SOLUTIONS FOR HOMEWORK 5
10.1.
(a)
a
n
= 1
/n
. 1 =
a
1
> a
2
> . . . >
0, hence this sequence is
nonincreasing and bounded.
(b)
b
n
= (
−
1)
n
/n
2
. This sequence is bounded, since
−
1
≤
b
n
≤
1 for
any
n
. However, it is not monotone:
b
1
< b
2
, yet
b
2
> b
3
.
(c)
c
n
=
n
5
.
This sequence is nondecreasing (
c
1
< c
2
< . . .
), and
unbounded (by Example 9.7 and Theorem 9.10, lim
c
n
= +
∞
).
(d)
d
n
= sin(
πn/
7). This sequence is bounded:
−
1
≤
d
n
≤
1 for every
n
. However, this sequence is not monotone:
d
n
=
d
n
+14
for every
n
,
hence
d
1
< d
2
, yet
d
2
> d
15
.
(e)
e
n
= (
−
2)
n
. The sequence is not monotone (
e
1
< e
2
,
e
3
< e
2
), and
unbounded (lim

e
n

= +
∞
).
(f)
f
n
=
n/
3
n
. We prove that this sequence is nonincreasing. Indeed,
f
n
+1
f
n
=
1
3
·
n
+ 1
n
=
1
3
parenleftBig
1 +
1
n
parenrightBig
≤
1
3
parenleftBig
1 + 1
parenrightBig
<
1
.
Thus, 1
/
3 =
f
1
> f
2
> f
3
> . . . >
0, and our sequence is bounded.
10.6.
(a) For every
ε >
0 there exists
N
∈
N
s.t.
1
/
2
N
< ε
(see
Example 9.7). We shall show that

s
n
−
s
m

< ε
whenever
n, m > N
.
Indeed, suppose
m > n > N
. By the triangle inequality,

s
m
−
s
n

=
vextendsingle
vextendsingle
vextendsingle
m

1
summationdisplay
k
=
n
(
s
k
+1
−
s
k
)
vextendsingle
vextendsingle
vextendsingle
≤
m

1
summationdisplay
k
=
n

s
k
+1
−
s
k

<
m

1
summationdisplay
k
=
n
1
2
k
=
1
/
2
n
−
1
/
2
m
1
−
1
/
2
<
1
2
n

1
≤
1
2
N
< ε.
(b) For
n
∈
N
let
s
n
=
n
summationdisplay
k
=1
1
k
.
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 Winter '08
 staff
 Logic, Inductive Reasoning, Limit, Metric space, lim sn

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