Math 310 
hw
4 solutions
8.8a, 8.10, 9.1b;
Wednesday, 30 Sept 2009
8.8 (a) Prove the following:
lim
n
→∞
h
p
n
2
+ 1

n
i
= 0
.
We first observe that
p
n
2
+ 1

n
=
p
n
2
+ 1

n
√
n
2
+ 1 +
n
√
n
2
+ 1 +
n
=
(
n
2
+ 1)

n
2
√
n
2
+ 1 +
n
=
1
√
n
2
+ 1 +
n
.
Let
ε >
0 and set
N
:=
1
ε
. Let
n
∈
N
and suppose that
n > N
. Then we have (note
√
n
2
+ 1 +
n >
0):
p
n
2
+ 1

n

0 =
1
√
n
2
+ 1 +
n
=
1
√
n
2
+ 1 +
n
<
1
n
< ε.
Hence,
√
n
2
+ 1

n
→
0.
8.10 Let (
s
n
)
n
be a convergent sequence and suppose that lim
n
→∞
s
n
> a
. Prove that there exists a number
N
such
that
n > N
implies
s
n
> a
.
Set
s
denote the limit of the convergent sequence (
s
n
)
n
. Then by assumption,
s > a
and so
ε
:=
s

a >
0. By
definition of limit there exists a number
N
such that for all
n
∈
N
,
n > N
implies

s
n

s

< ε
. Let
n
∈
N
and
suppose that
n > N
, then we have
a
=
s

ε < s
n
< s
+
ε.
Hence,
s
n
> a
as desired.
9.1 (b) Using the limit theorems 9.29.6 and 9.7, prove the following. Justify all steps.
lim
n
→∞
3
n
+ 7
6
n

5
=
1
2
.
We will need the facts, the first was proved in class (see also Example 9.7(a)) and the second is obvious,
lim
n
→∞
1
n
= 0
lim
n
→∞
c
=
c
where
c
is a constant. Observe that our sequence may be rewritten as follows:
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 Spring '10
 janeday
 lim, Limit of a sequence, Sn

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