1
Let
N
= max
{
N
1
, N
2
}
. Then

s
n
t
n

st

=

s
n
t
n

s
n
t
+
s
n
t

st
 ≤ 
s
n

t
n

t

+

t

s
n

s
 ≤
M

t
n

t

+

t

s
n

s

<
2
M
+
2

t

=
for all
n > N
. Note:
if
t
= 0,
then

s
n
t
n

st

<
/
2
<
for all
n > N
.
5. (1)
Let
M
be a real number. Since
s
n
→
+
∞
there is a positive integer
N
such that
s
n
> M
for all
n > N
. Since
t
n
≥
s
n
for all
n
,
t
n
> M
for all
n > M
so
t
n
diverges to
+
∞
.
(2)
Modify the proof of (1).
6. Suppose
s
n
→
L, u
n
→
L
and
s
n
≤
t
n
≤
u
n
for all
n
. Let
>
0. There exists a positive
integer
N
1
such that

s
n

L

<
for all
n > N
1
;
that is
L

< s
n
< L
+
when
n > N
1
.
There exists a positive integer
N
2
such that

t
n

L

<
for all
n > N
2
;
that is
L

< t
n
< L
+
when
n > N
2
.
Let
N
= max
{
N
1
, N
2
}
. Then, for all
n > N
,
L

< s
n
≤
t
n
≤
u
n
< L
+
which implies

t
n

L

<
.
Therefore
t
n
→
L
.
Exercises 2.3
1.
(a) True.
Theorem 8.
(b) True.
Theorem 8.
(c) True. A convergent sequence is bounded, monotone or not!!
(d) False.
s
n
=
(

1)
n
n
is convergent (
→
0) but not monotone.
2.
(a)
(

1)
n
n
Cauchy but not monotone.
(b)
{
1
,
2
,
3
,
4
,
· ··}
monotone but not Cauchy.
(c)
{
1
,
0
,
1
,
0
,
· ··}
bounded but not Cauchy.
2
5.
Bounded:
by induction. Claim
0
< s
n
<
3
for all
n
. Let
S
be the set of positive
integers for which the claim is true.
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 Fall '08
 Staff
 Tn, Natural number, Limit of a sequence, Sn