Since z
n
= inf A
n
≤
sup A
n
= y
n
, we have z
n
≤
y
n
for all n
≥
1. Then lim z
n
≤
lim y
n
by
Theorem 2.3.4(ii). Thus lim inf a
n
≤
lim sup a
n
. If a
n
= (1)
n
, we clearly have z
n
= 1, y
n
= 1 for
all n
≥
1. Thus lim inf a
n
¡ lim sup a
n
in this case.
(d)
Show that lim inf a
n
= lim sup a
n
if and only if lim a
n
exists. In this case, all three share
the same value.
We have z
n
≤
a
n
≤
y
n
for all n
≥
1. If lim inf a
n
= lim sup a
n
, we have lim z
n
= lim y
n
=
‘
and
then lim a
n
exists and is equal to
‘
by the Squeeze Theorem.
Exercise 2.4.8
For each series, find an explicit formula for the sequence of partial sums and determine if the
series converges.
(b)
∞
∑
n
=
1
1
n
(
n
+
1
)
Let (s
m
) denote the sequence of partial sums of
∞
∑
n
=
0
1
2
n
Then compute s
m
=
1
1
x
1
2
+
1
2
x
1
3
+
1
3
x
1
4
+...+
1
m
x
1
m
+
1
= (
1
1

1
2
) + (
1
2

1
3
) + (
1
3

1
4
) + . . .
+ (
1
m

1
m
+
1
) = 1 
1
m
+
1
. Therefore the Algebraic Limit Theorem implies that (s
m
) converges
and lim
m
→
∞
s
m
= lim
m
→
∞
(1
1
m
+
1
) = 1.
Exercise 2.4.9
Complete the proof of Theorem 2.4.6 by showing that if the series
∞
∑
n
=
1
2
n
b
2
n
diverges, then so does
∞
∑
n
=
1
b
n
Because
∞
∑
n
=
1
2
n
b
2
n
2
Hunter College
MATH 351 Mathematical Analysis I
diverges, its monotone sequence of partial sums (t
k
) must be unbounded. To show that (s
m
) is
unbounded it is enough to show that for all k
∈
N
, there is term s
m
satisfying s
m
≥
t
k
/2. Given
an arbitrary k, we focus our attention on s
2
k
and observe that
s
2
k
= b
1
+ b
2
+ (b
3
+ b
4
) + (b
5
+ b
6
+ b
7
+ b
8
) + ... + (b
2
k

1
+
1
+ ... + b
2
k
)
≥
b
1
+ b
2
+ (b
4
+ b
4
) + (b
8
+ b
8
+ b
8
+ b
8
) + ... + (b
2
k
+ ... + b
2
k
)
= b
1
+ b
2
+ 2b
4
+ 4b
8
+ ... + 2
k

1
b
2
k
= 1/2(2b1+ 2b2+ 4b4 + 8b8 + ... + 2
k
b
2
k
)
=b
1
/2 +t
k
/2.
Because (t
k
) is unbounded, the sequence (s
m
) must also be unbounded and cannot converge.
Therefore,
∞
∑
n
=
1
b
n
diverges.
Exercise 2.5.1
Give an example of each of the following, or argue that such a request is impossible.
(a)
A sequence that has a subsequence that is bounded but contains no subsequence that
converges.
Impossible. Let (b
n
) be a bounded sequence of (a
n
) then by BolzanoWeierstrass Theorem (b
n
)
contains a subsequence that converges.
(b)
A sequence that does not contain 0 or 1 as a term but contains subsequences converging
to each of these values.
Consider a
n
= (0.1, 0.9, 0.01, 0.99, 0.001, 0.999, . . .)
a
n
=
(
1

(
0.1
)
m
,
i f n
=
2
m
,
m
∈
N
,
(
0.1
)
m
,
i f n
=
2
m

1,
m
∈
N
(1)
(2)
(c)
A sequence that contains subsequences converging to every point in the infinite set
{
1,
1/2, 1/3, 1/4, 1/5, . . .
}
.
Consider a
n
= (1, 1, (1/2), 1, (1/2), (1/3), 1, (1/2), (1/3), (1/4), . . .).
(d)
A sequence that contains subsequences converging to every point in the infinite set
{
1,
1/2, 1/3, 1/4, 1/5, . . .
}
, and no subsequences converging to points outside of this set.
Impossible because we can build a subsequence approaching 0 which is outside the set. Choose
a
n
1
from the subsequence approaching 1 with

a
n
1
1

<
1, (i.e.
ε
1
= 1). For k
≥
2, choose term
a
nk
from the subsequence approaching 1/k with

a
nk
(1/k)

<
(1/k), (i.e.
ε
k
= 1/k), and also
require n
k
>
n
k
1
. Then (a
n
k
) is a subsequence with

a
nk

1
k

<
1
k
,
1
k

1
k
<
a
nk
<
1
k
+
1
k
, so

a
nk
0

<
2
k
, which implies lim a
nk
= 0.
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 Spring '11
 HaHe
 Mathematical analysis, Limit of a sequence