. Completely justify all your claims.
Solution:
First we prove by induction that 2
≤
x
n
<
3 for any
n
∈
N
.
For
n
= 1 this is given by hypothesis. Assume that 2
≤
x
n
<
3 for a given
n
. Then 0
≤
x
n

2
<
1, thus
0
≤
√
x
n

2
<
1, so 2
≤
x
n
+1
= 2 +
√
x
n

2
<
3. Thus 2
≤
x
n
<
3 for any
n
∈
N
, so we proved that
the sequence is bounded.
Next we show that the sequence is increasing. This can be done again by induction, but it is also possible
to do it directly, using the bounds we obtained in the ﬁrst step.
Recall that if
y
∈
R
,
0
≤
y <
1, we have
√
y
≥
y
. Since we showed 2
≤
x
n
<
3, we thus have
0
≤
x
n

2
<
1, so
√
x
n

2
≥
x
n

2. This implies that
x
n
+1
= 2 +
√
x
n

2
≥
x
n
, for an arbitrary
n
∈
N
, thus our sequence is increasing. Note also that if
x
1
= 2, then
x
n
= 2 for any
n
(can be shown
immediately by induction), so the sequence is constant, hence trivially convergent to 2. If
x
1
>
2, then
all inequalities are strict, so our sequence is strictly increasing.
We showed that
{
x
n
}
n
is bounded and increasing, so by the monotone convergence theorem, the sequence
is convergent. Let’s denote the limit with
L
. Taking the limit as
n
→ ∞
in the recursive relation, we get
L
= 2 +
√
L

2. Solving we get two solutions
L
1
= 2 and
L
2
= 3. As we noted above, if
x
1
= 2, the
sequence is constant 2, so the limit is 2. If
x
1
>
2, then 2
< x
1
< x
2
< ... < x
n
< ... <
3, so the limit is
the supremum of the sequence, and the supremum cannot be 2. Thus the limit is 3.
2
Observation:
Another solution, perhaps even shorter, can be obtained by showing from the recursive
relation that
x
n
= 2 + (
x
1

2)
1
/
(2
n
)
,
∀
n
∈
N
. Try to show this and then get the rest of the problem
based on this observation (look also at the Example 2.21, page 45 textbook).
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View Full Document6.
(20 pts) Show that a sequence
{
x
n
}
n
is not bounded from above if and only if there exists a subsequence
{
x
n
k
}
k
such that
x
n
k
→ ∞
as
k
→ ∞
. (Note that you have to prove both implications.)
Solution:
First of all,
{
x
n
}
n
is bounded from above, by deﬁnition, if and only if there exists a constant
M
∈
R
, such that
x
n
≤
M,
∀
n
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 Fall '12
 Draghic
 Mathematical analysis, Order theory, Xn, 1 2k, Dominated convergence theorem, a2 b2

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