14. Let
vextendsingle
vextendsingle
vextendsingle
vextendsingle
1
n

1
n
+ 1

0
vextendsingle
vextendsingle
vextendsingle
vextendsingle
=
vextendsingle
vextendsingle
vextendsingle
vextendsingle
1
n

1
n
+ 1
vextendsingle
vextendsingle
vextendsingle
vextendsingle
=
vextendsingle
vextendsingle
vextendsingle
vextendsingle
n
+ 1

n
n
(
n
+ 1)
vextendsingle
vextendsingle
vextendsingle
vextendsingle
=
1
n
2
+
n
<
1
n
2
<
1
n
. Since
1
n
→
0 as
n
→ ∞
,
the result follows by Theorem 16.8.
1
17. Let
s
n
=
k
n
/n
!.
(a)Consider the sequence of ratios:
18. Given that
(
s
n
)
is a convergent sequence. Suppose
s
n
→
s
.
Let
(
t
n
)
be
the sequence
(
t
n
) = (
b, b, b, b, . . .
).
Then
t
n
→
b
.
Since
s
n
≤
b
for all
n
,
it follows that
s
≤
b
by Theorem 17.4.
Now let
(
u
n
)
be the sequence
(
u
n
) = (
a, a, a, a, . . .
). Then
u
n
→
a
. Since
a
≤
s
n
for all
n
,
it follows that
a
≤
s
,
again by Theorem 17.4.
Section 18.
3.
(a)
s
1
= 1 and
s
n
+1
=
1
4
(
s
n
+ 5)
for all
n >
1.
Note:
s
2
=
3
2
, s
3
=
13
8
, s
4
=
53
32
. . .
.
(
s
n
) is increasing: By induction –
1
∈
S
:
s
2
=
1
4
(
s
1
+ 5) =
1
4
(6) =
6
4
=
3
2
>
1 =
s
1
. Therefore
1
∈
S
.
Assume that
k
∈
S
. That is, assume
s
k
+1
≥
s
k
.
Prove that
k
+ 1
∈
S
. That is, prove that
s
k
+2
≥
s
k
+1
.
You've reached the end of your free preview.
Want to read all 5 pages?
 Fall '08
 Staff
 Tn, Natural number, Sn