2
1
Indeed, the sequence (0
,...,
0
,
0
,
2
,
2
,
2
) represents the geometric series 2
∑
∞
i
=
n
+2
3

i
which converges to 3

(
n
+2)
2
1

1
3
= 3

(
n
+1)
which is also represented by (0
0
,
1
,
0
,
0
,
0
).
2
This result is
surprising
. At the
k
th step in constructing
C
, we removed 2
k

1
intervals of
length 3

k
from [0
,
1], and so the total length removed is
∑
∞
k
=1
2
k

1
3
k
=
1
3
∑
∞
k
=0
2
k
3
k
= 1. So, even
though we removed almost all of [0
,
1] in the construction, the resulting “dust”
C
can still be put
into 11 correspondence with [0
,
1].
1
MATH 117 HOMEWORK #6 SOLUTIONS
15.3:
(a) First,
d
2
((
x
1
,y
1
)
,
(
x
2
2
))
≥
0 since both

x
1

x
2
 ≥
0 and

y
1

y
2
 ≥
0.
Furthermore,
d
2
((
x
1
1
)
,
(
x
2
2
)) = 0 if and only if max
{
x
1

x
2

,

y
1

y
2
}
= 0, which happens if and only if

x
1

x
2

= 0 and

y
1

y
2

= 0, or
equivalently (
x
1
1
) = (
x
2
2
).
d
2
is symmetric since taking absolute values is symmetric.
Finally, we show
d
2
satisﬁes the triangle inequality.
d
2
((
x
1
1
)
,
(
x
3
3
)) =
max
{
x
1

x
3

,

y
1

y
3
}
= max
{
x
1

x
2
+
x
2

x
3

,

y
1

y
2
+
y
2

y
3
} ≤
max
{
x
1

x
2

+

x
2

x
3

,

y
1

y
2

+

y
2

y
3
} ≤
max
{
x
1

x
2

,

y
1

y
2
}
+max
{
x
2

x
3

,

y
2

y
3
}
=
d
2
((
x
1
1
)
,
(
x
2
2
))+
d
2
((
x
2
2
)
,
(
x
3
3
)).
Hence,
d
2
is a metric on
R
2
.
(b)

1

1
1
1
Figure 1
16.6:
(a) Let
² >
0. Then

²
k

>
0. Let
N
be suﬃciently large so that
1
N
<

²
k

. Then
for any
n > N
,

k
n

< ²
. Hence, lim
n
→∞
k
n
= 0.
(b) Let
² >
0. Then
²
1
/k
>
0. Let
N
be suﬃciently large so that
1
N
< ²
1
/k
.
Then for
n > N
,
1
n
k
<
1
N
k
< ²
k/k
=
²
. Hence, lim
n
→∞
1
n
k
= 0.
(c) Let
² >
0 and let
N >
1
²
. Then for
n > N
,
ﬂ
ﬂ
ﬂ
ﬂ
3
n
+ 1
n
+ 2

3
ﬂ
ﬂ
ﬂ
ﬂ
<
ﬂ
ﬂ
ﬂ
ﬂ
3
n
+ 1
n

3
ﬂ
ﬂ
ﬂ
ﬂ
=
1
n
<
1
N
< ²
. Hence, lim
n
→∞
3
n
+1
n
+2
= 3.
(d) Let
² >
0 and let
N >
1
²
. Then for
n > N
,
ﬂ
ﬂ
ﬂ
ﬂ
sin
n
n
ﬂ
ﬂ
ﬂ
ﬂ
≤
ﬂ
ﬂ
ﬂ
ﬂ
1
n
ﬂ
ﬂ
ﬂ
ﬂ
<
1
N
< ²
. Hence,
lim
n
→∞
sin
n
n
= 0.
(e) Let
² >
0 and let
N
be suﬃciently large so that
1
N

2
< ²
. Then for
n > N
,
ﬂ
ﬂ
ﬂ
ﬂ
n
+ 2
n
2

3
ﬂ
ﬂ
ﬂ
ﬂ
<
ﬂ
ﬂ
ﬂ
ﬂ
n
+ 2
n
2

4
ﬂ
ﬂ
ﬂ
ﬂ
=
ﬂ
ﬂ
ﬂ
ﬂ
1
n

2
ﬂ
ﬂ
ﬂ
ﬂ
<
1
N

2
< ²
. Hence, lim
n
→∞
n
+2
n
2

3
= 0.
16.8:
(a) The sequence
a
n
= 2
n
is unbounded and thus divergent.
(b) Let
²
=
1
2
and let
b
∈
R
. The
²
=
1
2
neighborhood of
b
may contain either
1 or

1 but it will not contain both. Hence, for any
N
∈
N
, there will be
terms of the sequence
b
n
, with
n > N
, not contained in
N
(
b,
1
2
). Therefore,
b
n
does not converge to any
b
∈
R
which is to say that
b
n
diverges.
(c) Let
²
=
1
4
and use a similar argument as in part (b).
(d) The sequence
d
n
is unbounded and thus divergent.
16.11:
Suppose the sequence
s
n
converges to
s
. Let
² >
0. Since
s
n
→
s
, there
is
N
s
∈
N
such that for
n > N
s
,

s
n

s

< ²
. Let
N
t
> N
s

k
. Then for all
n > N
t
,

t
n

s

=

s
n
+
k

s

< ²
since if
n > N
t
then
n
+
k > N
s
. Therefore,
t
n
→
s
.
The proof of the converse is similar (let
N
s
> N
t
+
k
).
1
MATH 117 HOMEWORK #7 SOLUTIONS
17.3:
(a)
lim
n
→∞
3
n
2
+ 4
n
7
n
2

5
n
= lim
n
→∞
3 +
4
n
7

5
n
=
lim
n
→∞
(3 +
4
n
)
lim
n
→∞
(7

5
n
)
=
3 + 4 lim
n
→∞
1
n
7

5 lim
n
→∞
1
n
=
3
7
(b) Similar to (a). Answer is 0.
You've reached the end of your free preview.
Want to read all 11 pages?
 Fall '08
 Akhmedov,A
 Math