Proof.
Let
S
=
{
n
2
N

P
(
n
) is false.
}
. To derive a contradiction, we suppose that
S
is not
empty. By the Well Ordering Principle (Corollary 1.13),
S
has a minimal element. Call the
minimal element (
k
+ 1), so (
k
+ 1)
2
S
⇢
N
but for any
n
k
,
n
62
S
. Therefore,
P
(
k
) is
true. But, by (2),
P
(
k
) true implies
P
(
k
+1) is also true, and this contradicts (
k
+1)
2
S
.
From the proof, we can see that the following version of induction is also verified:
If we can show both:
(1)
P
(1)
is true; and
(2) For every
k
2
N
, if
P
(
n
)
is assumed to be true for all
n
k
, then
P
(
k
+ 1)
is true,
then
P
(
n
)
is true for all
n
2
N
.
Sometimes this form of induction is needed (for example, for sequences defined by iteration
involving several previous terms.) This is discussed in section 1.2 of the textbook.
22
2.3
Divergent Sequences and Subsequences
Although we prefer sequences which converge, there are also many interesting things to learn
about divergent sequences.
Sequences can diverge in various ways, some more interesting
than others.
The simplest kind of divergence is called
proper divergence
in the book.
Definition 2.15
(Properly divergent sequences)
.
Let
(
x
n
)
n
2
N
be a sequence in
R
.
(a) We say the sequence properly diverges to
+
1
if:
8
a >
0
9
K
2
N
so that
x
n
> a
8
n
≥
K.
We write
lim
n
!1
x
n
= +
1
or
x
n
!
n
!1
+
1
, even though the sequence does not converge.
(b) We say the sequence properly diverges to
1
if:
8
b <
0
9
K
2
N
so that
x
n
< b
8
n
≥
K.
We write
lim
n
!1
x
n
=
1
or
x
n
!
n
!1
1
, even though the sequence does not converge.
So a sequence diverges to infinity if, for any fixed number
a >
0,
x
n
is
eventually always
larger than
a
.
Example 2.16.
Let
x
n
=
p
n
(3 + sin(
n
)),
n
2
N
. Then
x
n
!
n
!1
+
1
.
Choose any
a >
0. Since sin(
n
)
≥ 
1 for any
n
, we have
x
n
≥
p
n
(3

1) = 2
p
n > a
provided
n >
a
2
4
. So, take
K
2
N
with
K >
a
2
4
. If
n
≥
K >
a
2
4
, by the above calculation,
x
n
> a
, and by definition
x
n
!
n
!1
+
1
.
Proposition 2.17.
If
(
x
n
)
n
2
N
is monotone increasing and not
bounded above, then
x
n
!
n
!1
+
1
(properly divergent).
If
(
x
n
)
n
2
N
is monotone decreasing and not
bounded below, then
x
n
!
n
!1
1
(properly
divergent).
Proof.
Take any
a >
0. If (
x
n
)
n
2
N
is unbounded above, then in particular
a
is not an upper
bound for the sequence, so there exists
K
2
N
so that
a < x
K
.
Since the sequence is
monotone increasing,
x
K
x
n
for all
n
≥
K
, and therefore
a < x
K
< x
n
8
n
≥
K.
By definition,
x
n
!
n
!1
+
1
.
The second statement is similar, and is left as an exercise.
23
K
is
a
natural
every a
o
there
exists
ke
natural
so
that k
da
every
n
K
Xa
a
Xn
Lrn
a
n
I
4
n
K I
I
Example 2.18.
Let
r >
1 and define the sequence
x
n
=
r
n
,
n
2
N
. Then,
x
n
+1
=
r
n
+1
=
r x
n
> x
n
8
n
2
N
, so (
x
n
)
n
2
N
is (strictly) monotone increasing. In particular, notice that
x
n
>
1
8
n
.
We claim that the sequence is not bounded above, in which case we can apply Propo
sition 2.17 to conclude
x
n
!
n
!1
+
1
.
We argue by contradiction, and suppose that (
x
n
)
is bounded above. In that case, by the Monotone Sequence Theorem,
x
n
=
r
n
!
n
!1
x
for
some
x
2
R
. But we pass to the limit in the equation
x
n
+1
=
r x
n
, to obtain the equation
x
=
r x
. Since
r >
1, we must have
x
= 0. But, by Theorem 2.4,
x
≥
1, a contradiction.