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. . , for whichxn = (−1)n , has no limit.3.1.2 Properties of the Limit of a Sequencea. General PropertiesWe assign to this group the properties possessed not only by numerical sequences,but by other kinds of sequences as well, as we shall see below, although at presentwe shall study these properties only for numerical sequences.A sequence assuming only one value will be called a constant sequence.Definition 4 If there exists a number A and an index N such that xn = A for alln > N , the sequence {xn } will be called ultimately constant.Definition 5 A sequence {xn } is bounded if there exists M such that |xn | < M forall n ∈ N.823LimitsTheorem 1 a) An ultimately constant sequence converges.b) Any neighborhood of the limit of a sequence contains all but a finite numberof terms of the sequence.c) A convergent sequence cannot have two different limits.d) A convergent sequence is bounded.Proof a) If xn = A for n > N , then for any neighborhood V (A) of A we havexn ∈ V (A) when n > N , that is, limn→∞ xn = A.b) This assertion follows immediately from the definition of a convergent sequence.c) This is the most important part of the theorem.
Let limn→∞ xn = A1 andlimn→∞ xn = A2 . If A1 = A2 , we fix nonintersecting neighborhoods V (A1 )and V (A2 ) of A1 and A2 . These neighborhoods might be, for example, the δneighborhoods of A1 and A2 for δ < 12 |A1 − A2 |. By definition of limit we findindices N1 and N2 such that xn ∈ V (A1 ) for all n > N1 and xn ∈ V (A2 ) for alln > N2 .
But then for N = max{N1 , N2 } we have xn ∈ V (A1 ) ∩ V (A2 ). But this isimpossible, since V (A1 ) ∩ V (A2 ) = ∅.d) Let limn→∞ xn = A. Setting ε = 1 in the definition of a limit, we find N suchthat |xn − A| < 1 for all n > N . Then for n > N we have |xn | < |A| + 1. If we nowtake M > max{|x1 |, . . . , |xn |, |A| + 1} we find that |xn | < M for all n ∈ N.b.
Passage to the Limit and the Arithmetic OperationsDefinition 6 If {xn } and {yn } are two numerical sequences, their sum, product, andquotient (in accordance with the general definition of sum, product, and quotient offunctions) are the sequences xn.(xn + yn ) ,(xn · yn ) ,ynThe quotient, of course, is defined only when yn = 0 for all n ∈ N.Theorem 2 Let {xn } and {yn } be numerical sequences. If limn→∞ xn = A andlimn→∞ yn = B, thena) limn→∞ (xn + yn ) = A + B;b) limn→∞ (xn · yn ) = A · B;A, provided yn = 0 (n = 1, 2, .
. .) and B = 0.c) limn→∞ yxnn = BProof As an exercise we use the estimates for the absolute errors that arise underarithmetic operations with approximate values of quantities, which we already know(Sect. 2.2.4).Set |A − xn | = Δ(xn ), |B − yn | = Δ(yn ). Then for case a) we have(A + B) − (xn + yn ) ≤ Δ(xn ) + Δ(yn ).3.1 The Limit of a Sequence83Suppose ε > 0 is given. Since limn→∞ xn = A, there exists N such that Δ(xn ) <ε/2 for all n > N .
Similarly, since limn→∞ yn = B, there exists N such thatΔ(yn ) < ε/2 for all n > N . Then for n > max{N , N } we shall have(A + B) − (xn + yn ) < ε,which, by definition of limit, proves assertion a).b) We know that(A · B) − (xn · yn ) ≤ |xn |Δ(yn ) + |yn |Δ(xn ) + Δ(xn ) · Δ(yn ).Given ε > 0 find numbers N and N such thatΔ(xn ) < min 1,∀n > N∀n > NΔ(yn ) < min 1,ε3(|B| + 1)ε3(|A| + 1),.Then for n > N = max{N , N } we shall have|xn | < |A| + Δ(xn ) < |A| + 1,|yn | < |B| + Δ(yn ) < |B| + 1,εεε· min 1,≤ .Δ(xn ) · Δ(yn ) < min 1,333Hence for n > N we have|xn |Δ(yn ) < |A| + 1 ·εε< ,3(|A| + 1)3εε|yn |Δ(xn ) < |B| + 1 ·< ,3(|B| + 1)3εΔ(xn ) · Δ(yn ) < ,3and therefore |AB − xn yn | < ε for n > N .c) We use the estimate A xn |xn |Δ(yn ) + |yn |Δ(xn )|1 − ≤,·B y 21−δ(yynn)nn)where δ(yn ) = Δ(y|yn | .For a given ε > 0 we find numbers N and N such thatε|B|Δ(xn ) < min 1,,∀n > N8843∀n > N Limitsε · B2|B|Δ(yn ) < min,.4 16(|A| + 1)Then for n > max{N , N } we shall have|xn | < |A| + Δ(xn ) < |A| + 1,|yn | > |B| − Δ(yn ) > |B| −|B| |B|>,4212<,|yn | |B|0 < δ(yn ) =Δ(yn ) |B|/4 1<= ,|yn ||B|/2 211 − δ(yn ) > ,2and therefore 4ε1ε · B2= ,Δ(y)<|A|+1··n22ynB 16(|A| + 1) 4 1 Δ(xn ) < 2 · ε|B| = ε ,y |B|84n|xn | ·0<and consequently1< 2,1 − δ(yn ) A xn − <εB y nwhen n > N.Remark The statement of the theorem admits another, less constructive method ofproof that is probably known to the reader from the high-school course in the rudiments of analysis.
We shall mention this method when we discuss the limit of anarbitrary function. But here, when considering the limit of a sequence, we wishedto call attention to the way in which bounds on the errors in the result of an arithmetic operations can be used to set permissible bounds on the errors in the values ofquantities on which an operation is carried out.c. Passage to the Limit and InequalitiesTheorem 3 a) Let {xn } and {yn } be two convergent sequences with limn→∞ xn = Aand limn→∞ yn = B. If A < B, then there exists an index N ∈ N such that xn < ynfor all n > N .3.1 The Limit of a Sequence85b) Suppose the sequences {xn }, {yn }, and {zn } are such that xn ≤ yn ≤ zn for alln > N ∈ N. If the sequences {xn } and {zn } both converge to the same limit, then thesequence {yn } also converges to that limit.Proof a) Choose a number C such that A < C < B.
By definition of limit, we canfind numbers N and N such that |xn − A| < C − A for all n > N and |yn − B| <B − C for all n > N . Then for n > N = max{N , N } we shall have xn < A + C −A = C = B − (B − C) < yn .b) Suppose limn→∞ xn = limn→∞ zn = A. Given ε > 0 choose N and N suchthat A − ε < xn for all n > N and zn < A + ε for all n > N . Then for n > N =max{N , N } we shall have A − ε < xn ≤ yn ≤ zn < A + ε, which says |yn − A| < ε,that is A = limn→∞ yn .Corollary Suppose limn→∞ xn = A and limn→∞ yn = B.
If there exists N suchthat for all n > N we havea)b)c)d)xn > yn , then A ≥ B;xn ≥ yn , then A ≥ B;xn > B, then A ≥ B;xn ≥ B, then A ≥ B.Proof Arguing by contradiction, we obtain the first two assertions immediatelyfrom part a) of the theorem. The third and fourth assertions are the special casesof the first two obtained when yn ≡ B.It is worth noting that strict inequality may become equality in the limit. Forexample n1 > 0 for all n ∈ N, yet limn→∞ n1 = 0.3.1.3 Questions Involving the Existence of the Limit of a Sequencea. The Cauchy CriterionDefinition 7 A sequence {xn } is called a fundamental or Cauchy sequence2 if forany ε > 0 there exists an index N ∈ N such that |xm − xn | < ε whenever n > N andm > N.Theorem 4 (Cauchy’s convergence criterion) A numerical sequence converges ifand only if it is a Cauchy sequence.2 Bolzano introduced Cauchy sequences in an attempt to prove, without having at his disposal aprecise concept of a real number, that a fundamental sequence converges.
Cauchy gave a proof,taking the nested interval principle, which was later justified by Cantor, as obvious.863LimitsProof Suppose limn→∞ xn = A. Given ε > 0, we find an index N such that |xn −A| < 2ε for n > N . Then if m > N and n > N , we have |xm − xn | ≤ |xm − A| + |xn −A| < 2ε + 2ε = ε, and we have thus verified that the sequence is a Cauchy sequence.Now let {xk } be a fundamental sequence. Given ε > 0, we find an index N suchthat |xm − xk | < 3ε when m ≥ N and k ≥ N . Fixing m = N , we find that for anyk>NεεxN − < xk < xN + ,(3.1)33but since only a finite number of terms of the sequence have indices not largerthan N , we have shown that a fundamental sequence is bounded.For n ∈ N we now set an := infk≥n xk , and bn := supk≥n xk .It is clear from these definitions that an ≤ an+1 ≤ bn+1 ≤ bn (since the greatestlower bound does not decrease and the least upper bound does not increase when wepass to a smaller set).
By the nested interval principle, there is a point A common toall of the closed intervals [an , bn ].Sincean ≤ A ≤ bnfor any n ∈ N andan = inf xk ≤ xk ≤ sup xk = bkk≥nk≥nfor k ≥ n, it follows that|A − xk | ≤ bn − an .(3.2)But it follows from Eq. (3.1) thatxN −εε≤ inf xk = an ≤ bn = sup xk ≤ xN +3 k≥n3k≥nfor n > N , and therefore2ε<ε3for n > m. Comparing Eqs. (3.2) and (3.3), we find thatbn − an ≤(3.3)|A − xk | < ε,for any k > N , and we have proved that limk→∞ xk = A.Example 8 The sequence (−1)n (n = 1, 2, . .
.) has no limit, since it is not a Cauchysequence. Even though this fact is obvious, we shall give a formal verification. Thenegation of the statement that {xn } is a Cauchy sequence is the following:∃ε > 0 ∀N ∈ N ∃n > N ∃m > N |xm − xn | ≥ ε ,3.1 The Limit of a Sequence87that is, there exists ε > 0 such that for any N ∈ N two numbers n, m larger than Nexist for which |xm − xn | ≥ ε.In our case it suffices to set ε = 1. Then for any N ∈ N we shall have |xN +1 −xN +2 | = |1 − (−1)| = 2 > 1 = ε.Example 9 Letx1 = 0.α1 ,x2 = 0.α1 α2 ,x3 = 0.α1 α2 α3 ,...,xn = 0.α1 α2 .
. . αn ,...be a sequence of finite binary fractions in which each successive fraction is obtainedby adjoining a 0 or a 1 to its predecessor. We shall show that such a sequence alwaysconverges. Let m > n. Let us estimate the difference xm − xn : αn+1αm |xm − xn | = n+1 + · · · + m ≤22≤12n+1+ ··· +( 12 )n+1 − ( 12 )m+111=< n.2m21 − 12Thus, given ε > 0, if we choose N so that 21N < ε, we obtain the estimate |xm −xn | < 21n < 21N < ε for all m > n > N , which proves that the sequence {xn } is aCauchy sequence.Example 10 Consider the sequence {xn }, wherexn = 1 +11+ ··· + .2nSince|x2n − xn | =1111+ ··· +>n·= ,n+1n+n2n 2for all n ∈ N, the Cauchy criterion implies immediately that this sequence does nothave a limit.b. A Criterion for the Existence of the Limit of a Monotonic SequenceDefinition 8 A sequence {xn } is increasing if xn < xn+1 for all n ∈ N, nondecreasing if xn ≤ xn+1 for all n ∈ N, nonincreasing if xn ≥ xn+1 for all n ∈ N, and decreasing if xn > xn+1 for all n ∈ N.
Sequences of these four types are called monotonicsequences.Definition 9 A sequence {xn } is bounded above if there exists a number M suchthat xn < M for all n ∈ N.883LimitsTheorem 5 (Weierstrass) In order for a nondecreasing sequence to have a limit itis necessary and sufficient that it be bounded above.Proof The fact that any convergent sequence is bounded was proved above undergeneral properties of the limit of a sequence.
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