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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