Download Chapter 2. Real Numbers §1. Rational Numbers A commutative ring

Chapter 2. Real Numbers
§1. Rational Numbers
A commutative ring is called a field if its nonzero elements form a group under
multiplication. Let (F, +, ·) be a filed with 0 as its additive identity element and 1 as its
multiplicative identity element. We always assume that 1 6= 0. For b ∈ F , its additive
inverse is denoted by −b. The subtraction in F is defined as follows:
a − b := a + (−b),
a, b ∈ F.
For b ∈ F \ {0}, its multiplicative inverse is denoted by b−1 . The division in F is defined
as follows:
a/b := a ÷ b := ab−1 , a ∈ F, b ∈ F \ {0}.
For n ∈ IN0 and a ∈ F , na is defined recursively: na := 0 for n = 0 and (n + 1)a := na + a.
Recall that (ZZ, +, ·) is a commutative ring with 1 as its (multiplicative) identity element. But any integer m ∈ ZZ \ {0, 1, −1} has no multiplicative inverse in ZZ. Thus, it is
desirable to extend the ring (ZZ, +, ·) to a field.
Let us introduce a relation in the set ZZ × IN. For two elements (m, n) and (p, q)
in ZZ × IN, we define (m, n) ∼ (p, q) if mq = np. We claim that it is an equivalence
relation. Clearly, this relation is reflexive and symmetric. Suppose (m, n) ∼ (p, q) and
(p, q) ∼ (r, s). Then mq = np and ps = rq. Hence, mqs = nps = nrq. Since q 6= 0, it
follows that ms = nr, that is, (m, n) ∼ (r, s). This shows that the relation is transitive.
For (m, n) ∈ ZZ × IN, we use m
n or m/n to denote the equivalence class of (m, n):
m
:= {(r, s) ∈ ZZ × IN : rn = sm}.
n
Then m/n = m0 /n0 if and only if mn0 = nm0 . Let Q
Q be the set of equivalence classes. An
element of Q
Q is called a rational number.
The addition in Q
Q is defined by the rule
m p
mq + np
+ :=
.
n
q
nq
The addition is well defined. Indeed, if m/n = m0 /n0 and p/q = p0 /q 0 , then mn0 = m0 n
and pq 0 = p0 q. It follows that (mq + np)(n0 q 0 ) = (m0 q 0 + n0 p0 )(nq). It is easily seen that
the addition is associative and commutative. Moreover, 0/1 is the identity element for
addition. Each element m/n ∈ Q
Q has an addictive inverse:
−m m
m −m
0
+
=
+
= .
n
n
n
n
1
1
Thus, (Q
Q, +) is an abelian group.
The multiplication in Q
Q is defined by the rule
mp
mp
:=
.
nq
nq
If m/n = m0 /n0 and p/q = p0 /q 0 , then mn0 = m0 n and pq 0 = p0 q. It follows that
(mp)(n0 q 0 ) = (m0 p0 )(nq). So the multiplication is well defined. It is easily seen that the
multiplication is associative and commutative. Moreover, the multiplication is distributive
with respect to addition:
mp r m p m r
+
+
.
=
n q
s
nq
ns
Clearly, 1/1 is the identity element for multiplication. If m/n 6= 0/1 and p/q 6= 0/1, then
(m/n)(p/q) 6= 0/1. Furthermore, if m/n ∈ Q
Q and m > 0, then n/m is its multiplicative
inverse; if m/n ∈ Q
Q and m < 0, then (−n)/(−m) is its multiplicative inverse. We conclude
that (Q
Q, +, ·) is a field.
Let ϕ be the mapping from ZZ to Q
Q given by ϕ(m) = m/1, m ∈ ZZ. Then ϕ is injective.
We have ϕ(m + n) = ϕ(m) + ϕ(n) and ϕ(mn) = ϕ(m)ϕ(n) for all m, n ∈ ZZ. Thus, the
ring (ZZ, +, ·) is embedded into the field (Q
Q, +, ·) via the mapping ϕ. We identify ϕ(ZZ)
with ZZ and write m for m/1. In particular, 0 = 0/1 is the additive identity element, and
1 = 1/1 is the multiplicative identity element.
For two elements m/n and p/q in Q
Q, we write m/n ≤ p/q if mq ≤ np. Then ≤
is a linear ordering in Q
Q. Indeed, it is obvious that ≤ is reflexive and antisymmetric.
Suppose m/n ≤ p/q and p/q ≤ r/s, where q, s ∈ IN. Then mq ≤ np and ps ≤ qr; hence
mqs ≤ nps ≤ nqr. Since mqs ≤ nqr and q > 0, we obtain ms ≤ nr, i.e., m/n ≤ r/s.
This shows that ≤ is transitive. Moreover, either mq ≤ np or np ≤ mq. So ≤ is a linear
ordering. Furthermore, if a, b ∈ Q
Q and a ≤ b, then a + c ≤ b + c for all c ∈ Q
Q, and ac ≤ bc
for all c ≥ 0.
A field (F, +, ·) with a linear ordering ≤ is called an ordered field if the following
two conditions are satisfied for all x, y, z ∈ F :
(a) x ≤ y implies x + z ≤ y + z;
(b) if z ≥ 0, then x ≤ y implies xz ≤ yz.
Thus, (Q
Q, +, ·) with its natural ordering ≤ is an ordered field.
Let F be an ordered field. An element a of F is said to be positive if a > 0, and it is
said to be non-negative if a ≥ 0. Suppose x, y ∈ F . Then x < y implies x + z < y + z for
all z ∈ F and xz < yz for all z > 0. Moreover, if x 6= 0, then x2 = x · x > 0. Indeed, this is
true if x > 0. If x < 0, then 0 = (−x)+x < (−x)+ 0 = −x, and hence x2 = (−x)(−x) > 0.
In particular, 1 = 12 > 0.
2
An ordered field F is said to be archimedean if for every pair of positive elements
x and y in F , there is a positive integer k such that kx > y. The ordered field Q
Q is an
archimedean field. Indeed, if x and y are two positive elements in Q
Q, then x = m/n and
y = p/q for some m, n, p, q ∈ IN. There exists a positive integer k such that kmq > np.
Consequently, kx = km/n > p/q = y.
Although the rational numbers form a rich algebraic system, they are inadequate for
the purpose of analysis because they are, in a sense, incomplete.
For example, there is no rational number r such that r2 = 2. In order to prove
this statement, it suffices to show that m2 6= 2n2 for any pair of positive integers m
and n. Write m = 2j p and n = 2k q, where j, k ∈ IN0 and p and q are odd numbers.
Then we have m2 = 22j p2 and 2n2 = 22k+1 q 2 . Note that p2 and q 2 are odd numbers.
Moreover, 2j 6= 2k + 1, since one is an even number and the other is an odd number.
There are two possibilities: 2j > 2k + 1 or 2j < 2k + 1. If 2j > 2k + 1, then 22j−2k−1 p2
is an even number and q 2 is an odd number. Hence, 22j−2k−1 p2 6= q 2 . It follows that
m2 = 22j p2 6= 22k+1 q 2 = 2n2 . If 2j < 2k + 1, then p2 6= 22k+1−2j q 2 , and thereby
m2 6= 2n2 .
Let A be the set {r ∈ Q
Q : r2 ≤ 2}. Then A has no least upper bound in Q
Q.
§2. Real Numbers
The absolute value of a rational number r is defined by
n
r
if r ≥ 0,
|r| :=
−r if r < 0.
For r, s ∈ Q
Q, we have |rs| = |r||s| and |r + s| ≤ |r| + |s|.
A sequence (rn )n=1,2,... of rational numbers is said to converge to a rational number r
if, for every rational number ε > 0, there exists a positive integer N (ε) such that |rn −r| < ε
for all n ≥ N (ε). In this case, we say that r is the limit of the sequence and write
lim rn = r.
n→∞
If a sequence of rational numbers converges, then it has a unique limit. Also, a
convergent sequence is bounded.
A sequence (rn )n=1,2,... of rational numbers is called fundamental if, for every rational number ε > 0, there exists a positive integer N (ε) such that |rm − rn | < ε whenever
m, n ≥ N (ε).
A sequence of rational numbers that converges (to a rational number) is fundamental.
Also, a fundamental sequence of rational numbers is bounded.
3
Two fundamental sequences (rn )n=1,2... and (sn )n=1,2... are said to be equivalent if
limn→∞ (rn − sn ) = 0. We write (rn )n=1,2... ∼ (sn )n=1,2... when (rn )n=1,2... and (sn )n=1,2...
are equivalent. It is easy to verify that this is indeed an equivalence relation. An equivalence class of a fixed fundamental sequence of rational numbers is called a real number.
Thus, a real number is represented by a fundamental sequence (rn )n=1,2... of rational numbers. Such a representation is not unique. If (sn )n=1,2... is another representation for the
same real number, then we must have limn→∞ (rn − sn ) = 0.
Let IR denote the set of real numbers. We define addition and multiplication on
IR as follows. Suppose x and y are real numbers represented by fundamental sequences
(rn )n=1,2,... and (sn )n=1,2,... of rational numbers, respectively. Then x + y is defined to
be the real number represented by (rn + sn )n=1,2,... and xy is the real number represented by (rn sn )n=1,2,... . If (rn )n=1,2... ∼ (rn0 )n=1,2... and (sn )n=1,2... ∼ (s0n )n=1,2... , then
limn→∞ (rn − rn0 ) = 0 and limn→∞ (sn − s0n ) = 0. It follows that
lim (rn + sn ) − (rn0 + s0n ) = 0 and
n→∞
lim (rn sn − rn0 s0n ) = 0.
n→∞
Therefore, the addition and multiplication are well defined. Evidently, the addition and
multiplication are associative and commutative.
For r ∈ Q
Q, let ϕ(r) be the real number represented by the sequence (rn )n=1,2... , where
rn = r for all n ∈ IN. Then we have ϕ(r + s) = ϕ(r) + ϕ(s) and ϕ(rs) = ϕ(r)ϕ(s) for
r, s ∈ Q
Q. Clearly, ϕ is an injective mapping from Q
Q to IR. We identify ϕ(r) with r. Thus,
Q
Q is embedded into IR via the mapping ϕ.
It is easily seen that 0 is the identity for addition. Suppose x is a real number
represented by a fundamental sequence (rn )n=1,2,... of rational numbers. Let −x be the
real number represented by (−rn )n=1,2,... . Then (−x) + x = 0. This shows that (IR, +) is
an abelian group.
For three real numbers x, y, z ∈ IR, we have (x + y)z = xz + yz. In other words, the
multiplication is distributive with respect to addition. Thus, (IR, +, ·) is a commutative
ring with 1 as its (multiplicative) identity. Suppose x ∈ IR \ {0} and x is represented
by a fundamental sequence (rn )n=1,2,... of rational numbers. Then there exists a rational
number ε > 0 and a positive integer N such that |rn | ≥ ε for all n ≥ N . Indeed, if this is
not true, then there exists a sequence (nk )k=1,2,... of positive integers such that nk < nk+1
and |rnk | < 1/k for all k ∈ IN. It follows that limk→∞ rnk = 0 and thereby x = 0. Thus,
|rn | ≥ ε for all n ≥ N . We define x−1 to be the real number represented by the sequence
(sn )n=1,2,... , where sn = 1 for n < N and sn = rn−1 for n ≥ N . Clearly, (sn )n=1,2,... is a
fundamental sequence. We have x−1 x = 1. This shows that (IR, +, ·) is a field.
4
A sequence (rn )n=1,2,... of rational numbers is said to be eventually non-negative (nonpositive) if there exists a positive integer N such that xn ≥ 0 (xn ≤ 0) for all n ≥ N .
A real number z is said to be non-negative and written z ≥ 0, if z is represented by a
fundamental sequence of rational numbers that is eventually non-negative. For two real
numbers x and y, we write x ≤ y if y − x ≥ 0. The relation ≤ is a linear ordering.
Indeed, this relation is reflexive, since x ≤ x for all x ∈ IR. Let x, y, z ∈ IR. If x ≤ y
and y ≤ z, then y − x ≥ 0 and z − y ≥ 0. It follows that z − x = (z − y) + (y − x) ≥ 0.
Hence, ≤ is transitive. Moreover, if x ≤ y and y ≤ x, then both y − x and x − y are
non-negative. Consequently, y − x and x − y are represented by fundamental sequences
(rn )n=1,2... and (sn )n=1,2,... of rational numbers respectively such that both sequences are
eventually non-negative. In particular, (rn )n=1,2... and (−sn )n=1,2,... are equivalent. It
follows that limn→∞ (rn + sn ) = limn→∞ (rn − (−sn )) = 0. There exists a positive integer
N such that 0 ≤ rn ≤ rn + sn for all n ≥ N . Hence, limn→∞ rn = 0. This shows x − y = 0,
i.e., x = y. Thus, the relation ≤ is antisymmetric. It remains to prove that any two real
numbers x and y are comparable. Suppose that y − x is represented by a fundamental
sequence (rn )n=1,2... of rational numbers. If the sequence is eventually non-negative, then
x ≤ y; if the sequence is eventually non-positive, then y ≤ x. If the sequence is neither
eventually non-negative nor eventually non-positive, then both the sets {n ∈ IN : rn ≥ 0}
and {n ∈ IN : rn ≤ 0} are infinite. Since (rn )n=1,2... is fundamental, for every rational
number ε > 0, there exists a positive integer N such that rm − ε < rn < rm + ε for all
m, n > N . But there exist m1 , m2 > N such that rm1 ≥ 0 and rm2 ≤ 0. Therefore,
−ε ≤ rm1 − ε < rn < rm2 + ε ≤ ε. Thus, −ε < rn < ε for all n > N . It follows that
y − x = 0 and x = y. We have proved that the relation ≤ is a linear ordering.
Theorem 2.1. The field (IR, +, ·) with the ordering ≤ is an Archimedean ordered field.
Proof. Let x, y, z ∈ IR. If x ≤ y, then (y + z) − (x + z) = y − x ≥ 0; hence x + z ≤ y + z.
If, in addition, z ≥ 0, then yz − xz = (y − x)z ≥ 0. Hence, xz ≤ yz for all z ≥ 0. This
shows that (IR, +, ·) with the ordering ≤ is an ordered field.
Suppose x > 0 and y > 0. Then u := x/y > 0. The real number u is represented by a
fundamental sequence (rn )n=1,2,... of rational numbers. By what has been proved before,
there exists a rational number ε > 0 and a positive integer N such that |rn | ≥ ε for all
n ≥ N . Since u > 0, the set {n ∈ IN : rn < 0} is finite. Hence, the sequence (rn − ε)n=1,2,...
is eventually non-negative. This shows u ≥ ε. For the rational number ε > 0, there exists
a positive integer k such that kε > 1. Consequently, ku ≥ kε > 1. It follows that kx > y.
Therefore, (IR, +, ·) with the ordering ≤ is an Archimedean ordered field.
As a corollary of the above theorem, there exists a rational number between any
5
two distinct real numbers. Let x and y be two real numbers such that x < y. By the
Archimedean property, there exists a positive integer k such that k(y − x) > 2. It follows
that y−x > 2/k. Let m be the least integer such that m ≥ kx. Then x ≤ m/k < (m+1)/k.
On the other hand, (m − 1)/k < x implies (m + 1)/k < x + 2/k < y. Thus, r := (m + 1)/k
is a rational number such that x < r < y.
Theorem 2.2. Every nonempty subset of IR bounded above has the least upper bound.
Proof. Suppose that X is a nonempty subset of IR bounded above. Let
B := {b ∈ Q
Q : b ≥ x for all x ∈ X}
and A := Q
Q \ B.
Then both A and B are nonempty. Moreover, a < b whenever a ∈ A and b ∈ B. We
define two sequences (an )n=1,2,... and (bn )n=1,2,... recursively as follows. Choose a1 ∈ A
and b1 ∈ B. Suppose n ∈ IN. If (an + bn )/2 ∈ A, set an+1 := (an + bn )/2 and bn+1 := bn ;
if (an + bn )/2 ∈ B, set an+1 := an and bn+1 := (an + bn )/2. Then an ∈ A, bn ∈ B, and
bn+1 − an+1 = (bn − an )/2. By induction on n, we obtain bn − an = (b1 − a1 )/2n−1 for all
n ∈ IN. Consequently, limn→∞ (bn − an ) = 0. Moreover, we have |an+1 − an | ≤ (bn − an )/2.
It follows that |an+1 − an | ≤ (b1 − a1 )/2n . Hence, for n, k ∈ IN we have
|an+k − an | ≤
k−1
X
|an+j+1 − an+j | ≤
j=0
k−1
X
j=0
b1 − a1
b1 − a1
< n−1 .
n+j
2
2
For every rational number ε > 0, there exists some N ∈ IN such that (b1 − a1 )/2N −1 < ε.
Hence, for n > N and k ≥ 1 we obtain |an+k − an | < ε. This shows that (an )n=1,2,... is a
fundamental sequence. For the same reason, (bn )n=1,2,... is also a fundamental sequence.
These two sequences are equivalent. Let u be the (unique) real number represented by
these two sequences of rational numbers. We claim that u is the least upper bound of X.
To justify our claim, it suffices to prove the following two statements: (1) u ≥ x for all
x ∈ X; (2) if y is an upper bound for X, then u ≤ y.
Suppose that (1) is not true. Then u < x for some x ∈ X. There exists a rational
number r such that u < r < x. We have r ≤ bn for all n ∈ IN. Hence, (bn − r)n=1,2,... is
a non-negative sequence of rational numbers. Consequently, u − r ≥ 0, i.e., u ≥ r. This
contradiction verifies statement (1).
Suppose that (2) is not true. Then u > y for some upper bound y for X. There
exists a rational number s such that y < s < u. We have an ≤ s for all n ∈ IN. Hence,
(s − an )n=1,2,... is a non-negative sequence of rational numbers. Consequently, s − u ≥ 0,
i.e., u ≤ s. This contradiction verifies statement (2).
In light of (1) and (2), we conclude that u is the least upper bound of X.
6
For a pair of real numbers a and b, we define
(a, b) := {x ∈ IR : a < x < b},
[a, b] := {x ∈ IR : a ≤ x ≤ b},
[a, b) := {x ∈ IR : a ≤ x < b},
(a, b] := {x ∈ IR : a < x ≤ b}.
The set (a, b) is clled an open interval, the set [a, b] is called a closed interval, and the
sets [a, b) and (a, b] are called half-open (or half-closed) intervals.
We introduce two symbols ∞ and −∞. The ordering ≤ in IR can be extended to
IR := IR ∪ {−∞, ∞} by defining
−∞ < a < ∞ for all a ∈ IR.
Then we have (−∞, ∞) = IR and
(a, ∞) = {x ∈ IR : x > a},
[a, ∞] = {x ∈ IR : x ≥ a},
(−∞, b) = {x ∈ IR : x < b},
(−∞, b] = {x ∈ IR : x ≤ b}.
§3. Limits of Sequences
The absolute value of a real number x is defined by
|x| :=
n
x
if x ≥ 0,
−x if x < 0.
For x, y ∈ IR, we have |xy| = |x||y| and |x + y| ≤ |x| + |y|.
A sequence (xn )n=1,2,... of real numbers is said to converge to a real number x if for
every real number ε > 0 there exists a positive integer N (ε) such that
|xn − x| < ε for all n > N (ε).
The real number x is called the limit of the sequence and we write
lim xn = x.
n→∞
If a sequence does not converge to a real number, it is said to diverge.
A sequence (xn )n=1,2,... of real numbers is said to diverge to ∞, and we write
limn→∞ xn = ∞ provided that for every M ∈ IR there exists a positive integer N such
that xn > M for all n > N . Similarly, A sequence (xn )n=1,2,... of real numbers is said to
diverge to −∞, and we write limn→∞ xn = −∞ provided that for every M ∈ IR there
exists a positive integer N such that xn < M for all n > N .
7
If a sequence converges, its limit is unique. Moreover, a convergent sequence is
bounded.
Suppose that (xn )n=1,2,... and (yn )n=1,2,... are convergent sequences of real numbers
with limn→∞ xn = x and limn→∞ yn = x. Then
(1) limn→∞ (xn + yn ) = x + y;
(2) limn→∞ (axn ) = ax for all a ∈ IR;
(3) limn→∞ (xn yn ) = xy;
(4) limn→∞ (xn /yn ) = x/y, provided yn 6= 0 for all n ∈ IN and y 6= 0;
(5) If xn ≤ yn for all n ∈ IN, then x ≤ y;
(6) If x = y and xn ≤ zn ≤ yn for all n ∈ IN, then limn→∞ zn = x.
A sequence (xn )n=1,2,... of real numbers is said to be increasing if xn ≤ xn+1 for all
n ∈ IN, and decreasing if xn+1 ≤ xn for all n ∈ IN. A monotone sequence is either
increasing or decreasing.
Theorem 3.1. Every monotone bounded sequence of real numbers is convergent.
Proof. Let (xn )n=1,2,... be a bounded and increasing sequence of real numbers. By Theorem 2.2, the set {xn : n ∈ IN} has a least upper bound. Let x := sup{xn : n ∈ IN}. We
claim that limn→∞ xn = x.
For ε > 0, x − ε is not an upper bound for the set {xn : n ∈ IN}, because x is the
least upper bound. Hence, there exists a positive integer N such that x − ε < xN . Since
the sequence (xn )n=1,2,... is increasing, we have
|xn − x| = x − xn ≤ x − xN < ε for all n > N.
This shows limn→∞ xn = x.
The proof for the decreasing case is similar.
Let (xn )n=1,2,... be a bounded sequence of real numbers. For k = 1, 2, . . ., let
sk := sup{xn : n ≥ k} and tk := inf{xn : n ≥ k}.
Then (sk )k=1,2,... is a decreasing sequence and (tk )k=1,2,... is an increasing sequence. We
define
lim sup xn := lim sk and lim inf xn := lim tk .
n→∞
n→∞
k→∞
Clearly, sk ≥ tk for all k ∈ IN; hence,
lim sup xn ≥ lim inf xn .
n→∞
n→∞
8
k→∞
For example, the sequence ((−1)n )n=1,2,... is bounded. For k = 1, 2, . . ., we have
sk = sup{(−1)n : n ≥ k} = 1 and tk = inf{(−1)n : n ≥ k} = −1.
Consequently,
lim sup (−1)n = 1
lim inf (−1)n = −1.
and
n→∞
n→∞
Theorem 3.2. Let (xn )n=1,2,... be a bounded sequence of real numbers. Then the sequence converges to a real number x if and only if
lim inf xn = lim sup xn = x.
n→∞
n→∞
Proof. For k = 1, 2, . . ., let tk := inf{xn : n ≥ k} and sk := sup{xn : n ≥ k}. Suppose
limn→∞ xn = x. Then for every real number ε > 0 there exists a positive integer N (ε)
such that |xn − x| < ε for n > N (ε), that is,
x − ε < xn < x + ε
for all n > N (ε).
It follows that
x − ε ≤ tk ≤ sk ≤ x + ε
for all k > N (ε).
Hence, limk→∞ tk = limk→∞ sk = x.
Conversely, suppose lim inf n→∞ xn = lim supn→∞ xn = x. Then we have
lim tk = lim sk = x.
k→∞
k→∞
Consequently, for every ε > 0, there exists a positive integer N (ε) such that
x − ε < tk ≤ sk < x + ε
for all k > N (ε).
It follows that
x − ε < xn < x + ε
for all n > N (ε),
because tk ≤ xn ≤ sk for n ≥ k. Therefore, limn→∞ xn = x.
A sequence (xn )n=1,2,... of real numbers is said to be a Cauchy sequence (or a
fundamental sequence) if for every ε > 0 there exists a positive integer N (ε) such that
|xn − xm | < ε
whenever m, n > N (ε).
Clearly, a Cauchy sequence is bounded. Moreover, a convergent sequence is a Cauchy
sequence.
9
Theorem 3.3. Every Cauchy sequence of real numbers converges.
Proof. Let (xn )n=1,2,... be a Cauchy sequence of real numbers. It is bounded. Let
sk := sup{xn : n ≥ k} and tk := inf{xn : n ≥ k},
k = 1, 2, . . . .
For ε > 0, there exists a positive integer N (ε) such that
|xn − xm | < ε
whenever m, n > N (ε).
Moreover, for k > N (ε), there exist m ≥ k and n ≥ k such that
sk < xm + ε
and tk > xn − ε.
Hence,
0 ≤ sk − tk < (xm − xn ) + 2ε < 3ε
for k > N (ε).
This shows that limk→∞ (sk − tk ) = 0. Consequently,
lim inf xn = lim tk = lim sk = lim sup xn .
n→∞
k→∞
k→∞
n→∞
By Theorem 3.2, the sequence (xn )n=1,2,... converges.
§4. Infinite Series
Given a sequence (an )n=1,2,... of real numbers, we define the sequence (sn )n=1,2,...
recursively by s1 := a1 and sn+1 := sn + an+1 , n ∈ IN. Then sn is called the nth partial
sum of the sequence (an )n=1,2,... and we write
sn =
n
X
ak = a1 + · · · + an ,
n ∈ IN.
k=1
P∞
We also refer to the sequence (sn )n=1,2,... of partial sums as the infinite series
n=1 an .
P∞
If (sn )n=1,2,... converges to a real number s, we say that the series n=1 an converges and
we write
∞
X
an = s.
n=1
P∞
The real number s is called the sum of the infinite series
n=1 an . If the sequence
P∞
(sn )n=1,2,... diverges, then we say that the series n=1 an diverges. If limn→∞ sn = ∞, we
10
P∞
P∞
say that the series n=1 an diverges to ∞ and write n=1 an = ∞. If limn→∞ sn = −∞,
P∞
P∞
we say that the series n=1 an diverges to −∞ and write n=1 an = −∞.
P∞
It is easily seen that an = sn − sn−1 for n ≥ 2. If the series
n=1 an converges, then
(sn )n=1,2,... converges to a real number s. It follows that
lim an = lim (sn − sn−1 ) = s − s = 0.
n→∞
n→∞
Thus, if a sequence (an )n=1,2,... diverges or limn→∞ an 6= 0, then the series
diverges.
If a, r ∈ IR and an = arn−1 for n ∈ IN, then the series
∞
X
an =
n=1
∞
X
P∞
n=1
an
arn−1
n=1
is called a geometric series. The case a = 0 is trivial. Suppose a 6= 0. If |r| ≥ 1, then
the sequence (arn−1 )n=1,2,... diverges or converges to a nonzero real number. Hence, the
P∞
geometric series n=1 arn−1 diverges for |r| ≥ 1. Suppose |r| < 1. Then
sn =
n
X
ark−1 = a(1 + r + · · · + rn−1 ) =
k=1
a(1 − rn )
,
1−r
n ∈ IN.
For |r| < 1 we have limn→∞ rn = 0. Consequently,
lim sn =
n→∞
Therefore, for |r| < 1, the geometric series
P∞
a
.
1−r
n=1
arn−1 converges and its sum is a/(1 − r).
Theorem 4.1. Let (an )n=1,2,... be a sequence of real numbers with an ≥ 0 for all n ∈ IN.
P∞
Then the series n=1 an converges if and only if the sequence (sn )n=1,2,... of partial sums
is bounded.
Proof. We have sn = a1 + · · · + an . Since an ≥ 0 for all n ∈ IN, sn+1 ≥ sn for all
n ∈ IN. Thus, (sn )n=1,2,... is an increasing sequence. If this sequence is bounded, then it
P∞
converges, by Theorem 3.1. Hence, the series n=1 an converges if (sn )n=1,2,... is bounded.
P∞
If (sn )n=1,2,... is unbounded, then the sequence diverges. Consequently, the series n=1 an
diverges.
Theorem 4.2. Let (an )n=1,2,... and (bn )n=1,2,... be two sequences of real numbers such
P∞
P∞
that 0 ≤ an ≤ bn for all n ∈ IN. If the series n=1 bn converges, then the series n=1 an
converges.
Proof. For n ∈ IN, let sn := a1 + · · · + an and tn := b1 + · · · + bn . Since 0 ≤ an ≤ bn for all
P∞
n ∈ IN, we have sn ≤ tn for all n ∈ IN. If the series n=1 bn converges, then the sequence
11
(tn )n=1,2,... is bounded. Consequently, the sequence (sn )n=1,2,... is bounded. Therefore,
P∞
the series n=1 an converges, by Theorem 4.1.
Let us investigate convergence or divergence of the p-series
∞
X
1
,
np
n=1
where p is a real number. For n ∈ IN, let an := 1/np and sn := a1 + · · · + an . Suppose
p > 1. For n ∈ IN have
k
s2n −1 =
n
2X
−1
X
k=1 m=2k−1
n
n
k=1
k=1
X 2k−1
X
1
1
≤
=
(21−p )k−1 <
.
p
k−1
p
m
(2
)
1 − 21−p
This shows that the sequence (sn )n=1,2,... is bounded. Hence, the p-series converges for
p > 1. For p = 1 and n ∈ IN we have
s2n = 1 +
n
X
k
2
X
n
k=1 m=2k−1 +1
X 2k−1
1
n
≥1+
=1+ .
k
m
2
2
k=1
This shows that the sequence (sn )n=1,2,... is unbounded. Hence, the harmonic series
P∞
p
n=1 1/n diverges. For p ≤ 1, we have 1/n ≤ 1/n for all n ∈ IN. By Theorem 4.2,
the p-series diverges for p ≤ 1.
For a real number a, let a+ := max{a, 0} and a− := max{−a, 0}. We call a+ the
positive part of a and a− the negative part of a, respectively. Evidently, |a| = a+ + a−
and a = a+ − a− .
P∞
Theorem 4.3. Let (an )n=1,2,... be a sequence of real numbers. If the series n=1 |an |
P∞
converges, then the series n=1 an converges.
−
Proof. We observe that 0 ≤ a+
n ≤ |an | and 0 ≤ an ≤ |an | for all n ∈ IN. If the series
P∞ −
P∞
P∞ +
n=1 an converge, by Theorem 4.2. But
n=1 |an | converges, then both
n=1 an and
P∞
+
−
an = an − an for all n ∈ IN. We conclude that the series n=1 an converges.
P∞
P∞
If n=1 |an | converges, then we say that the series n=1 an converges absolutely.
P∞
P∞
P∞
If
n=1 |an | diverges, then we say that
n=1 an converges
n=1 an converges, but
conditionally. For example, the alternating harmonic series
∞
X
(−1)n
n
n=1
converges conditionally.
12
Theorem 4.4. The set IR of real numbers is uncountable.
Proof. We use {0, 1}IN to denote the set of all mappings from IN to {0, 1}. For u ∈ {0, 1}IN ,
we define ϕ(u) to be the ternary series
∞
X
u(n)
.
3n
n=1
Since 0 ≤ u(n)/3n ≤ 1/3n for all n ∈ IN, the above series converges. Let us show that ϕ
is an injective mapping. Suppose that u, v ∈ {0, 1}IN and u 6= v. Let k be the least integer
such that u(k) 6= v(k). Without loss of any generality we may assume that u(k) = 0 and
v(k) = 1. Then
∞
X
1
1
ϕ(v) − ϕ(u) ≥ k −
> 0.
3
3n
n=k+1
This shows that ϕ is an injective mapping from {0, 1}IN to IR.
If IR were countable, then there would exist an injective mapping ψ from IR to IN.
Consequently, ψ ◦ ϕ would be an injective mapping from {0, 1}IN to IN. But {0, 1}IN is
uncountable. Hence, there is no injective mapping from {0, 1}IN to IN. This contradiction
demonstrates that IR is uncountable.
13