Sequences PDF A sequence of real numbers

A sequence (xn) of real numbers converges if and only if it is a Cauchy sequence. Proof: Assume that (xn) converges to a real number XO' Let the.

1) Cauchy Convergence Criterion: A sequence (xn) is Cauchy if and

1) Cauchy Convergence Criterion: A sequence (xn) is Cauchy if and only if it is convergent. Proof. Suppose (xn) is a convergent sequence and lim(xn) = x.

Lecture summary

A sequence (an) is said to be a Cauchy sequence iff for any ? > 0 there of real numbers in terms of operations on the underlying Cauchy sequences ...

1.4 Cauchy Sequence in R

Every Cauchy sequence in R converges to an element in [ab]. Proof. Cauchy seq. A real vector space V is a set of elements called vectors

ANALYSIS I 9 The Cauchy Criterion

We say that (an) is a Cauchy sequence if for all ? > 0 Every real Cauchy sequence is convergent. ... n=0 be a sequence of real (complex) numbers.

Chapter 2. Sequences §1. Limits of Sequences Let A be a nonempty

Theorem 4.2. A sequence of real numbers is convergent if and only if it is a Cauchy sequence. Proof. Suppose that (xn)n=

Sequences

A sequence of real numbers converges if and only if it is a Cauchy sequence. Proof. First suppose that (xn) converges to a limit x ? R. Then for every ? > 0.

1 Review and Practice Problems for Exam # 2 - MATH 401/501

Know the ? N definition of Cauchy sequences and of convergent sequences in R to A sequence of real numbers converges if and only if the limsup and the ...

1 Review and Practice Problems for Exam # 2 - MATH 401/501

Know the ? N definition of Cauchy sequences and of convergent sequences in R to A sequence of real numbers converges if and only if the limsup and the ...

1 Sequence and Series of Real Numbers

(ii) If (an) has a convergent subsequence with limit a then the sequence (an) itself will converge to a. Proof. Let (an) be Cauchy sequence and ? > 0 be given.

Chapter 3 Sequences

In this chapter, we discuss sequences. We say what it means for a sequence to converge, and dene the limit of a convergent sequence. We begin with some preliminary results about the absolute value, which can be used to dene a distance function, or metric, onR. In turn, convergence is dened in terms of this metric.

3.1. The absolute value

Denition 3.1.The absolute value ofx2Ris dened by

jxj=( xifx0, xifx <0. Some basic properties of the absolute value are the following.

Proposition 3.2.For allx;y2R:

(a)jxj 0 andjxj= 0 if and only ifx= 0; (b)j xj=jxj; (c)jx+yj jxj+jyj(triangle inequality); (d)jxyj=jxjjyj; Proof.Parts (a), (b) follow immediately from the denition. Part (c) remains valid if we change the signs of bothxandyor exchangexandy. Therefore we can assume thatx0 andjxj jyjwithout loss of generality, in which casex+y0. Ify0, corresponding to the case whenxandyhave the same sign, then jx+yj=x+y=jxj+jyj: Ify <0, corresponding to the case whenxandyhave opposite signs andx+y >0, then jx+yj=x+y=jxj jyj363. Sequenceswhich proves (c). Part (d) remains valid if we changextoxorytoy, so we can assume thatx;y0 without loss of generality. Thenxy0 andjxyj=xy= jxjjyj. One useful consequence of the triangle inequality is the following reverse trian- gle inequality.

Proposition 3.3.Ifx;y2R, then

jjxj jyjj jxyj:

Proof.By the triangle inequality,

jxj=jxy+yj jxyj+jyj sojxjjyj jxyj. Similarly, exchangingxandy, we getjyjjxj jxyj, which proves the result. We can give an equivalent condition for the boundedness of a set by using the absolute value instead of upper and lower bounds as in Denition 2.9. Proposition 3.4.A setARis bounded if and only if there exists a real number

M0 such that

jxj Mfor everyx2A: Proof.If the condition in the proposition holds, thenMis an upper bound of AandMis a lower bound, soAis bounded. Conversely, ifAis bounded from above byM0and from below bym0, thenjxj Mfor everyx2AwhereM= maxfjm0j;jM0jg. A third way to say that a set is bounded is in terms of its diameter.

Denition 3.5.LetAR. The diameter ofAis

diamA= supfjxyj:x;y2Ag: Then a set is bounded if and only if its diameter is nite. Example 3.6.IfA= (a;a), then diamA= 2a, andAis bounded. IfA= (1;a), then diamA=1, andAis unbounded.

3.2. Sequences

A sequence (xn) of real numbers is an ordered list of numbersxn2R, called the terms of the sequence, indexed by the natural numbersn2N. We often indicate a sequence by listing the rst few terms, especially if they have an obvious pattern. Of course, no nite list of terms is, on its own, sucient to dene a sequence.

3.2. Sequences3705101520253035402

2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3 n xnFigure 1.A plot of the rst 40 terms in the sequencexn= (1 + 1=n)n, illustrating that it is monotone increasing and converges toe2:718, whose value is indicated by the dashed line.

Example 3.7.Here are some sequences:

1;8;27;64; :::; xn=n3;

1;12 ;13 ;14 ; ::: xn=1n

1;1;1;1; ::: xn= (1)n+1;

(1 + 1); 1 +12 2 1 +13 3 ; ::: x n= 1 +1n n Note that unlike sets, where elements are not repeated, the terms in a sequence may be repeated. The formal denition of a sequence is as a function onN, which is equivalent to its denition as a list. Denition 3.8.A sequence (xn) of real numbers is a functionf:N!R, where x n=f(n). We can consider sequences of many dierent types of objects (for example, sequences of functions) but for now we only consider sequences of real numbers, and we will refer to them as sequences for short. A useful way to visualize a sequence (xn) is to plot the graph ofxn2Rversusn2N. (See Figure 1 for an example.)

383. SequencesIf we want to indicate the range of the indexn2Nexplicitly, we write the

sequence as (xn)1n=1. Sometimes it is convenient to start numbering a sequence from a dierent integer, such asn= 0 instead ofn= 1. In that case, a sequence (xn)1n=0is a functionf:N0!Rwherexn=f(n) andN0=f0;1;2;3;:::g, and similarly for other starting points. Every functionf:N!Rdenes a sequence, corresponding to an arbitrary choice of a real numberxn2Rfor eachn2N. Some sequences can be dened explicitly by giving an expression for thenth terms, as in Example 3.7; others can be dened recursively. That is, we specify the value of the initial term (or terms) in the sequence, and denexnas a function of the previous terms (x1;x2;:::;xn1). A well-known example of a recursive sequence is the Fibonacci sequence (Fn)

1;1;2;3;5;8;13; :::;

which is dened byF1=F2= 1 and F n=Fn1+Fn2forn3: That is, we add the two preceding terms to get the next term. In general, we cannot expect to solve a recursion relation to get an explicit expression for thenth term in a sequence, but the recursion relation for the Fibonacci sequence is linear with constant coecients, and it can be solved to give an expression forFncalled the

Euler-Binet formula.

Proposition 3.9(Euler-Binet formula).Thenth term in the Fibonacci sequence is given by F n=1p5 n 1 n ; =1 +p5 2 Proof.The terms in the Fibonacci sequence are uniquely determined by the linear dierence equation F nFn1Fn2= 0; n3; with the initial conditions F

1= 1; F2= 1:

We see thatFn=rnis a solution of the dierence equation ifrsatises r

2r1 = 0;

which gives r=or1 ; =1 +p5 2

1:61803:

By linearity,

F n=An+B 1 n is a solution of the dierence equation for arbitrary constantsA,B. This solution satises the initial conditionsF1=F2= 1 if A=1p5 ; B=1p5 which proves the result.

3.3. Convergence and limits39Alternatively, once we know the answer, we can prove Proposition 3.9 by in-

duction. The details are left as an exercise. Note that although the right-hand side of the equation forFninvolves the irrational numberp5, its value is an integer for everyn2N. The numberappearing in Proposition 3.9 is called the golden ratio. It has the property that subtracting 1 from it gives its reciprocal, or 1 =1 Geometrically, this property means that the removal of a square from a rectangle whose sides are in the ratioleaves a rectangle whose sides are in the same ratio. The numberwas originally dened in Euclid's Elements as the division of a line in \extreme and mean ratio," and Ancient Greek architects arguably used rectangles with this proportion in the Parthenon and other buildings. During the Renaissance, was referred to as the \divine proportion." The rst use of the term \golden section" appears to be by Martin Ohm, brother of the physicist Georg Ohm, in a book published in 1835.

3.3. Convergence and limits

Roughly speaking, a sequence (xn) converges to a limitxif its termsxnget arbi- trarily close toxfor all suciently largen. Denition 3.10.A sequence (xn) of real numbers converges to a limitx2R, written x= limn!1xn;orxn!xasn! 1; if for every >0 there existsN2Nsuch that jxnxj< for alln > N: A sequence converges if it converges to some limitx2R, otherwise it diverges. Although we don't show it explicitly in the denition,Nis allowed to depend on. Typically, the smaller we choose, the larger we have to makeN. One way to view a proof of convergence is as a game: If I give you an >0, you have to come up with anNthat \works." Also note thatxn!xasn! 1means the same thing asjxnxj !0 asn! 1. It may appear obvious that a limit is unique if one exists, but this fact requires proof. Proposition 3.11.If a sequence converges, then its limit is unique. Proof.Suppose that (xn) is a sequence such thatxn!xandxn!x0asn! 1.

Let >0. Then there existN;N02Nsuch that

jxnxj<2 for alln > N; jxnx0j<2 for alln > N0:

403. SequencesChoose anyn >maxfN;N0g. Then, by the triangle inequality,

jxx0j jxxnj+jxnx0j<2 +2 Since this inequality holds for all >0, we must havejxx0j= 0 (otherwise the inequality would be false for=jxx0j=2>0), sox=x0. The following notation for sequences that \diverge to innity" is convenient.

Denition 3.12.If (xn) is a sequence then

lim n!1xn=1; orxn! 1asn! 1, if for everyM2Rthere existsN2Rsuch that x n> Mfor alln > N: Also lim n!1xn=1; orxn! 1asn! 1, if for everyM2Rthere existsN2Rsuch that x n< Mfor alln > N: That is,xn! 1asn! 1means the terms of the sequence (xn) get arbitrarily large and positive for all suciently largen, whilexn! 1asn! 1means that the terms get arbitrarily large and negative for all suciently largen. The notationxn! 1doesnotmean that the sequence converges. To illustrate these denitions, we discuss the convergence of the sequences in

Example 3.7.

Example 3.13.The terms in the sequence

1;8;27;64; :::; xn=n3

eventually exceed any real number, son3! 1asn! 1and this sequence does not converge. Explicitly, letM2Rbe given, and chooseN2Nsuch that N > M

1=3. (If1< M <1, we can chooseN= 1.) Then for alln > N, we have

3> N3> M, which proves the result.

Example 3.14.The terms in the sequence

1;12 ;13 ;14 ; ::: xn=1n get closer to 0 asngets larger, and the sequence converges to 0: lim n!11n = 0: To prove this limit, let >0 be given. ChooseN2Nsuch thatN >1=. (Such a number exists by the Archimedean property ofRstated in Theorem 2.18.) Then for alln > N1n 0=1n <1N which proves that 1=n!0 asn! 1.

3.3. Convergence and limits41Example 3.15.The terms in the sequence

1;1;1;1; ::: xn= (1)n+1;

oscillate back and forth innitely often between 1 and1, but they do not approach any xed limit, so the sequence does not converge. To show this explicitly, note that for everyx2Rwe have eitherjx1j 1 orjx+ 1j 1. It follows that there is noN2Nsuch thatjxnxj<1 for alln > N. Thus, Denition 3.10 fails if = 1 however we choosex2R, and the sequence does not converge.

Example 3.16.The convergence of the sequence

(1 + 1); 1 +12 2 1 +13 3 ;::: x n= 1 +1n n illustrated in Figure 1, is less obvious. Its terms are given by x n=ann; an= 1 +1n Asnincreases, we take larger powers of numbers that get closer to one. Ifa >1 is any xed real number, thenan! 1asn! 1so the sequence (an) does not converge (see Proposition 3.31 below for a detailed proof). On the other hand, if a= 1, then 1n= 1 for alln2Nso the sequence (1n) converges to 1. Thus, there are two competing factors in the sequence with increasingn:an!1 butn! 1. It is not immediately obvious which of these factors \wins." In fact, they are in balance. As we prove in Proposition 3.32 below, the sequence converges with lim n!1 1 +1n n =e; where 2< e <3. This limit can be taken as the denition ofe2:71828.

For comparison, one can also show that

lim n!1 1 +1n 2 n = 1;limn!1 1 +1n n2 =1: In the rst case, the rapid approach ofan= 1+1=n2to 1 \beats" the slower growth in the exponentn, while in the second case, the rapid growth in the exponentn2 \beats" the slower approach ofan= 1 + 1=nto 1. An important property of a sequence is whether or not it is bounded. Denition 3.17.A sequence (xn) of real numbers is bounded from above if there existsM2Rsuch thatxnMfor alln2N, and bounded from below if there existsm2Rsuch thatxnmfor alln2N. A sequence is bounded if it is bounded from above and below, otherwise it is unbounded. An equivalent condition for a sequence (xn) to be bounded is that there exists

M0 such that

jxnj Mfor alln2N:

423. SequencesExample 3.18.The sequence (n3) is bounded from below but not from above,

while the sequences (1=n) and(1)n+1are bounded. The sequence

1;2;3;4;5;6; ::: xn= (1)n+1n

is not bounded from below or above. We then have the following property of convergent sequences. Proposition 3.19.A convergent sequence is bounded. Proof.Let (xn) be a convergent sequence with limitx. There existsN2Nsuch that jxnxj<1 for alln > N:

The triangle inequality implies that

jxnj jxnxj+jxj<1 +jxjfor alln > N:

Dening

M= maxfjx1j;jx2j;:::;jxNj;1 +jxjg;

we see thatjxnj Mfor alln2N, so (xn) is bounded. Thus, boundedness is a necessary condition for convergence, and every un- bounded sequence diverges; for example, the unbounded sequence in Example 3.13 diverges. On the other hand, boundedness is not a sucient condition for conver- gence; for example, the bounded sequence in Example 3.15 diverges. The boundedness, or convergence, of a sequence (xn)1n=1depends only on the behavior of the innite \tails" (xn)1n=Nof the sequence, whereNis arbitrarily

large. Equivalently, the sequence (xn)1n=1and the shifted sequences (xn+N)1n=1have the same convergence properties and limits for everyN2N. As a result,

changing a nite number of terms in a sequence doesn't alter its boundedness or convergence, nor does it alter the limit of a convergent sequence. In particular, the existence of a limit gives no information about how quickly a sequence converges to its limit. Example 3.20.Changing the rst hundred terms of the sequence (1=n) from 1=n ton, we get the sequence

1;2;3; :::;99;100;1101

;1102 ;1103 which is still bounded (although by 100 instead of by 1) and still convergent to

0. Similarly, changing the rst billion terms in the sequence doesn't change its

boundedness or convergence. We introduce some convenient terminology to describe the behavior of \tails" of a sequence, Denition 3.21.LetP(x) denote a property of real numbersx2R. If (xn) is a real sequence, thenP(xn) holds eventually if there existsN2Nsuch thatP(xn) holds for alln > N; andP(xn) holds innitely often if for everyN2Nthere exists n > Nsuch thatP(xn) holds.

3.4. Properties of limits43For example, (xn) is bounded if there existsM0 such thatjxnj M

eventually; and (xn) does not converge tox2Rif there exists0>0 such that jxnxj 0innitely often. Note that if a propertyPholds innitely often according to Denition 3.21, then it does indeed hold innitely often: IfN= 1, then there existsn1>1 such thatP(xn1) holds; ifN=n1, then there existsn2> n1such thatP(xn2) holds; then there existsn3> n2such thatP(xn3) holds, and so on.

3.4. Properties of limits

In this section, we prove some order and algebraic properties of limits of sequences.

3.4.1. Monotonicity.Limits of convergent sequences preserve (non-strict) in-

equalities. Theorem 3.22.If (xn) and (yn) are convergent sequences andxnynfor all n2N, then lim n!1xnlimn!1yn: Proof.Suppose thatxn!xandyn!yasn! 1. Then for every >0 there existsP;Q2Nsuch that jxxnj<2 for alln > P; jyynj<2 for alln > Q:

Choosingn >maxfP;Qg, we have

x=xn+xxn< yn+2 =y+yny+2 < y+:

Sincex < y+for every >0, it follows thatxy.

This result, of course, remains valid if the inequalityxnynholds only for all suciently largen. Limits need not preserve strict inequalities. For example,

1=n >0 for alln2Nbut limn!11=n= 0.

It follows immediately that if (xn) is a convergent sequence withmxnM for alln2N, then mlimn!1xnM:quotesdbs_dbs6.pdfusesText_11

[PDF] Sequences A sequence of real numbers

Chapter 3

Sequences

3.1. The absolute value

Denition 3.1.The absolute value ofx2Ris dened by

Proposition 3.2.For allx;y2R:

Proposition 3.3.Ifx;y2R, then

Proof.By the triangle inequality,

M0 such that

Denition 3.5.LetAR. The diameter ofAis

3.2. Sequences

3.2. Sequences3705101520253035402

Example 3.7.Here are some sequences:

1;8;27;64; :::; xn=n3;

1;1;1;1; ::: xn= (1)n+1;

383. SequencesIf we want to indicate the range of the indexn2Nexplicitly, we write the

1;1;2;3;5;8;13; :::;

Euler-Binet formula.

1= 1; F2= 1:

2r1 = 0;

1:61803:

By linearity,

3.3. Convergence and limits39Alternatively, once we know the answer, we can prove Proposition 3.9 by in-

3.3. Convergence and limits

Let >0. Then there existN;N02Nsuch that

403. SequencesChoose anyn >maxfN;N0g. Then, by the triangle inequality,

Denition 3.12.If (xn) is a sequence then

Example 3.7.

Example 3.13.The terms in the sequence

1;8;27;64; :::; xn=n3

1=3. (If1< M <1, we can chooseN= 1.) Then for alln > N, we have

3> N3> M, which proves the result.

Example 3.14.The terms in the sequence

3.3. Convergence and limits41Example 3.15.The terms in the sequence

1;1;1;1; ::: xn= (1)n+1;

Example 3.16.The convergence of the sequence

For comparison, one can also show that

M0 such that

423. SequencesExample 3.18.The sequence (n3) is bounded from below but not from above,

1;2;3;4;5;6; ::: xn= (1)n+1n

The triangle inequality implies that

Dening

M= maxfjx1j;jx2j;:::;jxNj;1 +jxjg;

1;2;3; :::;99;100;1101

0. Similarly, changing the rst billion terms in the sequence doesn't change its

3.4. Properties of limits43For example, (xn) is bounded if there existsM0 such thatjxnj M

3.4. Properties of limits

3.4.1. Monotonicity.Limits of convergent sequences preserve (non-strict) in-

Choosingn >maxfP;Qg, we have

Sincex < y+for every >0, it follows thatxy.

1=n >0 for alln2Nbut limn!11=n= 0.