SEQUENCES AND SERIES
Section 3.4 Subsequences and the Bolzano-Weierstrass Theorem
0<b<1, thenxnþ1 ¼bnþ1<bn ¼xnso that the sequenceðxnÞis decreasing. It is also clear that 0xn 1, so it follows from the Monotone Convergence Theorem 3.3.2 that the sequence is convergent. Letx:¼limxn. Sinceðx2nÞis a subsequence ofðxnÞit follows from Theorem 3.4.2 thatx¼limðx2nÞ. Moreover, it follows from the relationx2n¼b2n ¼ ðbnÞ2¼x2n and Theorem 3.2.3 that
x¼limðx2nÞ ¼
limðxnÞ 2¼x2:
Therefore we must have eitherx¼0 orx¼1. Since the sequenceðxnÞis decreasing and bounded above byb<1, we deduce thatx ¼0.
(b) limðc1=nÞ ¼1 forc>1.
This limit has been obtained in Example 3.1.11(c) forc>0, using a rather ingenious argument. We give here an alternative approach for the casec>1. Note that ifzn:¼c1=n; then zn>1 andznþ1<zn for all n2N. (Why?) Thus by the Monotone Convergence Theorem, the limitz:¼limðznÞexists. By Theorem 3.4.2, it follows thatz¼limðz2nÞ. In addition, it follows from the relation
z2n¼c1=2n¼ ðc1=nÞ1=2¼z1n=2 and Theorem 3.2.10 that
z¼limðz2nÞ ¼
limðznÞ 1=2¼z1=2:
Therefore we havez2 ¼zwhence it follows that eitherz¼0 orz¼1. Sincezn >1 for all n2N, we deduce thatz¼1.
We leave it as an exercise to the reader to consider the case 0<c<1. &
The following result is based on a careful negation of the definition of limðxnÞ ¼x. It leads to a convenient way to establish the divergence of a sequence.
3.4.4 Theorem Let X¼ ðxnÞ be a sequence of real numbers, Then the following are equivalent:
(i) The sequence X¼ ðxnÞdoes not converge to x2R.
(ii) There exists an e0>0 such that for any k2N, there exists nk2N such that nkk andjxnkxj e0.
(iii) There exists ane0>0and a subsequence X0¼ ðxnkÞof X such thatjxnkxj e0for all k2N.
Proof. ðiÞ ) ðiiÞ IfðxnÞdoes not converge tox, then for somee0>0 it is impossible to find a natural numberksuch that for allnkthe termsxnsatisfyjxnxj<e0. That is, for eachk2N it isnot true that forall nk the inequalityjxnxj<e0 holds. In other words, for eachk2N there exists a natural numbernkksuch thatjxnkxj e0.
ðiiÞ ) ðiiiÞ Lete0be as in (ii) and letn12Nbe such thatn11 andjxn1xj e0. Now let n22N be such that n2>n1andjxn2xj e0; let n32N be such that n3>n2andjxn3xj e0. Continue in this way to obtain a subsequence X0¼ ðxnkÞof Xsuch thatjxnkxj e0 for allk2N.
ðiiiÞ ) ðiÞ SupposeX¼ ðxnÞhas a subsequenceX0¼ ðxnkÞsatisfying the condition in (iii). ThenXcannot converge tox; for if it did, then, by Theorem 3.4.2, the subsequence X0would also converge tox. But this is impossible, since none of the terms ofX0belongs to
the e0-neighborhood ofx. Q.E.D.
3.4 SUBSEQUENCES AND THE BOLZANO-WEIERSTRASS THEOREM 79
Since all subsequences of a convergent sequence must converge to the same limit, we have part (i) in the following result. Part (ii) follows from the fact that a convergent sequence is bounded.
3.4.5 Divergence Criteria If a sequence X¼ ðxnÞof real numbers has either of the following properties, then X is divergent.
(i) X has two convergent subsequences X0¼ ðxnkÞand X00¼ ðxrkÞwhose limits are not equal.
(ii) X is unbounded.
3.4.6 Examples (a) The sequence X:¼ ðð1ÞnÞis divergent.
The subsequenceX0:¼ ðð1Þ2nÞ ¼ ð1; 1;. . .Þconverges to 1, and the subsequence X00:¼ ðð1Þ2n1Þ ¼ ð1;1;. . .Þconverges to1. Therefore, we conclude from Theo- rem 3.4.5(i) thatXis divergent.
(b) The sequence 1 ;12;3;14;. . .
is divergent.
This is the sequenceY¼ ðynÞ, whereyn¼nifnis odd, andyn¼1=nifnis even. It can easily be seen that Yis not bounded. Hence, by Theorem 3.4.5(ii), the sequence is divergent.
(c) The sequenceS :¼(sin n) is divergent.
This sequence is not so easy to handle. In discussing it we must, of course, make use of elementary properties of the sine function. We recall that sinðp=6Þ ¼12¼sinð5p=6Þand that sinx>12 for x in the interval I1:¼ ðp=6; 5p=6Þ. Since the length of I1 is 5p=6p=6¼2p=3>2, there are at least two natural numbers lying inside I1; we let n1be the first such number. Similarly, for eachk2N, sinx>12forx in the interval
Ik:¼
p=6þ2pðk1Þ;5p=6þ2pðk1Þ :
Since the length ofIkis greater than 2, there are at least two natural numbers lying insideIk; we letnkbe the first one. The subsequenceS0:¼ ðsinnkÞofSobtained in this way has the property that all of its values lie in the interval12;1
. Similarly, ifk2N andJkis the interval
Jk:¼
7p=6þ2pðk1Þ;11p=6þ2pðk1Þ ;
then it is seen that sinx<12for allx2Jkand the length ofJkis greater than 2. Letmkbe the first natural number lying in Jk. Then the subsequenceS00:¼ ðsinmkÞof S has the property that all of its values lie in the interval1;12
.
Given any real numberc, it is readily seen that at least one of the subsequencesS0and S00lies entirely outside of the12-neighborhood ofc. Thereforeccannot be a limit ofS. Since
c2R is arbitrary, we deduce thatSis divergent. &
The Existence of Monotone Subsequences
While not every sequence is a monotone sequence, we will now show that every sequence has a monotone subsequence.
3.4.7 Monotone Subsequence Theorem If X¼ ðxnÞ is a sequence of real numbers, then there is a subsequence of X that is monotone.
Proof. For the purpose of this proof, we will say that themth termxmis a ‘‘peak’’ if xmxnfor allnsuch thatnm. (That is,xmis never exceeded by any term that follows it
in the sequence.) Note that, in a decreasing sequence, every term is a peak, while in an increasing sequence, no term is a peak.
We will consider two cases, depending on whetherXhas infinitely many, or finitely many, peaks.
Case 1: X has infinitely many peaks. In this case, we list the peaks by increasing subscripts:xm1;xm2;. . .;xmk; :. . .Since each term is a peak, we have
xm1xm2 xmk :
Therefore, the subsequenceðxmkÞof peaks is a decreasing subsequence ofX.
Case 2: Xhas a finite number (possibly zero) of peaks. Let these peaks be listed by increasing subscripts:xm1;xm2;. . .;xmr. Lets1:¼mrþ1 be the first index beyond the last peak. Sincexs1is not a peak, there existss2>s1such thatxs1<xs2. Sincexs2is not a peak, there existss3>s2 such thatxs2<xs3. Continuing in this way, we obtain an increasing
subsequenceðxskÞof X. Q.E.D.
It is not difficult to see that a given sequence may have one subsequence that is increasing, and another subsequence that is decreasing.
The Bolzano-Weierstrass Theorem
We will now use the Monotone Subsequence Theorem to prove the Bolzano-Weierstrass Theorem, which states that every bounded sequence has a convergent subsequence.
Because of the importance of this theorem we will also give a second proof of it based on the Nested Interval Property.
3.4.8 The Bolzano-Weierstrass Theorem A bounded sequence of real numbers has a convergent subsequence.
First Proof. It follows from the Monotone Subsequence Theorem that ifX¼ ðxnÞis a bounded sequence, then it has a subsequence X0¼ ðxnkÞ that is monotone. Since this subsequence is also bounded, it follows from the Monotone Convergence Theorem 3.3.2
that the subsequence is convergent. Q.E.D.
Second Proof. Since the set of valuesfxn :n2Ngis bounded, this set is contained in an intervalI1:¼ ½a;b . We taken1:¼1.
We now bisectI1into two equal subintervalsI10 andI100, and divide the set of indices fn2N:n>1g into two parts:
A1:¼ fn2N:n>n1;xn2I10g; B1¼ fn2N:n>n1;xn2I100g:
IfA1is infinite, we takeI2:¼I10and letn2be the smallest natural number inA1. IfA1is a finite set, thenBlmust be infinite, and we takeI2:¼I100and letn2be the smallest natural number inB1.
We now bisect I2 into two equal subintervals I20 and I200, and divide the set fn2N:n>n2g into two parts:
A2¼ fn2N:n>n2;xn;2I20g; B2:¼ fn2N:n>n2;xn2I200g
IfA2is infinite, we takeI3:¼I20and letn3be the smallest natural number inA2. IfA2is a finite set, thenB2must be infinite, and we takeI3:¼I200and letn3be the smallest natural number inB2.
3.4 SUBSEQUENCES AND THE BOLZANO-WEIERSTRASS THEOREM 81
We continue in this way to obtain a sequence of nested intervals I1I2 Ik and a subsequenceðxnkÞofXsuch thatxnk 2Ikfork2N. Since the length ofIkis equal toðbaÞ=2k1, it follows from Theorem 2.5.3 that there is a (unique) common point j2Ik for allk2N. Moreover, sincexnk andj both belong toIk, we have
jxnkjj ðbaÞ=2k1;
whence it follows that the subsequenceðxnkÞof Xconverges toj. Q.E.D.
Theorem 3.4.8 is sometimes called the Bolzano-Weierstrass Theoremfor sequences, because there is another version of it that deals with bounded sets inR(see Exercise 11.2.6).
It is readily seen that a bounded sequence can have various subsequences that converge to different limits or even diverge. For example, the sequenceðð1ÞnÞhas subsequences that converge to1, other subsequences that converge toþ1, and it has subsequences that diverge.
LetXbe a sequence of real numbers and letX0be a subsequence ofX. ThenX0is a sequence in its own right, and so it has subsequences. We note that ifX00is a subsequence of X0, then it is also a subsequence ofX.
3.4.9 Theorem Let X¼ ðxnÞbe a bounded sequence of real numbers and let x2Rhave the property that every convergent subsequence of X converges to x. Then the sequence X converges to x.
Proof. SupposeM>0 is a bound for the sequenceXso thatjxnj Mfor alln2N. IfX does not converge to x, then Theorem 3.4.4 implies that there exist e0>0 and a subsequenceX0¼ ðxnkÞofXsuch that
ð1Þ jxnkxj e0 for all k2N:
SinceX0is a subsequence ofX, the numberMis also a bound forX0. Hence the Bolzano- Weierstrass Theorem implies thatX0has a convergent subsequenceX00. SinceX00is also a subsequence ofX, it converges toxby hypothesis. Thus, its terms ultimately belong to the
e0-neighborhood ofx, contradicting (1). Q.E.D.
Limit Superior and Limit Inferior
A bounded sequence of real numbersðxnÞmay or may not converge, but we know from the Bolzano-Weierstrass Theorem 3.4.8 that there will be a convergent subsequence and possibly many convergent subsequences. A real number that is the limit of a subsequence ofðxnÞ is called asubsequential limit ofðxnÞ. We letSdenote the set of all subsequential limits of the bounded sequenceðxnÞ. The setSis bounded, because the sequence is bounded.
For example, ifðxnÞis defined by xn:¼ ð1Þnþ2=n, then the subsequence ðx2nÞ converges to 1, and the subsequenceðx2n1Þconverges to1. It is easily seen that the set of subsequential limits isS¼ f1; 1g. Observe that the largest member of the sequence itself isx2¼2, which provides no information concerning the limiting behavior of the sequence.
An extreme example is given by the set of all rational numbers in the interval [0, 1].
The set is denumerable (see Section 1.3) and therefore it can be written as a sequenceðrnÞ. Then it follows from the Density Theorem 2.4.8 that every number in [0, 1] is a subsequential limit ofðrnÞ. Thus we haveS¼ ½0;1.
A bounded sequence ðxnÞthat diverges will display some form of oscillation. The activity is contained in decreasing intervals as follows. The interval½t1;u1, wheret1 :¼
inffxn:n2Ng and u1 :¼supfxn:n2Ng, contains the entire sequence. If for each m¼1;2;. . ., we definetm:¼inffxn:nmgandum:¼supfxn:nmg, the sequences ðtmÞandðumÞare monotone and we obtain a nested sequence of intervals½tm;um where the mth interval contains them-tail of the sequence.
The preceding discussion suggests different ways of describing limiting behavior of a bounded sequence. Another is to observe that if a real numbervhas the property thatxn>v forat mosta finite number of values ofn, then no subsequence ofðxnÞcan converge to a limit larger than vbecause that would require infinitely many terms of the sequence be larger thanv. In other words, ifvhas the property that there existsNvsuch thatxnvfor all nNv, then no number larger thanvcan be a subsequential limit of ðxnÞ.
This observation leads to the following definition of limit superior. The accompanying definition of limit inferior is similar.
3.4.10 Definition LetX¼ ðxnÞbe a bounded sequence of real numbers.
(a) Thelimit superiorofðxnÞis the infimum of the setVofv2Rsuch thatv<xnfor at most a finite number ofn2N. It is denoted by
lim supðxnÞ or lim supX or limðxnÞ:
(b) Thelimit inferiorofðxnÞis the supremum of the set ofw2Rsuch thatxm<wfor at most a finite number ofm2N. It is denoted by
lim infðxnÞ or lim infX or limðxnÞ:
For the concept of limit superior, we now show that the different approaches are equivalent.
3.4.11 Theorem If ðxnÞ is a bounded sequence of real numbers, then the following statements for a real number xare equivalent.
(a) x¼lim supðxnÞ:
(b) Ife>0,there are at most a finite number of n2N such that xþe<xn,but an infinite number ofn2Nsuch that xe<xn.
(c) If um¼supfxn:nmg;then x¼inffum:m2Ng ¼limðumÞ. (d) If S is the set of subsequential limits ofðxnÞ,then x¼supS.
Proof. (a) implies (b). Ife>0, then the fact thatxis an infimum implies that there exists avinVsuch thatxv<xþe. Thereforexalso belongs toV, so there can be at most a finite number ofn2Nsuch thatxþe<xn. On the other hand,xeis not inV so there are an infinite number ofn2N such thatxe<xn.
(b) implies (c). If (b) holds, givene>0, then for all sufficiently largemwe have um<xþe. Therefore, inffum:m2Ng xþe. Also, since there are an infinite number of n2N such that xe<xn, then xe<um for all m2N and hence xe inffum:m2Ng. Since e>0 is arbitrary, we conclude that x¼inffum:m2Ng. Moreover, since the sequenceðumÞis monotone decreasing, we have infðumÞ ¼limðumÞ. (c) implies (d). Suppose that X0¼ ðxnkÞis a convergent subsequence of X¼ ðxnÞ:
Sincenkk, we havexnkukand hence limX0limðukÞ ¼x:Conversely, there exists n1 such thatu11xn1 u1. Inductively choosenkþ1>nksuch that
uk 1
kþ1<xnkþ1 uk:
Since limðukÞ ¼x, it follows that x¼limðxnkÞ, and hence x2S.
3.4 SUBSEQUENCES AND THE BOLZANO-WEIERSTRASS THEOREM 83
(d) implies (a). Letw¼supS. Ife>0 is given, then there are at most finitely manyn withwþe<xn. Thereforewþebelongs toV and lim supðxnÞ wþe. On the other hand, there exists a subsequence ofðxnÞconverging to some number larger thanwe, so that we is not in V, and hence welim supðxnÞ. Since e>0 is arbitrary, we
conclude thatw¼lim supðxnÞ. Q.E.D.
As an instructive exercise, the reader should formulate the corresponding theorem for the limit inferior of a bounded sequence of real numbers.
3.4.12 Theorem A bounded sequenceðxnÞis convergent if and only if lim supðxnÞ ¼ lim infðxnÞ.
We leave the proof as an exercise. Other basic properties can also be found in the exercises.
Exercises for Section 3.4
1. Give an example of an unbounded sequence that has a convergent subsequence.
2. Use the method of Example 3.4.3(b) to show that if 0<c<1, then limðc1=nÞ ¼1.
3. LetðfnÞ be the Fibonacci sequence of Example 3.1.2(d), and letxn:¼fnþ1=fn. Given that limðxnÞ ¼Lexists, determine the value ofL.
4. Show that the following sequences are divergent.
(a)
1 ð1Þnþ1=n ; (b) ðsinnp=4Þ:
5. LetX¼ ðxnÞandY¼ ðynÞbe given sequences, and let the ‘‘shuffled’’ sequenceZ¼ ðznÞbe defined byz1:¼x1;z2:¼y1;. . .;z2n1:¼xn;z2n:¼yn; :. . .Show thatZis convergent if and only if bothXandYare convergent and limX¼limY.
6. Letxn:¼n1=nforn2N.
(a) Show thatxnþ1<xnif and only ifð1þ1=nÞn<n, and infer that the inequality is valid for n3. (See Example 3.3.6.) Conclude thatðxnÞis ultimately decreasing and thatx:¼ limðxnÞexists.
(b) Use the fact that the subsequenceðx2nÞalso converges toxto conclude thatx¼1.
7. Establish the convergence and find the limits of the following sequences:
(a)
ð1þ1=n2Þn2
; (b)
ð1þ1=2nÞn ; (c)
ð1þ1=n2Þ2n2
; (d)
ð1þ2=nÞn : 8. Determine the limits of the following.
(a)
ð3nÞ1=2n ; (b)
ð1þ1=2nÞ3n :
9. Suppose that every subsequence ofX¼ ðxnÞhas a subsequence that converges to 0. Show that limX¼0.
10. LetðxnÞbe a bounded sequence and for eachn2Nletsn:¼supfxk:kngandS:¼inffsng.
Show that there exists a subsequence ofðxnÞthat converges toS.
11. Suppose thatxn0 for alln2N and that lim
ð1Þnxn exists. Show thatðxnÞconverges.
12. Show that if ðxnÞ is unbounded, then there exists a subsequence ðxnkÞ such that limð1=xnkÞ ¼0.
13. Ifxn:¼ ð1Þn=n, find the subsequence ofðxnÞthat is constructed in the second proof of the Bolzano-Weierstrass Theorem 3.4.8, when we takeI1:¼ ½1;1.
14. LetðxnÞbe a bounded sequence and lets:¼supfxn:n2Ng. Show that ifs2 fx= n:n 2Ng, then there is a subsequence ofðxnÞthat converges tos.
15. LetðInÞbe a nested sequence of closed bounded intervals. For eachn2N, letxn2In. Use the Bolzano-Weierstrass Theorem to give a proof of the Nested Intervals Property 2.5.2.
16. Give an example to show that Theorem 3.4.9 fails if the hypothesis thatXis a bounded sequence is dropped.
17. Alternate the terms of the sequences (1þ1=n) and (1=n) to obtain the sequenceðxnÞgiven by ð2;1;3=2;1=2;4=3;1=3;5=4;1=4;. . .Þ:
Determine the values of lim supðxnÞand lim infðxnÞ. Also find supfxngand inffxng.
18. Show that ifðxnÞis a bounded sequence, thenðxnÞconverges if and only if lim supðxnÞ ¼ lim infðxnÞ.
19. Show that ifðxnÞandðynÞare bounded sequences, then
lim supðxnþynÞ lim supðxnÞ þlim supðynÞ:
Give an example in which the two sides are not equal.