Limit Theorems

SEQUENCES AND SERIES

Section 3.2 Limit Theorems

We note that if Zis the sequence Z :¼

0;2;0;. . .;1þ ð1Þⁿ;. . .

;

then we can defineXþZ,X–ZandXZ, butX=Zis not defined since some of the terms of Zare zero.

We now show that sequences obtained by applying these operations to convergent sequences give rise to new sequences whose limits can be predicted.

3.2.3 Theorem (a) Let X ¼ (xn) and Y ¼ (yn) be sequences of real numbers that converge to x and y, respectively, and let c2R.Then the sequences XþY,X – Y,XY,and cX converge to xþy,x – y,xy,and cx,respectively.

(b) If X¼(xn)converges to x and Z¼(zn)is a sequence of nonzero real numbers that converges to z and if z6¼0, then the quotient sequence X=Z converges to x=z.

Proof. (a) To show that limðxnþy_nÞ ¼xþy, we need to estimate the magnitude of jðxnþy_nÞ ðxþyÞj. To do this we use the Triangle Inequality 2.2.3 to obtain

jðxnþy_nÞ ðxþyÞj ¼ jðxnxÞ þ ðy_nyÞj jxnx j þ jy_nyj:

By hypothesis, ife>0 there exists a natural numberK1such that ifnK1, thenjxnxj<

e=2; also there exists a natural numberK2such that ifnK2, thenjy_nyj<e=2. Hence if KðeÞ:¼supfK1;K2g, it follows that ifnKðeÞthen

jðxnþy_nÞ ðxþyÞj jxnxj þ jy_nyj

< 1 2eþ1

2e¼e:

Sincee>0 is arbitrary, we infer thatXþY¼ ðxnþy_nÞconverges toxþy.

Precisely the same argument can be used to show thatXY ¼ ðxny_nÞconverges to xy.

To show thatXY¼ ðxny_nÞconverges toxy, we make the estimate jx_ny_nxyj ¼ jðx_ny_nx_nyÞ þ ðx_nyxyÞj

jxnðy_nyÞj þ jðxnxÞyj

¼ jx_njjy_nyj þ jx_nxjjyj:

According to Theorem 3.2.2 there exists a real numberM1 >0 such thatjx_nj M1for all n2N and we setM :¼supfM1;jyjg. Hence we have

jxny_nxyj Mjy_nyj þMjxnxj:

From the convergence ofXandYwe conclude that ife>0 is given, then there exist natural numbers K1 and K2 such that if nK1 then jxnxj<e=2M, and if nK2 then jy_nyj<e=2M. Now letKðeÞ ¼supfK1; K2g;then, ifnKðeÞwe infer that

jxny_nxyj Mjy_nyj þMjxnxj

< Mðe=2MÞ þMðe=2MÞ ¼e:

Since e>0 is arbitrary, this proves that the sequenceXY¼ ðxny_nÞconverges toxy. The fact thatcX¼(cxn)converges tocxcan be proved in the same way; it can also be deduced by takingYto be the constant sequenceðc;c;c;. . .Þ. We leave the details to the reader.

(b) We next show that ifZ¼ ðznÞis a sequence of nonzero numbers that converges to a nonzero limitz, then the sequence (l=zn) of reciprocals converges to 1=z. First leta:¼¹₂jzj so that a>0. Since limðznÞ ¼z, there exists a natural number K1 such that if nK1 then jznzj<a. It follows from Corollary 2.2.4(a) of the Triangle Inequality thata jznzj jznj jzjfornK1, whence it follows that¹₂jzj ¼ jzj a jznj for nK1. Therefore 1=jznj 2=jzjfor nK1 so we have the estimate

1 zn1

z

¼ zzn

znz

¼ 1

jznzjjzznj 2

jzj²jzznj for all nK1:

Now, if e>0 is given, there exists a natural number K2 such that if nK2 then jznzj<¹₂ejzj². Therefore, it follows that ifKðeÞ ¼supfK1; K2g, then

1 zn1

z

<e for all n>KðeÞ:

Sincee>0 is arbitrary, it follows that lim 1

zn ¼1 z:

The proof of (b) is now completed by takingYto be the sequenceð1=znÞand using the fact thatXY ¼ ðxn=znÞconverges toxð1=zÞ ¼x=z. Q.E.D.

Some of the results of Theorem 3.2.3 can be extended, by Mathematical Induction, to a finite number of convergent sequences. For example, ifA¼ ða_nÞ; B¼ ðb_nÞ;. . .;Z¼ ðz_nÞ are convergent sequences of real numbers, then their sum AþBþ þZ¼ ða_nþb_n þ þz_nÞis a convergent sequence and

(1) limða_nþb_nþ þz_nÞ ¼limða_nÞ þlimðb_nÞ þ þlimðz_nÞ:

Also their productAB Z:¼ ða_nb_n z_nÞis a convergent sequence and (2) limðanbn znÞ ¼ ðlimðanÞÞ ðlimðbnÞÞ ðlimðznÞÞ:

Hence, ifk2N and ifA¼ ðanÞis a convergent sequence, then (3) limða^k_nÞ ¼ ðlimða_nÞÞ^k: We leave the proofs of these assertions to the reader.

3.2.4 Theorem If X¼(xn)is a convergent sequence of real numbers and if xn0for all n2N,then x¼limðx_nÞ 0.

Proof. Suppose the conclusion is not true and thatx<0; thene:¼ xis positive. SinceX converges tox, there is a natural numberKsuch that

xe<x_n <xþe for all nK:

In particular, we havex_K <xþe¼xþ ðxÞ ¼0. But this contradicts the hypothesis thatxn0 for alln2N. Therefore, this contradiction implies thatx0. Q.E.D.

We now give a useful result that is formally stronger than Theorem 3.2.4.

3.2.5 Theorem If X¼ ðxnÞand Y¼ ðy_nÞare convergent sequences of real numbers and if xny_n for all n2N,then limðxnÞ limðy_nÞ.

3.2 LIMIT THEOREMS 65

Proof. Letzn:¼y_nxnso thatZ :¼ ðznÞ ¼YXandzn0 for alln2N. It follows from Theorems 3.2.3 and 3.2.4 that

0limZ ¼limðy_nÞ limðxnÞ;

so that limðxnÞ limðy_nÞ: ^Q.E.D.

The next result asserts that if all the terms of a convergent sequence satisfy an inequality of the formax_nb, then the limit of the sequence satisfies the same inequality. Thus if the sequence is convergent, one may ‘‘pass to the limit’’ in an inequality of this type.

3.2.6 Theorem If X¼ ðx_nÞis a convergent sequence and if ax_n b for all n2N, then alimðx_nÞ b.

Proof. Let Y be the constant sequence ðb; b; b;. . .Þ. Theorem 3.2.5 implies that limXlimY ¼b. Similarly one shows thatalimX. Q.E.D.

The next result asserts that if a sequenceYis squeezed between two sequences that converge to thesame limit, then it must also converge to this limit.

3.2.7 Squeeze Theorem Suppose that X¼ ðxnÞ; Y ¼ ðy_nÞ,and Z¼ ðznÞare sequences of real numbers such that

x_ny_nz_n for all n2N;

and thatlimðx_nÞ ¼limðz_nÞ.Then Y ¼ ðy_nÞis convergent and limðx_nÞ ¼limðy_nÞ ¼limðz_nÞ:

Proof. Letw:¼limðx_nÞ ¼limðz_nÞ. Ife>0 is given, then it follows from the conver- gence ofXandZtowthat there exists a natural numberKsuch that if nK then

jxnwj<e and jznwj<e:

Since the hypothesis implies that

xnwy_nwznw for all n2N;

it follows (why?) that

e<y_nw<e

for allnK. Sincee>0 is arbitrary, this implies that limðy_nÞ ¼w. Q.E.D.

Remark Since any tail of a convergent sequence has the same limit, the hypotheses of Theorems 3.2.4, 3.2.5, 3.2.6, and 3.2.7 can be weakened to apply to the tail of a sequence.

For example, in Theorem 3.2.4, ifX¼ ðx_nÞis ‘‘ultimately positive’’ in the sense that there existsm2Nsuch thatxn0 for allnm, then the same conclusion thatx0 will hold.

Similar modifications are valid for the other theorems, as the reader should verify.

3.2.8 Examples (a) The sequence ðnÞis divergent.

It follows from Theorem 3.2.2 that if the sequenceX:¼ ðnÞis convergent, then there exists a real number M >0 such that n¼ jnj<M for all n2N. But this violates the Archimedean Property 2.4.3.

(b) The sequence ðð1ÞⁿÞis divergent.

This sequence X¼ ðð1ÞⁿÞ is bounded (take M :¼ 1), so we cannot invoke Theorem 3.2.2. However, assume that a:¼limX exists. Lete:¼1 so that there exists a natural numberK1such that

jð1Þⁿaj<1 for all nK1:

Ifnis an odd natural number withnK1this givesj 1aj<1, so that2<a<0. On the other hand, if n is an even natural number with nK1, this inequality gives j1aj<1 so that 0<a<2. Since a cannot satisfy both of these inequalities, the hypothesis that X is convergent leads to a contradiction. Therefore the sequence X is divergent.

(c) lim 2nþ1 n

¼2:

If we letX :¼ (2) andY :¼(1=n), then ðð2nþ1Þ=nÞ ¼XþY. Hence it follows from Theorem 3.2.3(a) that limðXþYÞ ¼limXþlimY ¼2þ0¼2:

(d) lim 2nþ1 nþ5

¼2:

Since the sequencesð2nþ1Þand (nþ5) are not convergent (why?), it is not possible to use Theorem 3.2.3(b) directly. However, if we write

2nþ1

nþ5 ¼2þ1=n 1þ5=n;

we can obtain the given sequence as one to which Theorem 3.2.3(b) applies when we takeX:¼ ð2þ1=nÞandZ :¼ ð1þ5=nÞ. (Check that all hypotheses are satisfied.) Since limX¼2 and limZ¼16¼0, we deduce that limðð2nþ1Þ=ðnþ5ÞÞ ¼2=1¼2.

(e) lim 2n n²þ1

¼0:

Theorem 3.2.3(b) does not apply directly. (Why?) We note that 2n

n²þ1¼ 2 nþ1=n;

but Theorem 3.2.3(b) does not apply here either, becauseðnþ1=nÞis not a convergent sequence. (Why not?) However, if we write

2n

n²þ1¼ 2=n 1þ1=n²;

then we can apply Theorem 3.2.3(b), since limð2=nÞ ¼0 and limð1þ1=n²Þ ¼16¼0.

Therefore limð2n=ðn²þ1ÞÞ ¼0=1¼0.

(f) lim sinn n

¼0:

We cannot apply Theorem 3.2.3(b) directly, since the sequence (n) is not convergent [neither is the sequence (sinn)]. It does not appear that a simple algebraic manipulation will enable us to reduce the sequence into one to which Theorem 3.2.3 will apply. However, if we note that1sinn1, then it follows that

1 nsinn

n 1

n for all n2N:

3.2 LIMIT THEOREMS 67

Hence we can apply the Squeeze Theorem 3.2.7 to infer that limðn¹sinnÞ ¼0. (We note that Theorem 3.1.10 could also be applied to this sequence.)

(g) Let X¼ ðxnÞ be a sequence of real numbers that converges tox2R. Letp be a polynomial; for example, let

pðtÞ:¼akt^kþak1t^k1þ þa1tþa0;

where k2N and aj2R for j¼0; 1;. . .; k. It follows from Theorem 3.2.3 that the sequenceðpðxnÞÞconverges topðxÞ. We leave the details to the reader as an exercise.

(h) Let X¼ ðxnÞ be a sequence of real numbers that converges to x2R. Let rbe a rational function (that is,rðtÞ:¼pðtÞ=qðtÞ, wherepandqare polynomials). Suppose that qðxnÞ 6¼0 for all n2N and that qðxÞ 6¼0. Then the sequence ðrðxnÞÞ converges to rðxÞ ¼pðxÞ=qðxÞ. We leave the details to the reader as an exercise. &

We conclude this section with several results that will be useful in the work that follows.

3.2.9 Theorem Let the sequence X¼ ðx_nÞconverge to x. Then the sequenceðjx_njÞof absolute values converges tojxj.That is, if x¼limðx_nÞ,thenjxj ¼limðjx_njÞ.

Proof. It follows from the Triangle Inequality (see Corollary 2.2.4(a)) that jx_nj jxj jx_nxj for all n2N:

The convergence ofðjxnjÞtojxjis then an immediate consequence of the convergence of

ðxnÞtox. Q.E.D.

3.2.10 Theorem Let X¼ ðxnÞbe a sequence of real numbers that converges to x and suppose that xn0. Then the sequenceðp Þffiffiffiffiffixn of positive square roots converges and limðp Þ ¼ffiffiffiffiffixn ffiffiffi

px .

Proof. It follows from Theorem 3.2.4 thatx¼limðx_nÞ 0 so the assertion makes sense.

We now consider the two cases: (i)x¼0 and (ii)x >0.

Case (i) Ifx¼0, lete>0 be given. Sincex_n!0 there exists a natural numberK such that ifnK then

0x_n¼x_n0<e²:

Therefore [see Example 2.1.13(a)], 0pffiffiffiffiffix_n <e for nK. Sincee>0 is arbitrary, this implies thatp !ffiffiffiffiffix_n 0.

Case (ii) Ifx>0, thenpffiffiffix>0 and we note that ffiffiffiffiffi

x_n p ffiffiffi

px

¼ðp ffiffiffiffiffix_n ffiffiffi px

Þðp þffiffiffiffiffix_n ffiffiffi px ffiffiffiffiffi Þ

xn

p þ ffiffiffi

px ¼ x_nx ffiffiffiffiffi xn

p þ ffiffiffi px Since pffiffiffiffiffixnþ ffiffiffi

px ffiffiffi

px

>0, it follows that pffiffiffiffiffix_n ffiffiffi

px 1 ffiffiffix p

jx_nxj:

The convergence ofp !ffiffiffiffiffixn ffiffiffi px

follows from the fact that xn!x. Q.E.D.

For certain types of sequences, the following result provides a quick and easy ‘‘ratio test’’ for convergence. Related results can be found in the exercises.

3.2.11 Theorem Let ðxnÞ be a sequence of positive real numbers such that L:¼ limðxnþ1=xnÞexists. If L<1,thenðxnÞconverges andlimðxnÞ ¼0.

Proof. By 3.2.4 it follows thatL0. Letr be a number such thatL<r<1, and let e:¼rL>0. There exists a numberK2N such that ifnK then

xnþ1

x_n L

<e:

It follows from this (why?) that ifnK, then xnþ1

x_n <Lþe¼Lþ ðrLÞ ¼r:

Therefore, ifnK, we obtain

0<x_nþ1<x_nr<x_n1r²< <x_Kr^nKþ1:

If we setC:¼x_K=r^K, we see that 0<x_nþ1<Cr^nþ1 for allnK. Since 0<r<1, it follows from 3.1.11(b) that limðrⁿÞ ¼0 and therefore from Theorem 3.1.10 that

limðxnÞ ¼0. Q.E.D.

As an illustration of the utility of the preceding theorem, consider the sequenceðxnÞ given byxn:¼n=2ⁿ. We have

xnþ1

x_n ¼nþ1 2^nþ1 2ⁿ

n ¼1 2 1þ1

n

;

so that limðx_nþ1=x_nÞ ¼¹₂. Since¹₂<1, it follows from Theorem 3.2.11 that limðn=2ⁿÞ ¼0.

Exercises for Section 3.2

1. Forxngiven by the following formulas, establish either the convergence or the divergence of the sequenceX¼ ðxnÞ.

(a) xn:¼ n

nþ1; (b) xn:¼ð1Þⁿn

nþ1 ; (c) xn:¼ n²

nþ1; (d) xn:¼2n²þ3

n²þ1 : 2. Give an example of two divergent sequencesXandYsuch that:

(a) their sumXþYconverges, (b) their productXYconverges.

3. Show that ifXandYare sequences such thatXandXþYare convergent, thenYis convergent.

4. Show that ifXandYare sequences such thatXconverges tox6¼0 andXYconverges, thenY converges.

5. Show that the following sequences are not convergent.

(a) ð2ⁿÞ; (b)

ð1Þⁿn² : 6. Find the limits of the following sequences:

(a) lim

ð2þ1=nÞ²

; (b) lim ð1Þⁿ

nþ2

; (c) lim

ffiffiffin p 1

ffiffiffin p þ1

; (d) lim nþ1

npffiffiffin

:

3.2 LIMIT THEOREMS 69

7. If ðbnÞ is a bounded sequence and limðanÞ ¼0, show that limðanbnÞ ¼0. Explain why Theorem 3.2.3cannotbe used.

8. Explain why the result in equation (3) before Theorem 3.2.4cannotbe used to evaluate the limit of the sequenceðð1þ1=nÞⁿÞ.

9. Lety_n:¼ ffiffiffiffiffiffiffiffiffiffiffi nþ1 p ffiffiffi

pn

forn2N. Show thatð ffiffiffi pn

y_nÞconverges. Find the limit.

10. Determine the limits of the following sequences.

(a) ð ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4n²þn

p 2nÞ; (b) ð ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

n²þ5n

p nÞ:

11. Determine the following limits.

(a) lim

ð3pffiffiffinÞ¹⁼²ⁿ

; (b) lim

ðnþ1Þ^1=lnðnþ1Þ . 12. If 0<a<b;determine lim a^nþ1þb^nþ1

aⁿþbⁿ

: 13. Ifa>0;b>0;show that lim ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

ðnþaÞðnþbÞ

p n

¼ ðaþbÞ=2.

14. Use the Squeeze Theorem 3.2.7 to determine the limits of the following,

(a) ðn¹⁼ⁿ²Þ; (b)

ðn!Þ¹⁼ⁿ² : 15. Show that ifzn:¼ ðaⁿþbⁿÞ¹⁼ⁿwhere 0<a<b, then limðznÞ ¼b.

16. Apply Theorem 3.2.11 to the following sequences, wherea,bsatisfy 0<a<1;b>1.

(a) ðaⁿÞ; (b) ðbⁿ=2ⁿÞ;

(c) ðn=bⁿÞ; (d) ð2³ⁿ=3²ⁿÞ:

17. (a) Give an example of a convergent sequenceðxnÞof positive numbers with limðxnþ1=xnÞ ¼1.

(b) Give an example of a divergent sequence with this property. (Thus, this property cannot be used as a test for convergence.)

18. LetX¼(xn) be a sequence of positive real numbers such that limðxnþ1=xnÞ ¼L>1. Show that Xis not a bounded sequence and hence is not convergent.

19. Discuss the convergence of the following sequences, wherea,bsatisfy 0<a<1;b>1.

(a) ðn²aⁿÞ; (b) ðbⁿ=n²Þ;

(c) ðbⁿ=n!Þ; (d) ðn!=nⁿÞ:

20. LetðxnÞbe a sequence of positive real numbers such that limðx¹_n⁼ⁿÞ ¼L<1. Show that there exists a numberrwith 0<r<1 such that 0<xn<rⁿfor all sufficiently largen2N. Use this to show that limðxnÞ ¼0.

21. (a) Give an example of a convergent sequence (x_n) of positive numbers with limðx¹⁼ⁿ_n Þ ¼1.

(b) Give an example of a divergent sequence (x_n) of positive numbers with limðx¹⁼ⁿ_n Þ ¼1.

(Thus, this property cannot be used as a test for convergence.)

22. Suppose thatðxnÞis a convergent sequence andðy_nÞis such that for anye>0 there existsMsuch thatjxny_nj<efor allnM. Does it follow that (y_n) is convergent?

23. Show that if (xn) and (yn) are convergent sequences, then the sequences (un) and (vn) defined by un:¼maxfxn;y_ngandvn:¼minfxn;y_ngare also convergent. (See Exercise 2.2.18.) 24. Show that if ðxnÞ;ðy_nÞ;ðznÞ are convergent sequences, then the sequence (wn) defined by

wn:¼midfxn;y_n;zngis also convergent. (See Exercise 2.2.19.)