Solving algebraically for s, we obtain
s
=
±
3. Since
s
n
≥
√
6 for all
n
,

3
cannot be the limit. We conclude that lim
s
n
=
s
= 3.
18.11 Finish the proof of Theorem 18.3 for a bounded decreasing sequence.
Proof.
Supposse that (
s
n
) is a bounded decreasing sequence. Let
S
denote
the nonempty bounded set
{
s
n
:
n
∈
N
}
. By the completeness axiom
S
has
a greatest lower bound, and we let
s
= inf
S
. We claim that lim
s
n
=
s
.
Given any
ε >
0
, s
+
ε
is not a lower bound for
S
. Thus there exists an
integer
N
such that
s
N
< s
+
ε
. Furthermore, since (
s
n
) is decreasing and
s
is an upper bound for
S
, we have
s
≤
s
n
≤
s
N
< s
+
ε
for all
n > N
. Hence (
s
n
) converges to s.
4
18.13 Prove Lemma 18.11
Lemma 18.11
Every Cauchy sequence is bounded.
Proof 1.
By Theorem 18.12 we know that every Cauchy sequence is con
vergent and by Theorem 16.13 every convergent sequence is bounded, then
every Cauchy sequence is bounded.
Proof 2.
Applying Definition 18.9 with
ε
= 1 we obtain
N
∈
N
so that
m, n > N
implies

s
n

s
m

<
1
In particular

s
n

s
N
+1

<
1 for all
n > N
, so

s
n

<

s
N
+1

+ 1 for
n > N
.
Let
M
= max
{
s
1

,

s
2

,

s
N

, . . . ,

s
N
+1

+ 1
}
then we have

s
n
 ≤
M
for all
n
∈
N
, so (
s
n
) is bounded.
5
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 Fall '08
 Akhmedov,A
 Mathematical analysis, Sn, Guillermo Cu´llar