Math 25: Advanced Calculus
Fall 2010
Homework # 7 – Solutions
7.1:
Prove that lim
n
→∞
(
√
n
+ 1

√
n
) = 0.
[Hint:
What would you have done in
calculus?]
Solution:
Multiplying by the conjugate of
√
n
+ 1

√
n
gives
lim
n
→∞
√
n
+ 1

√
n
=
lim
n
→∞
(
√
n
+ 1

√
n
)(
√
n
+ 1 +
√
n
)
√
n
+ 1 +
√
n
=
1
√
n
+ 1 +
√
n
.
Now we use the squeeze theorem. Since
√
n
+ 1+
√
n >
√
n
, it follows that
1
√
n
+1+
√
n
<
1
√
n
. Thus
0
<
1
√
n
+ 1 +
√
n
<
1
√
n
for all
n
∈
N
. We know that lim
n
→∞
1
√
n
= 0, and hence by the squeeze theorem,
lim
n
→∞
√
n
+ 1

√
n
= 0
,
as well.
1
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7.2:
Decide whether or not each of the following statements is true or false.
In either
case, you must justify your answer with a proof. For all of these scenarios, assume that
{
a
n
}
and
{
b
n
}
are sequences of real numbers.
(a). If
{
a
n
}
is unbounded, then
{
a
n
}
diverges to infinity.
FALSE: Let
{
a
n
}
be the sequence defined by
a
n
=
(
0
if
n
is odd
,
n
2
if
n
is even
.
This is the sequence:
0
,
1
,
0
,
2
,
0
,
3
,
0
,
4
,
0
,
5
,
0
,
6
,
. . . .
For any
M >
0, there is a
natural number
n > M
by the Archimedean principle, and hence
a
2
n
> M
. Thus
{
a
n
}
is unbounded. The sequence does not diverge to infinity because for any
n
,
there is a term
a
k
= 0 for some
k > n
.
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 Fall '10
 StevenKlee
 Calculus, Mathematical analysis

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