
is continuous at xo there is a > 0 such that
x 2 S and jx , xoj < jfN (x) , fN (xo )j < 3 :
)
+1
+1
Now it follows from the triangle inequality that
jf (x) , f (xo )j jf (x) , fN (x)j + jfN (x) , fN (xo )j + fN (xo ) , f (xo 3 + 3 + 3 = +1
+1
+1
+1
)
if x 2 S and jx , xo j < . This proves that f is continuous at xo. This theorem is extremely important to the application of power series. For example, formal power series arise naturally as formal solutions to dierential equations.
Although it is tempting at this point to say that, since the partial sums of a power
series are all continuous, that power series are continuous. This conclusion is a bit
premature, however. One problem to consider is that the sequence of partial sums
is continuous on all of R, but that the power series is guaranteed to converge only
within its radius of convergence.
114
To better understand this problem, we will use a technique we developed in our
study of sequences of real numbers.
Recall the notion of a Cauchy sequence: (sn ) is Cauchy if for every > 0 there
exists an N such that
)
n; m > N
jsn , sm j < :
The importance of this notion lied in the fact that it allowed one to test for
convergence without explicit knowledge of the limit; a sequence (sn ) converges if
and only if it is Cauchy.
Denition 25.3. A sequence (fn) of functions dened on a set
S R is uniformly Cauchy on S if
for each > 0, there exists a number N such that
jfn (x) , fm (x)j < ;
8 x 2 S ; 8 n; m > N :
Proposition . Let (fn) be a sequence of functions that converges
uniformly to a function f on a subset S R. Then (fn) is uniformly
Cauchy.
Proof. Suppose (fn) converges uniformly to f and let > 0. We then have an N
such that
jfn(x) , f (x)j < 2
;
8x2S ; 8n>N :
And so if n; m > N and x 2 S
jfn (x) , fm (x)j = jfn(x) , f (x) + f (x) , fm (x)j
jfn(x) , f (x)j + jfm (x) , f (x)j
< 2 + 2
=
Thus, (fn) is uniformly Cauchy. 115
Math 4023
Lecture 44
Recall
Denition 25.3. A sequence (fn) of functions dened on a set S R is
uniformly Cauchy on S if
for each > 0, there exists a number N such that
jfn (x) , fm (x)j < ; 8 x 2 S ; 8 n; m > N :
Proposition . Let (fn ) be a sequence of functions that converges
uniformly to a function f on a subset S R. Then (fn) is uniformly
Cauchy.
Proof. Suppose (fn) converges uniformly to f and let > 0. We then have an N
such that
jfn(x) , f (x)j < 2
;
8x2S ; 8n>N :
And so if n; m > N and x 2 S
jfn (x) , fm (x)j = jfn(x) , f (x) + f (x) , fm (x)j
jfn(x) , f (x)j + jfm (x) , f (x)j
< 2 + 2
=
Thus, (fn) is uniformly Cauchy. On the other hand, as I now intend to show, a sequence of functions that is
uniformly Cauchy sequence always converges uniformly to some function f . First,
I prove a lemma.
Lemma . Let (fn) be a sequence of functions dened on and uniformly Cauchy on a set S R. Then for every xo 2 S , the sequence
(fn(xo )) is Cauchy.
Proof. Choose > 0. Since (fn) is uniformly Cauchy, we have
jfn (x) , fm (x)j < ;
8 x 2 S ; 8 n; m > N :
116
In particular, for x = xo we have
jfn(xo ) , fm (xo )j < Thus, (f (xo )) is Cauchy. ; 8 n; m > N :
Theorem 25.6. Let (fn) be a sequence of functions dened on and
uniformly Cauchy on a set S R. Then there exists a function f on
S such that (fn ) converges uniformly to f on S .
Proof. First we have to construct f . By the preceding lemma, for any x 2 S , the
sequence (fn (x)) is Cauchy, hence convergent. We may thus dene f (x), x 2 S ,
by
f (x) = lim (fn (x)) :
By construction (fn (x)) converges pointwise to f . Our next task is to show that
(fn) actually converges uniformly to f .
(1)
Let > 0. Since (fn) is uniformly Cauchy, there is a number N such that
jfn(x) , fm (x)j < 2 ; 8 x 2 S ; 8 n; m > N :
This then implies
fn(x) 2 fm (x) , 2 ; fm (x) + 2
; 8n>N ;
hence
h
; f (x) + i :
f (x) = nlim
(
f
(
x
))
2
f
(
x
)
,
n
m
!1
2 m
2
And so
jf (x) , fm (x)j 2 ; 8 m > N
or
jf (x) , fm (x)j < ; 8 m > N :
This shows that (fm ) converges to f uniformly. This theorem is especially useful for handling series of functions , that is, expressions of the form
1
X
gn(x) :
n
=0
Such expressions make sense only if the sequence of partial sums converges pointwise on some subset S R. If the sequence of partial sums
fN (x) =
N
X
n
=0
!
gn(x)
117
actually converges uniformly on a set S , then we say the series
convergent on S .
P
gn is uniformly
Theorem 25.4. Let P1k gk be a series of functions such that each
gk is continuous on a subset S R and that the series
=0
n
X
k
gk
!
=0
converges uniformly on S . Then the series
continuous function.
P1
k
=0
gk represents a
Proof. Each partial sum
fn (x) =
n
X
k
gk (x)
=0
is a continuous, and by hypothesis the series (fn ) is uniformly convergent. By
Theorem 24.3 then, the limit function is continuous.
Theorem 25.7. (The Weierstrass M -test). Let (Mk ) be a sequence
of non-negative real numbers such that
1
X
k
Mk < 1 :
=0
If
jgk (x)j < Mk
for all k 2 N and all x 2 S , then
1
X
k
gk
=0
converges uniformly on S .
Proof. In view of Theorem 25.4 it suces to show that the sequence of partial
sums isPuniformly Cauchy on S . To verify the Cauchy criterion on S , let > 0.
Since Mk converges, it satises the Cauchy criterion. So there exists a number
N such that
nm>N
)
n
X
k m
=
Mk <
:
118
Hence, if n m > N , and x 2 S , we have
n
X
k m
=
gk (x)
n
X
k m
jgk (x)j =
n
X
m k
Mk < :
=
So the sequence of partial sums is uniformly Cauchy on S , hence the series converges uniformly on S .
Theorem 26.1. Let P1n anxn be a power series with radius of
convergence R > 0 (possibly R = 1). If 0 < r < R, then the power
series converges uniformly to a continuous function on [,r; r].
=0
Proof . Consider 0 < r < R. Because the radius of convergence of
1
X
k
is dened as
anxn
=0
1
lim sup jan j =n
P
itPis clear that the series jan jxn has the same radius of convergence as the series
an xn. Since r < R we have
R=
1
X
n
1
jan j rn < 1 :
=0
We also have
ak xk and so the series
jak j rk = Mk
1
X
k
;
x 2 [,r; r]
ak xk
=0
converges uniformly on [,r; r] by the Weierstrass M -test. The limit function is
continuous at each point of [,r; r] by Theorem 25.5.