(
⇒
) This is a bit harder. We assume that relation (1) is true and we’ll construct inductively a subsequence
{
x
n
k
}
k
which will approach
∞
as
k
→ ∞
.
To pick the first term of our subsequence, apply first (1) with
M
=
M
1
= 1. It follows that there exists
n
1
∈
N
such that
x
n
1
>
1. Now the idea is to apply (1) with a bigger and bigger
M
. But just choosing
M
= 2
, M
= 3, etc. may not work, because we have no guarantee that the ranks
n
2
,
n
3
, etc. which we
get from (1) will be in increasing order.
Here is how we pick the second term of the subsequence. Let
M
2
= max
{
2
, x
1
, x
2
, ..., x
n
1
}
and apply (1)
with
M
=
M
2
. It follows that there exists
n
2
∈
N
such that
x
n
2
> M
2
. From the choice of
M
2
we deduce
two things. First,
x
n
2
> M
2
≥
2. Secondly,
x
n
2
> M
2
≥
x
l
, for any
l
∈ {
1
,
2
, ..., n
1
}
; in particular, we get
that
n
2
> n
1
because otherwise
x
n
2
would be equal to one of the
x
l
,
l
∈ {
1
,
2
, ..., n
1
}
.
Suppose now that we picked
n
1
< n
2
< ... < n
k
, such that
x
n
l
> l
, for any
l
∈ {
1
,
2
, ..., k
}
and we’ll
construct the
k
+ 1th term of the subsequence. Let
M
k
+1
= max
{
2
, x
1
, x
2
, ..., x
n
k
}
and apply (1) with
M
=
M
k
+1
. It follows that there exists
n
k
+1
∈
N
such that
x
n
k
+1
> M
k
+1
. From the choice of
M
k
+1
it
follows as above that
n
k
+1
> n
k
and that
x
n
k
+1
> k
+ 1.
Thus by induction we construct the subsequence
{
x
n
k
}
k
of
{
x
n
}
n
, with the property that
x
n
k
> k
, for
any
k
∈
N
. The (extended) comparison Theorem implies immediately that
x
n
k
→ ∞
as
k
→ ∞
.
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 Fall '12
 Draghic
 Mathematical analysis, Order theory, Xn, 1 2k, Dominated convergence theorem, a2 b2

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