The Real And Complex Number Systems

(d) If U V, every subsequnce of a n converges to U.
Proof: By(a)and(b),given 0, then there exists a positive integer N 1 such that as
n ≥ N 1 ,wehave
a n U
and there exists a positive integer N 2 such that as n ≥ N 2 ,wehave
U − a n .
Hence, as n ≥ maxN 1 , N 2 ,wehave
U − a n U .
That is, a n is a convergent sequence with limit U. So, every subsequnce of a n
converges to U.
8.2 Given two real-valed sequence a n and b n bounded below. Prove hat
(a) lim sup n→ a n b n ≤ lim sup n→ a n lim sup n→ b n .
Proof: Note that a n and b n bounded below, we have lim sup n→ a n or is
finite. And lim sup n→ b n or is finite. It is clear if one of these limit superior is ,
sowemayassumethatbotharefinite.Leta lim sup n→ a n and b lim sup n→ b n . Then
given 0, there exists a positive integer N such that as n ≥ N, wehave
a n b n a b /2. *
In addition, let c lim sup n→ a n b n ,wherec by (*). So, for the same 0, and
given m N there exists a positive integer K such that as K ≥ N, wehave
c − /2 a K b K . **
By (*) and (**), we obtain that
c − /2 a K b K a b /2
which implies that
c ≤ a b
since is arbitrary. So,
lim sup a n b n ≤ lim sup a n lim sup b n.
n→ n→ n→
Remark: (1) The equality may NOT hold. For example,
a n −1 n and b n −1 n1 .
(2) The reader should noted that the finitely many terms does NOT change the relation
of order. The fact is based on process of proof.
(b) lim sup n→ a n b n ≤ lim sup n→ a n lim sup n→ b n if a n 0, b n 0 for all n, and
if both lim sup n→ a n and lim sup n→ b n are finite or both are infinite.
Proof: Let lim sup n→ a n a and lim sup n→ b n b. It is clear that we may assume
that a and b are finite. Given 0, there exists a positive integer N such that as n ≥ N,
we have
a n b n a b ab a b . *
In addition, let c lim sup n→ a n b n ,wherec by (*). So, for the same 0, and
given m N there exists a positive integer K such that as K ≥ N, wehave
c − a K b K . **
By (*) and (**), we obtain that
c − a K b K a b a b