E
xercise
1.57
(1) The sequence diverges to
1
, there is no convergent subsequence.
(2) The even subsequence converges to 2 and the odd subsequence converges to

2. By the
argument similar to Example 2.4.1, the limits of all convergent subsequences are 2 and

2.
(3) Since
(

1)
n
2
n
(
n
+ 1)
(
p
n
+ 2)
3
>
2
n
2
(2
p
n
)
3
>
p
n
4
and lim
p
n
4
=
1
, we see that lim
(

1)
n
2
n
(
n
+ 1)
(
p
n
+ 2)
3
=
1
. Consequently, all the subsequences
will diverge to
1
(see Exercise 1.31), and no subsequence converges.
(4) The even subsequence converges to 1 and the odd subsequence converges to

1. By the
argument similar to Example 1.5.3, the limits of all convergent subsequences are 1 and

1.
(5) The sequence is a combination of six sequences.
x
6
k
= 0
,
x
6
k
+1
=
p
3(6
k
+ 1)
4
,
x
6
k
+2
=
p
3(3
k
+ 1)
2((

1)
k
+1
(3
k
+ 1) + 1)
,
x
6
k
+3
= 0
,
x
6
k
+4
=

p
3(3
k
+ 2)
2((

1)
k
+1
(3
k
+ 2) + 1)
,
x
6
k
+5
=

p
3(6
k
+ 5)
4
.
Both
x
6
k
and
x
6
k
+3
converge to 0. Both
x
6
k
+1
and
x
6
k
++5
diverge to
1
. Both
x
6
k
+2
and
x
6
k
+4
are made up of two subsequences with limits
p
3
2
and

p
3
2
. Then by the argument similar to
Example 1.5.3, all the limits of convergent subsequences are 0,
p
3
2
and

p
3
2
.
(6) The even subsequence converges to 0 and the odd subsequence converges to 1.
The
limits of all convergent subsequences are 0 and 1.
(7) The even subsequence converges to 1 and the odd subsequence diverges to +
1
. The
only limit of all convergent subsequences is 1.
(8) 2 and 3.
E
xercise
1.58
Any
l
2
LIM
{
x
n
}
is the limit of a subsequence of
x
n
. This subsequence is also a subsequence
of
z
n
.
Therefore
l
2
LIM
{
z
n
}
.
This shows LIM
{
x
n
}
⇢
LIM
{
z
n
}
.
By the same reason,
LIM
{
y
n
}
⇢
LIM
{
z
n
}
.
On the other hand, any
l
2
LIM
{
z
n
}
is the limit of a subsequence
z
n
k
of
z
n
. By the con
struction of
z
n
, the subsequence
z
n
k
either contains infinitely many terms from
x
n
, or contains
infinitely many terms from
y
n
. If
z
n
k
contains infinitely many terms from
x
n
, then by picking
those terms of
z
n
k
that are from
x
n
, we get a sequence
z
n
kp
that is a subsequence of
z
n
k
and also
a subsequence of
x
n
. Therefore
l
= lim
k
!1
z
n
k
= lim
p
!1
z
n
kp
2
LIM
{
x
n
}
. The
2
LIM
{
x
n
}
part is due to
z
n
kp
being a subsequence of
x
n
. If
z
n
k
contains infinitely many terms from
y
n
,
then we get
l
2
LIM
{
y
n
}
. Therefore LIM
{
z
n
}
⇢
LIM
{
x
n
}
[
LIM
{
y
n
}
.
E
xercise
1.59
By lim
l
k
=
l
, for any
✏
>
0 and
N
, there is
k > N
, such that

l
k

l

<
✏
. By
l
k
= lim
x
n
p
for some subsequence
x
n
p
, there is
p > N
, such that

x
n
p

l
k

<
✏
. In particular, we proved
that there is
m
=
n
p
≥
p > N
, such that

x
m

l


x
n
p

l
k

+

l
k

l

<
2
✏
.
By Proposition 1.5.3, what we proved shows that
l
is the limit of a convergent subsequence.
E
xercise
1.60
We only need to find the maxima and minima of the limits found in Exercise 1.57.
(1)
lim = lim
3
p
n
=
1
.
(2)
lim
(

1)
n
2
n
+ 1
n
+ 2
= max
{

2
,
2
}
= 2, lim
(

1)
n
2
n
+ 1
n
+ 2
= min
{

2
,
2
}
=

2.
(3)
lim
(

1)
n
2
n
(
n
+ 1)
(
p
n
+ 2)
3
= +
1
, lim
(

1)
n
2
n
(
n
+ 1)
(
p
n
+ 2)
3
=
1
.
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 Spring '16
 Math, Limits, Limit, lim, Limit of a function, Limit superior and limit inferior, subsequence