Sequences and Series
2.3 Limit superior, limit inferior, and Bolzano–Weierstrass Note: 1–2 lectures, alternative proof of BW optional
In this section we study bounded sequences and their subsequences. In particular, we define the so-called limit superior and limit inferior of a bounded sequence and talk about limits of subsequences. Furthermore, we prove the Bolzano–Weierstrass theorem , which is an indispensable tool in analysis.
We have seen that every convergent sequence is bounded, although there exist many bounded divergent sequences. For example, the sequence{(−1)n}is bounded, but it is divergent. All is not lost however and we can still compute certain limits with a bounded divergent sequence.
2.3.1
Upper and lower limits
There are ways of creating monotone sequences out of any sequence, and in this fashion we get the so-calledlimit superiorandlimit inferior. These limits always exist for bounded sequences.
If a sequence{xn}is bounded, then the set{xk:k∈N}is bounded. Then for everynthe set
{xk:k≥n}is also bounded (as it is a subset).
Definition 2.3.1. Let{xn}be a bounded sequence. Define the sequences{an}and{bn}byan:= sup{xk:k≥n}andbn:=inf{xk:k≥n}. Define, if the limits exist,
limsup
n→∞ xn:=nlim→∞an, liminf
n→∞ xn:=nlim→∞bn.
For a bounded sequence, liminf and limsup always exist (see below). It is possible to define liminf and limsup for unbounded sequences if we allow∞and −∞. It is not hard to generalize the following results to include unbounded sequences, however, we first restrict our attention to bounded ones.
Proposition 2.3.2. Let{xn}be a bounded sequence. Let anand bnbe as in the definition above. (i) The sequence{an}is bounded monotone decreasing and{bn}is bounded monotone increas-
ing. In particular,liminfxnandlimsupxnexist. (ii) limsup
n→∞ xn=inf{an:n∈N}andliminfn→∞ xn=sup{bn:n∈N}. (iii) liminf
n→∞ xn≤limsupn→∞ xn.
Proof. Let us see why{an}is a decreasing sequence. Asanis the least upper bound for{xk:k≥n}, it is also an upper bound for the subset{xk:k≥(n+1)}. Thereforean+1, the least upper bound for {xk:k≥(n+1)}, has to be less than or equal toan, that is,an≥an+1. Similarly (an exercise),bn is an increasing sequence. It is left as an exercise to show that ifxnis bounded, thenanandbnmust be bounded.
The second item in the proposition follows as the sequences{an}and{bn}are monotone.
∗Named after the Czech mathematician (1781–1848), and the German mathematician (1815–1897).
For the third item, note thatbn≤an, as the inf of a nonempty set is less than or equal to its sup. The sequences {an} and {bn} converge to the limsup and the liminf respectively. Apply
to obtain lim n→∞bn≤nlim→∞an. limsup n→∞ xn liminf n→∞ xn ⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄ ⋄⋄⋄ ⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄⋄
Figure 2.5:First 50 terms of an example sequence. Termsxnof the sequence are marked with dots (•),
anare marked with circles (◦), andbnare marked with diamonds ().
Example 2.3.3: Let{xn}be defined by xn:=
(n+1
n ifnis odd, 0 ifnis even.
Let us compute the liminf and limsup of this sequence. See also . First the limit inferior: liminf
n→∞ xn=nlim→∞ inf{xk:k≥n}
= lim n→∞0=0. For the limit superior we write
limsup
n→∞ xn=nlim→∞ sup{xk:k≥n}
.
It is not hard to see that
sup{xk:k≥n}=
(n+1
n ifnis odd, n+2
n+1 ifnis even. We leave it to the reader to show that the limit is 1. That is,
limsup
n→∞ xn=1. Do note that the sequence{xn}is not a convergent sequence.
limsup n→∞ xn
liminf
n→∞ xn ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄
Figure 2.6:First 20 terms of the sequence in . The marking is the same as in .
We associate certain subsequences with limsup and liminf. It is important to notice that{an} and{bn}are not necessarily subsequences of{xn}, nor do they have to even consist of the same numbers. For example, for the sequence{1/n},bn=0 for alln∈N.
Theorem 2.3.4. If{xn}is a bounded sequence, then there exists a subsequence{xnk}such that lim
k→∞xnk =limsupn→∞ xn.
Similarly, there exists a (perhaps different) subsequence{xmk}such that lim
k→∞xmk =liminfn→∞ xn.
Proof. Definean:=sup{xk :k≥n}. Writex:=limsupxn=liman. Define the subsequence as follows. Pickn1:=1 and work inductively. Suppose we have defined the subsequence untilnkfor somek. Now pick somem>nksuch that
a(nk+1)−xm< k1 +1.
We can do this asa(nk+1) is a supremum of the set{xn:n≥nk+1}and hence there are elements
of the sequence arbitrarily close (or even possibly equal) to the supremum. Setnk+1:=m. The subsequence{xnk}is defined. Next we need to prove that it converges and has the right limit.
Note thata(nk−1+1)≥ank(why?) and thatank≥xnk. Therefore, for everyk≥2 we have |ank−xnk|=ank−xnk
≤a(nk−1+1)−xnk <1
Let us show that{xnk}converges tox. Note that the subsequence need not be monotone. Let
ε >0 be given. As{an}converges tox, then the subsequence {ank}converges tox. Thus there
exists anM1∈Nsuch that for allk≥M1we have |ank−x|<
ε
2. Find anM2∈Nsuch that
1 M2 ≤
ε
2.
TakeM:=max{M1,M2,2}and compute. For allk≥Mwe have |x−xnk|=|ank−xnk+x−ank| ≤ |ank−xnk|+|x−ank| < 1 k+ ε 2 ≤ M1 2+ ε 2 ≤ ε 2+ ε 2 =ε. We leave the statement for liminf as an exercise.
2.3.2
Using limit inferior and limit superior
The advantage of liminf and limsup is that we can always write them down for any (bounded) sequence. If we could somehow compute them, we could also compute the limit of the sequence if it exists, or show that the sequence diverges. Working with liminf and limsup is a little bit like working with limits, although there are subtle differences.
Proposition 2.3.5. Let{xn}be a bounded sequence. Then{xn}converges if and only if liminf
n→∞ xn=limsupn→∞ xn. Furthermore, if{xn}converges, then
lim
n→∞xn=liminfn→∞ xn=limsupn→∞ xn.
Proof. Letanandbnbe as in . In particular, for alln∈N,
bn≤xn≤an.
If liminfxn=limsupxn, then we know that{an}and{bn}have limits and that these two limits are the same. By the squeeze lemma ( ),{xn}converges and
lim
n→∞bn=nlim→∞xn=nlim→∞an.
Now suppose{xn}converges tox. We know by that there exists a subsequence {xnk}that converges to limsupxn. As{xn}converges tox, every subsequence converges toxand
Limit superior and limit inferior behave nicely with subsequences.
Proposition 2.3.6. Suppose{xn}is a bounded sequence and{xnk}is a subsequence. Then liminf
n→∞ xn≤liminfk→∞ xnk ≤limsupk→∞
xnk≤limsup n→∞ xn.
Proof. The middle inequality has been proved already. We will prove the third inequality, and leave the first inequality as an exercise.
We want to prove that limsupxnk ≤limsupxn. Defineaj :=sup{xk :k≥ j} as usual. Also
definecj:=sup{xnk:k≥ j}. It is not true thatcj is necessarily a subsequence ofaj. However, as nk≥kfor allk, we have that{xnk:k≥ j} ⊂ {xk:k≥ j}. A supremum of a subset is less than or
equal to the supremum of the set and therefore cj≤aj. We apply to conclude
lim
j→∞cj≤ jlim→∞aj, which is the desired conclusion.
Limit superior and limit inferior are the largest and smallest subsequential limits. If the subsequence in the previous proposition is convergent, then we have that liminfxnk =limxnk = limsupxnk. Therefore,
liminf
n→∞ xn≤klim→∞xnk≤limsupn→∞ xn.
Similarly, we get the following useful test for convergence of a bounded sequence. We leave the proof as an exercise.
Proposition 2.3.7. A bounded sequence{xn}is convergent and converges to x if and only if every convergent subsequence{xnk}converges to x.
2.3.3
Bolzano–Weierstrass theorem
While it is not true that a bounded sequence is convergent, the Bolzano–Weierstrass theorem tells us that we can at least find a convergent subsequence. The version of Bolzano–Weierstrass that we present in this section is the Bolzano–Weierstrass for sequences.
Theorem 2.3.8(Bolzano–Weierstrass). Suppose a sequence{xn}of real numbers is bounded. Then there exists a convergent subsequence{xni}.
Proof. We use . It says that there exists a subsequence whose limit is limsupxn. The reader might complain right now that is strictly stronger than the Bolzano– Weierstrass theorem as presented above. That is true. However, only applies to the real line, but Bolzano–Weierstrass applies in more general contexts (that is, inRn) with pretty much
the exact same statement.
As the theorem is so important to analysis, we present an explicit proof. The following proof generalizes more easily to different contexts.
Alternate proof of Bolzano–Weierstrass. As the sequence is bounded, then there exist two numbers a1<b1such thata1≤xn≤b1for alln∈N.
We will define a subsequence{xni}and two sequences{ai}and{bi}, such that{ai}is monotone
increasing, {bi} is monotone decreasing, ai≤xni ≤bi and such that limai=limbi. That xni
converges follows by the .
We define the sequences inductively. We will always have thatai<bi, and thatxn∈[ai,bi]for infinitely manyn∈N. We have already defineda1andb1. We taken1:=1, that isxn1 =x1.
Suppose that up to some k∈N we have defined the subsequence xn1,xn2, . . . ,xnk, and the
sequences a1,a2, . . . ,ak and b1,b2, . . . ,bk. Let y:= ak+2bk. Clearly ak <y< bk. If there exist infinitely many j∈Nsuch thatxj∈[ak,y], then setak+1:=ak,bk+1:=y, and picknk+1>nk such thatxnk+1 ∈[ak,y]. If there are not infinitely many jsuch thatxj∈[ak,y], then it must be true that
there are infinitely many j∈Nsuch thatxj∈[y,bk]. In this case pickak+1:=y,bk+1:=bk, and picknk+1>nk such thatxnk+1 ∈[y,bk].
We now have the sequences defined. What is left to prove is that limai=limbi. The limits exist as the sequences are monotone. In the construction,bi−aiis cut in half in each step. Therefore, bi+1−ai+1= bi−2ai. By ,
bi−ai=b1−a1 2i−1 . Letx:=limai. As{ai}is monotone,
x=sup{ai:i∈N}.
Lety:=limbi=inf{bi:i∈N}. Sinceai<bifor alli, theny≤x. As the sequences are monotone, then for anyiwe have (why?)
y−x≤bi−ai= b12−i−1a1. Becauseb1−a1
2i−1 is arbitrarily small andy−x≥0, we havey−x=0. Finish by the .
Yet another proof of the Bolzano–Weierstrass theorem is to show the following claim, which is left as a challenging exercise. Claim: Every sequence has a monotone subsequence.
2.3.4
Infinite limits
Just as for infima and suprema, it is possible to allow certain limits to be infinite. That is, we write limxn=∞or limxn=−∞for certain divergent sequences.
Definition 2.3.9. We say{xn}diverges to infinity if for everyK∈R, there exists anM∈Nsuch that for alln≥Mwe havexn>K. In this case we write
lim
n→∞xn:=∞.
Similarly, if for everyK∈Rthere exists anM∈Nsuch that for alln≥Mwe havexn<K, we say
{xn}diverges to minus infinityand we write lim
n→∞xn:=−∞.
∗Sometimes it is said that
With this definition and allowing∞and−∞, we can write limxnfor any monotone sequence. Proposition 2.3.10. Suppose{xn}is a monotone unbounded sequence. Then
lim n→∞xn=
(
∞ if{xn}is increasing, −∞ if{xn}is decreasing.
Proof. The case of monotone increasing follows from part c) below. Let us do monotone decreasing. Suppose{xn}is decreasing and unbounded, that is, for every K∈R,
there is anM ∈N such that xM <K. By monotonicity xn≤xM <K for all n≥M. Therefore, limxn=−∞. Example 2.3.11: lim n→∞n=∞, nlim→∞n 2=∞, lim n→∞−n=−∞. We leave verification to the reader.
We may also allow liminf and limsup to take on the values∞and−∞, so that we can apply liminf and limsup to absolutely any sequence, not just a bounded one. Unfortunately, the sequences {an}and{bn}are not sequences of real numbers but of extended real numbers. In particular,an can equal∞for somen, andbncan equal−∞. So we have no definition for the limits. But since the extended real numbers are still an ordered set, we can take suprema and infima.
Definition 2.3.12. Let {xn} be an unbounded sequence of real numbers. Define sequences of extended real numbers byan:=sup{xk:k≥n}andbn:=inf{xk:k≥n}. Define
limsup
n→∞ xn:=inf{an:n∈N}, and liminfn→∞ xn:=sup{bn:n∈N}.
This definition agrees with the definition for bounded sequences whenever liman or limbn makes sense including possibly∞and−∞.
Proposition 2.3.13. Let{xn}be an unbounded sequence. Define{an}and{bn}as above. Then {an}is decreasing, and{bn}is increasing. If anis a real number for every n, then limsupxn= liman. If bnis a real number for every n, thenliminfxn=limbn.
Proof. As before, an=sup{xk :k ≥n} ≥sup{xk :k ≥n+1}= an+1. So {an} is decreasing. Similarly,{bn}is increasing.
If the sequence{an}is a sequence of real numbers then liman=inf{an:n∈N}. This follows
from if {an} is bounded and if {an} is unbounded. We proceed similarly with{bn}.
The definition behaves as expected with limsup and liminf, see exercises and . Example 2.3.14: Supposexn:=0 for oddnandxn:=nfor evenn. Thenan=∞for everyn, since for anyM, there exists an evenksuch thatxk=k≥M. On the other hand,bn=0 for alln, as for anyn,{bk:k≥n}consists of 0 and nonnegative numbers. So,
lim
2.3.5
Exercises
Exercise2.3.1: Suppose{xn}is a bounded sequence. Define an and bn as in . Show that
{an}and{bn}are bounded.
Exercise2.3.2: Suppose{xn}is a bounded sequence. Define bnas in . Show that{bn}is an
increasing sequence.
Exercise2.3.3: Finish the proof of . That is, suppose{xn}is a bounded sequence and{xnk}
is a subsequence. Proveliminf
n→∞ xn≤liminfk→∞ xnk. Exercise2.3.4: Prove . Exercise2.3.5:
a) Let xn:=(−1) n
n , findlimsupxnandliminfxn. b) Let xn:=(n−1)(−1)
n
n , findlimsupxnandliminfxn.
Exercise2.3.6: Let{xn}and{yn}be bounded sequences such that xn≤ynfor all n. Then show that
limsup
n→∞ xn≤limsupn→∞ yn
and
liminf
n→∞ xn≤liminfn→∞ yn. Exercise2.3.7: Let{xn}and{yn}be bounded sequences.
a) Show that{xn+yn}is bounded.
b) Show that
(liminf
n→∞ xn) + (liminfn→∞ yn)≤liminfn→∞ (xn+yn).
Hint: Find a subsequence{xni+yni}of{xn+yn}that converges. Then find a subsequence{xnmi}of
{xni}that converges. Then apply what you know about limits.
c) Find an explicit{xn}and{yn}such that
(liminf
n→∞ xn) + (liminfn→∞ yn)<liminfn→∞ (xn+yn).
Hint: Look for examples that do not have a limit.
Exercise2.3.8: Let{xn}and{yn}be bounded sequences (from the previous exercise we know that{xn+yn}
is bounded). a) Show that
(limsup
n→∞ xn) + (limsupn→∞ yn)≥limsupn→∞ (xn+yn).
Hint: See previous exercise.
b) Find an explicit{xn}and{yn}such that
(limsup
n→∞ xn) + (limsupn→∞ yn)>limsupn→∞ (xn+yn).
Exercise2.3.9: If S⊂Ris a set, then x∈Ris acluster pointif for everyε>0, the set(x−ε,x+ε)∩S\ {x} is not empty. That is, if there are points of S arbitrarily close to x. For example, S:={1/n:n∈N}has a
unique (only one) cluster point0, but0∈/S. Prove the following version of the Bolzano–Weierstrass theorem:
Theorem. LetS⊂Rbe a bounded infinite set, then there exists at least one cluster point ofS.
Hint: If S is infinite, then S contains a countably infinite subset. That is, there is a sequence{xn}of