Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

688 Mathematical Comments

A.2 Sequences

The following is a short review of sequences of real numbers. In particular the

liminf and limsup of sequences are discussed. As a supplement to this text, the

authors offer a mathematical primer which can be downloaded at the site listed in

the Preface. In addition to the following review of sequences, it contains a brief

review of infinite series, and differentiable and integrable calculus including double

integration. Students that need a review of these concepts can freely download this

supplement.

Let{an}be a sequence of real numbers. Recall from calculus thatan→a

(limn→∞an=a) if and only if

for every>0, there exists anN 0 such thatn≥N 0 =⇒|an−a|<. (A.2.1)

LetAbe a set of real numbers that is bounded from above; that is, there exists
anM∈Rsuch thatx≤Mfor allx∈A. Recall thatais thesupremumofAif
ais the least of all upper bounds ofA. From calculus, we know that the supremum
of a set bounded from above exists. Furthermore, we know thatais the supremum
ofAif and only if, for all>0, there exists anx∈Asuch thata−<x≤a.
Similarly, we can define theinfimumofA.
We need three additional facts from calculus. The first is the Sandwich Theorem.

Theorem A.2.1 (Sandwich Theorem). Suppose for sequences{an}, {bn},and
{cn}thatcn≤an≤bn, for alln, and thatlimn→∞bn= limn→∞cn=a.Then
limn→∞an=a.

Proof:Let>0 be given. Because both{bn}and{cn}converge, we can chooseN 0
so large that|cn−a|<and|bn−a|<,forn≥N 0. Becausecn≤an≤bn,itis
easy to see that
|an−a|≤max{|cn−a|,|bn−a|},

for alln. Hence, ifn≥N 0 ,then|an−a|<.

The second fact concerns subsequences. Recall that{ank}is a subsequence of
{an}if the sequencen 1 ≤n 2 ≤ ···is an infinite subset of the positive integers.
Note thatnk≥k.

Theorem A.2.2.The sequence{an}converges toaif and only if every subsequence
{ank}converges toa.

Proof: Suppose the sequence{an}converges toa.Let{ank}be any subsequence.
Let>0 be given. Then there exists anN 0 such that|an−a|<,forn≥N 0.
For the subsequence, takek′to be the first index of the subsequence beyondN 0.
Because for allk,nk≥k,wehavethatnk≥nk′ ≥k′≥N 0 , which implies that
|ank−a|<.Thus,{ank}converges toa. The converse is immediate because a
sequence is also a subsequence of itself.

Finally, the third theorem concerns monotonic sequences.