Sequences and Their Limits

SEQUENCES AND SERIES

Section 3.1 Sequences and Their Limits

A sequence in a setSis a function whose domain is the setNof natural numbers, and whose range is contained in the setS. In this chapter, we will be concerned with sequences inR and will discuss what we mean by the convergence of these sequences.

3.1.1 Definition Asequence of real numbers(or asequence inR) is a function defined on the setN¼ f1;2;. . .gof natural numbers whose range is contained in the setRof real numbers.

In other words, a sequence in R assigns to each natural number n ¼ 1, 2, . . . a uniquely determined real number. IfX:N!Ris a sequence, we will usually denote the value ofXatnby the symbolxnrather than using the function notationXðnÞ. The valuesxn

are also called thetermsor theelementsof the sequence. We will denote this sequence by the notations

X; ðx_nÞ; ðx_n:n2NÞ:

Of course, we will often use other letters, such asY¼ ðy_kÞ;Z ¼ ðziÞ, and so on, to denote sequences.

We purposely use parentheses to emphasize that the ordering induced by the natural order of N is a matter of importance. Thus, we distinguish notationally between the sequenceðxn :n2NÞ, whose infinitely many terms have an ordering, and the set of values fxn:n2Ngin the range of the sequence that are not ordered. For example, the sequence

X:¼ ðð1Þⁿ:n2NÞhas infinitely many terms that alternate between1 and 1, whereas

the set of valuesfð1Þⁿ:n2Ngis equal to the set {1, 1}, which has only two elements.

Sequences are often defined by giving a formula for thenth termx_n. Frequently, it is convenient to list the terms of a sequence in order, stopping when the rule of formation seems evident. For example, we may define the sequence of reciprocals of the even numbers by writing

X:¼ 1 2;1

4;1 6;1

8;

;

though a more satisfactory method is to specify the formula for the general term and write X:¼ 1

2n:n2N

or more simplyX¼ ð1=2nÞ.

Another way of defining a sequence is to specify the value ofx1and give a formula for x_nþ1ðn1Þin terms of xn. More generally, we may specifyx1and give a formula for obtaining xnþ1 from x1;x2;. . .;xn. Sequences defined in this manner are said to be inductively(orrecursively) defined.

3.1.2 Examples (a) Ifb2R, the sequenceB:¼ ðb;b;b;. . .Þ, all of whose terms equal b, is called the constant sequence b. Thus the constant sequence 1 is the sequence ð1; 1; 1;. . .Þ, and the constant sequence 0 is the sequenceð0;0;0;. . .Þ.

3.1 SEQUENCES AND THEIR LIMITS 55

(b) Ifb2R, thenB:¼ ðbⁿÞis the sequenceB¼ ðb;b²;b³;. . .;bⁿ;. . .Þ. In particular, if b¼¹₂, then we obtain the sequence

1

2ⁿ:n2N

¼ 1 2;1

4;1 8;. . .;1

2ⁿ;. . .

:

(c) The sequence ofð2n:n2NÞof even natural numbers can be defined inductively by x1 :¼2; xnþ1:¼xnþ2;

or by the definition

y1:¼2; y_nþ1:¼y1þy_n:

(d) The celebrated Fibonacci sequenceF:¼ ðf_nÞis given by the inductive definition f1:¼1; f2:¼1; f_nþ1:¼f_n1þf_n ðn2Þ:

Thus each term past the second is the sum of its two immediate predecessors. The first ten terms ofFare seen to beð1; 1; 2; 3; 5; 8; 13;21; 34;55;. . .Þ. _&

The Limit of a Sequence

There are a number of different limit concepts in real analysis. The notion of limit of a sequence is the most basic, and it will be the focus of this chapter.

3.1.3 Definition A sequenceX¼ ðx_nÞinRis said toconvergetox2R, orxis said to be alimitofðx_nÞ, if for everye>0 there exists a natural numberKðeÞsuch that for all nKðeÞ, the termsxnsatisfyjx_nxj<e.

If a sequence has a limit, we say that the sequence isconvergent; if it has no limit, we say that the sequence isdivergent.

Note The notationKðeÞis used to emphasize that the choice ofKdepends on the value ofe. However, it is often convenient to writeKinstead ofKðeÞ. In most cases, a ‘‘small’’

value ofewill usually require a ‘‘large’’ value ofKto guarantee that the distancejxnxj betweenxnandxis less thanefor allnK¼KðeÞ.

When a sequence has limitx, we will use the notation limX¼x or limðx_nÞ ¼x:

We will sometimes use the symbolismx_n!x, which indicates the intuitive idea that the valuesxn‘‘approach’’ the numberxas n! 1.

3.1.4 Uniqueness of Limits A sequence inR can have at most one limit.

Proof. Suppose thatx⁰andx⁰⁰are both limits ofðx_nÞ. For eache>0 there existK⁰such thatjx_nx⁰j<e=2 for allnK⁰, and there existsK⁰⁰such thatjx_nx⁰⁰j<e=2 for all nK⁰⁰. We let Kbe the larger of K⁰ and K⁰⁰. Then for nK we apply the Triangle Inequality to get

jx⁰x⁰⁰j ¼ jx⁰xnþxnx⁰⁰j

jx⁰x_nj þ jx_nx⁰⁰j<e=2þe=2¼e:

Since e>0 is an arbitrary positive number, we conclude thatx⁰x⁰⁰¼0. Q.E.D.

Forx2R ande>0, recall that thee-neighborhood ofx is the set V_eðxÞ:¼ fu2R:juxj<eg:

(See Section 2.2.) Sinceu2VeðxÞis equivalent tojuxj<e, the definition of conver- gence of a sequence can be formulated in terms of neighborhoods. We give several different ways of saying that a sequencexnconverges tox in the following theorem.

3.1.5 Theorem Let X ¼ (x_n) be a sequence of real numbers, and let x2R. The following statements are equivalent.

(a) X converges to x.

(b) For everye>0,there exists a natural number K such that for all nK,the terms xn

satisfyjx_nxj<e.

(c) For everye>0,there exists a natural number K such that for all nK,the terms xn

satisfy xe<xn <xþe.

(d) For everye-neighborhood VeðxÞof x, there exists a natural number K such that for all nK,the terms xnbelong to VeðxÞ.

Proof. The equivalence of (a) and (b) is just the definition. The equivalence of (b), (c), and (d) follows from the following implications:

juxj<e () e<ux<e () xe<u<xþe () u2VeðxÞ:

Q.E.D.

With the language of neighborhoods, one can describe the convergence of the sequenceX¼ ðxnÞto the number x by saying:for eache-neighborhood Ve(x)of x, all but a finite number of terms of X belong to Ve(x). The finite number of terms that may not belong to thee-neighborhood are the termsx1;x2;. . .;xK1.

Remark The definition of the limit of a sequence of real numbers is used to verify that a proposed valuexis indeed the limit. It doesnotprovide a means for initially determining what that value ofxmight be. Later results will contribute to this end, but quite often it is necessary in practice to arrive at a conjectured value of the limit by direct calculation of a number of terms of the sequence. Computers can be helpful in this respect, but since they can calculate only a finite number of terms of a sequence, such computations do not in any way constitute a proof of the value of the limit.

The following examples illustrate how the definition is applied to prove that a sequence has a particular limit. In each case, a positivee is given and we are required to find aK, depending one, as required by the definition.

3.1.6 Examples (a) lim(l=n)¼0.

Ife>0 is given, then 1=e>0. By the Archimedean Property 2.4.3, there is a natural numberK¼KðeÞsuch that 1=K<e. Then, ifnK, we have 1=n1=K<e. Conse- quently, ifnK, then

1 n0

¼1 n<e:

Therefore, we can assert that the sequence (1=n) converges to 0.

3.1 SEQUENCES AND THEIR LIMITS 57

(b) lim(l=(n²þ1))¼0.

Lete>0 be given. To findK, we first note that ifn2N, then 1

n²þ1< 1 n² 1

n:

Now chooseKsuch that 1=K<e, as in (a) above. ThennKimplies that 1=n<e, and therefore

1 n²þ10

¼ 1 n²þ1<1

n<e:

Hence, we have shown that the limit of the sequence is zero.

(c) lim 3nþ2 nþ1

¼3:

Givene>0, we want to obtain the inequality

ð1Þ 3nþ2

nþ1 3

<e

when nis sufficiently large. We first simplify the expression on the left:

3nþ2 nþ1 3

¼ 3nþ23n3 nþ1

¼ 1 nþ1

¼ 1 nþ1<1

n:

Now if the inequality 1=n<eis satisfied, then the inequality (1) holds. Thus if 1=K<e, then for anynK, we also have 1=n<eand hence (1) holds. Therefore the limit of the sequence is 3.

(d) limð ffiffiffiffiffiffiffiffiffiffiffi nþ1 p ffiffiffi

pn Þ ¼0.

We multiply and divide by ffiffiffiffiffiffiffiffiffiffiffi nþ1 p þ ffiffiffi

pn to get ð ffiffiffiffiffiffiffiffiffiffiffi

nþ1 p ffiffiffi

pn

Þð ffiffiffiffiffiffiffiffiffiffiffi nþ1 p þ ffiffiffi

pn ffiffiffiffiffiffiffiffiffiffiffi Þ

nþ1 p þ ffiffiffi

pn ¼ nffiffiffiffiffiffiffiffiffiffiffiþ1n nþ1 p þ ffiffiffi

pn

¼ 1

ffiffiffiffiffiffiffiffiffiffiffi nþ1 p þ ffiffiffi

p n 1 ffiffiffin p

For a givene>0, we obtain 1=pffiffiffin<eif and only if 1=n<e²orn>1=e². Thus if we take K>1=e², then ffiffiffiffiffiffiffiffiffiffiffi

nþ1 p ffiffiffi

pn

<efor alln>K. (For example, if we are givene¼1=10, thenK>100 is required.)

(e) If 0<b<1, then limðbⁿÞ ¼0.

We will use elementary properties of the natural logarithm function. Ife>0 is given, we see that

bⁿ<e () nlnb<lne () n>lne=lnb:

(The last inequality is reversed because lnb<0.) Thus if we chooseKto be a number such thatK>lne=lnb, then we will have 0<bⁿ<efor allnK. Thus we have limðbⁿÞ ¼0.

For example, if b ¼ .8, and if e ¼ .01 is given, then we would need K>

ln:01=ln:820:6377. ThusK¼21 would be an appropriate choice fore¼.01. ^&

Remark TheKðeÞGame In the notion of convergence of a sequence, one way to keep in mind the connection between theeand theKis to think of it as a game called theKðeÞ Game. In this game, Player A asserts that a certain numberxis the limit of a sequence (xn).

Player B challenges this assertion by giving Player A a specific value fore>0. Player A must respond to the challenge by coming up with a value ofKsuch thatjxnxj<efor all n>K. If Player A can always find a value ofKthat works, then he wins, and the sequence is convergent. However, if Player B can give a specific value ofe>0 for which Player A

cannot respond adequately, then Player B wins, and we conclude that the sequence does not converge tox.

In order to show that a sequenceX¼ ðxnÞdoesnotconverge to the numberx, it is enough to produce one numbere0 >0 such that no matter what natural numberKis chosen, one can find a particular nK satisfying nK K such that jxnKxj e0. (This will be discussed in more detail in Section 3.4.)

3.1.7 Example The sequence ð0; 2; 0; 2;. . .; 0; 2;. . .Þ does not converge to the number 0.

If Player A asserts that 0 is the limit of the sequence, he will lose theKðeÞGame when Player B gives him a value ofe<2. To be definite, let Player B give Player A the value e0¼1. Then no matter what value Player A chooses for K, his response will not be adequate, for Player B will respond by selecting an even number n>K. Then the corresponding value isx_n¼2 so thatjx_n0j ¼2>1¼e0. Thus the number 0 is not

the limit of the sequence. &

Tails of Sequences

It is important to realize that the convergence (or divergence) of a sequence X¼ ðxnÞ depends only on the ‘‘ultimate behavior’’ of the terms. By this we mean that if, for any natural numberm, we drop the firstmterms of the sequence, then the resulting sequenceXm

converges if and only if the original sequence converges, and in this case, the limits are the same. We will state this formally after we introduce the idea of a ‘‘tail’’ of a sequence.

3.1.8 Definition IfX¼ ðx1;x2;. . .;xn;. . .Þis a sequence of real numbers and ifmis a given natural number, then them-tailofXis the sequence

Xm:¼ ðxmþn:n2NÞ ¼ ðxmþ1;xmþ2;. . .Þ

For example, the 3-tail of the sequenceX¼ ð2;4;6;8;10;. . .;2n;. . .Þ, is the sequence X3¼ ð8; 10;12;. . .;2nþ6;. . .Þ.

3.1.9 Theorem Let X¼ ðxn:n2NÞ be a sequence of real numbers and let m2N. Then the m-tail Xm¼ ðxmþn:n2NÞof X converges if and only if X converges. In this case, limXm¼limX.

Proof. We note that for anyp2N, the pth term of Xmis the (p þ m)th term of X. Similarly, ifq>m, then theqth term ofXis the ðqmÞth term ofXm.

AssumeXconverges tox. Then given anye>0, if the terms ofXfornKðeÞsatisfy jxnxj<e, then the terms ofXmforkKðeÞ msatisfyjxkxj<e. Thus we can take KmðeÞ ¼KðeÞ m, so thatXmalso converges tox.

Conversely, if the terms ofXmforkKmðeÞsatisfyjxkxj<e, then the terms ofX for nKðeÞ þmsatisfyjxnxj<e. Thus we can take KðeÞ ¼KmðeÞ þm.

Therefore,Xconverges toxif and only ifXmconverges tox. Q.E.D.

We shall sometimes say that a sequence X ultimately has a certain property if some tail of X has this property. For example, we say that the sequence ð3;4;5;5;5;. . .;5;. . .Þ is ‘‘ultimately constant.’’ On the other hand, the sequence ð3;5;3;5;. . .;3;5;. . .Þ is not ultimately constant. The notion of convergence can be stated using this terminology: A sequenceXconverges toxif and only if the terms of X are ultimately in every e-neighborhood of x. Other instances of this ‘‘ultimate terminology’’ will be noted later.

3.1 SEQUENCES AND THEIR LIMITS 59

Further Examples

In establishing that a numberxis the limit of a sequence (xn), we often try to simplify the differencejx_nxjbefore considering ane>0 and finding aKðeÞas required by the definition of limit. This was done in some of the earlier examples. The next result is a more formal statement of this idea, and the examples that follow make use of this approach.

3.1.10 Theorem Let (xn) be a sequence of real numbers and let x2R. If (an) is a sequence of positive real numbers withlimðanÞ ¼0and if for some constant C>0and some m2N we have

jx_nxj Ca_n for all nm;

then it follows that limðx_nÞ ¼x.

Proof. Ife>0 is given, then since limða_nÞ ¼0, we know there existsK¼Kðe=CÞsuch thatnK implies

an¼ jan0j<e=C:

Therefore it follows that if bothnK andnm, then jxnxj Can<Cðe=CÞ ¼e:

Since e>0 is arbitrary, we conclude thatx¼limðxnÞ. Q.E.D.

3.1.11 Examples (a) Ifa>0, then lim 1 1þna

¼0.

Sincea>0, then 0<na<1þna, and therefore 0<1=(1þna)<1=(na). Thus we have

1 1þna0

1

a 1

n for all n 2 N:

Since limð1=nÞ ¼0, we may invoke Theorem 3.1.10 withC¼1=aandm¼1 to infer that limð1=ð1þnaÞÞ ¼0.

(b) If 0<b<1, then limðbⁿÞ ¼0.

This limit was obtained earlier in Example 3.1.6(e). We will give a second proof that illustrates the use of Bernoulli’s Inequality (see Example 2.1.13(c)).

Since 0<b<1, we can writeb¼1=(1þa), wherea:¼(1=b)1 so thata>0. By Bernoulli’s Inequality, we haveð1þaÞⁿ1þna. Hence

0<bⁿ¼ 1

ð1þaÞⁿ 1 1þna< 1

na: Thus from Theorem 3.1.10 we conclude that limðbⁿÞ ¼0.

In particular, ifb¼.8, so thata¼.25, and if we are givene¼.01, then the preceding inequality gives usKðeÞ ¼4=ð:01Þ ¼400. Comparing with Example 3.1.6(e), where we obtainedK¼21, we see this method of estimation does not give us the ‘‘best’’ value ofK. However, for the purpose of establishing the limit, the size ofKis immaterial.

(c) Ifc>0, then limðc¹⁼ⁿÞ ¼1.

The casec¼1 is trivial, since then (c^1/n) is the constant sequenceð1; 1;. . .Þ, which evidently converges to 1.

Ifc>1, thenc^1/n¼1þdnfor somedn>0. Hence by Bernoulli’s Inequality 2.1.13(c), c¼ ð1þdnÞⁿ 1þndn for n2N:

Therefore we have c1nd_n, so thatd_n ðc1Þ=n. Consequently we have jc¹⁼ⁿ1j ¼d_n ðc1Þ1

n for n2N:

We now invoke Theorem 3.1.10 to infer that limðc¹⁼ⁿÞ ¼1 whenc>1.

Now suppose that 0<c<1; then c^1/n ¼ 1=(1 þ hn) for some hn > 0. Hence Bernoulli’s Inequality implies that

c¼ 1

ð1þh_nÞⁿ 1

1þnh_n< 1 nh_n;

from which it follows that 0<h_n<1=ncfor n2N. Therefore we have 0<1c¹⁼ⁿ¼ h_n

1þhn<h_n< 1 nc so that

jc¹⁼ⁿ1j< 1 c

1

n for n2N:

We now apply Theorem 3.1.10 to infer that limðc¹⁼ⁿÞ ¼1 when 0<c<1.

(d) limðn¹⁼ⁿÞ ¼1

Sincen¹⁼ⁿ>1 forn>1, we can writen¹⁼ⁿ¼1þk_n for somek_n>0 when n>1.

Hencen¼ ð1þk_nÞⁿ for n>1. By the Binomial Theorem, ifn >1 we have n¼1þnk_nþ¹₂nðn1Þk²_nþ 1þ¹₂nðn1Þk²_n; whence it follows that

n1¹₂nðn1Þk²_n:

Hencek²_n2=nforn>1. Ife>0 is given, it follows from the Archimedean Property that there exists a natural numberNesuch that 2=Ne<e². It follows that ifnsupf2;Negthen 2=n <e², whence

0<n¹⁼ⁿ1¼k_n ð2=nÞ¹⁼²<e:

Sincee>0 is arbitrary, we deduce that limðn¹⁼ⁿÞ ¼1. _&

Exercises for Section 3.1

1. The sequence (xn) is defined by the following formulas for thenth term. Write the first five terms in each case:

(a) xn:¼1þ ð1Þⁿ; (b) xn:¼ ð1Þⁿ=n;

(c) xn:¼ 1

nðnþ1Þ; (d) x:¼ 1

n²þ2:

3.1 SEQUENCES AND THEIR LIMITS 61

2. The first few terms of a sequence (xn) are given below. Assuming that the ‘‘natural pattern’’

indicated by these terms persists, give a formula for thenth termx_n.

(a) 5, 7, 9, 11, . . . , (b) 1=2, –1=4, 1=8, –1=16, . . . , (c) 1=2, 2=3, 3=4, 4=5, . . . , (d) 1, 4, 9, 16, . . . .

3. List the first five terms of the following inductively defined sequences.

(a) x1:¼1; xnþ1:¼3xnþ1;

(b) y1:¼2; ynþ1:¼¹2ðynþ2=ynÞ;

(c) z1:¼1; z2:¼2;znþ2:¼ ðznþ1þznÞ=ðznþ1znÞ;

(d) s1:¼3; s2:¼5;snþ2:¼snþsnþ1: 4. For anyb2R, prove that limðb=nÞ ¼0.

5. Use the definition of the limit of a sequence to establish the following limits.

(a) lim n n²þ1

¼0; (b) lim 2n

nþ1

¼2;

(c) lim 3nþ1 2nþ5

¼3

2; (d) lim n²1

2n²þ3

¼1 2: 6. Show that

(a) lim 1

ffiffiffiffiffiffiffiffiffiffiffi nþ7 p

¼0; (b) lim 2n

nþ2

¼2;

(c) lim ffiffiffin p nþ1

¼0; (d) lim ð1Þⁿn n²þ1

¼0:

7. Letxn:¼1=lnðnþ1Þforn2N.

(a) Use the definition of limit to show that limðxnÞ ¼0.

(b) Find a specific value ofK(e) as required in the definition of limit for each of (i)e¼1=2, and (ii)e¼1=10.

8. Prove that limðxnÞ ¼0 if and only if limðjxnjÞ ¼0. Give an example to show that the convergence ofðjxnjÞneed not imply the convergence ofðxnÞ.

9. Show that ifxn0 for alln2N and limðxnÞ ¼0, then limðp Þ ¼ffiffiffiffiffixn 0.

10. Prove that if limðxnÞ ¼xand ifx>0, then there exists a natural numberMsuch thatxn>0 for allnM.

11. Show that lim 1 n 1

nþ1

¼0:

12. Show that limð ffiffiffiffiffiffiffiffiffiffiffiffiffi n²þ1

p nÞ ¼0:

13. Show that limð1=3ⁿÞ ¼0.

14. Letb2Rsatisfy 0<b<1. Show that limðnbⁿÞ ¼0. [Hint: Use the Binomial Theorem as in Example 3.1.11(d).]

15. Show that lim

ð2nÞ¹⁼ⁿ

¼1.

16. Show that limðn²=n!Þ ¼0.

17. Show that limð2ⁿ=n!Þ ¼0. [Hint: Ifn3, then 0<2ⁿ=n!2 ²_{3 n}2

.]

18. If limðxnÞ ¼x>0, show that there exists a natural number K such that if nK, then

1

2x<xn<2x.