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Math 310 
hw
5 solutions
10.2, 10.6a, 10.10;
Wednesday, 14 Oct 2009
10.2 Prove Theorem 10.2 for bounded nonincreasing sequences.
Suppose that (
s
n
)
n
is a bounded nonincreasing sequence and set
s
:= inf
{
s
n
:
n
∈
N
}
(Note that
s
∈
R
). We will
prove that
s
n
→
s
. Let
ε >
0. Since
s
+
ε
is not an lower bound, there is
N
∈
N
so that
s
N
< s
+
ε
. Since (
s
n
)
n
is a nonincreasing sequence, for all
n
∈
N
,
n
≥
N
implies
s
n
≤
s
N
. Let
n
∈
N
and suppose
n
≥
N
, then since
s
≤
s
n
we have
s

ε < s
≤
s
n
≤
s
N
< s
+
ε
;
hence,

s
n

s

< ε
and so
s
n
→
s
.
±
10.6a Let (
s
n
)
n
be a sequence such that

s
n
+1

s
n

<
2

n
for all
n
∈
N
.
Prove that (
s
n
)
n
is a Cauchy sequence and hence a convergent sequence.
We will need the following straightforward fact: For
a
∈
(0
,
1) we have
1 +
a
+
···
+
a
k
=
1

a
k
+1
1

a
<
1
1

a
.
Let
ε >
0. By Example 9.7(a), 2

n
→
0 (note that for all
n
∈
N
, 0
<
2

n
<
1). Hence, there is
N
0
such that for
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This note was uploaded on 03/26/2010 for the course MATHEMARIC 131a taught by Professor Janeday during the Spring '10 term at San Jose State.
 Spring '10
 janeday

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