•
If (
s
n
) does not converge to any real number then we say that it
diverges
.
•
The simplest example of a sequence is a
constant sequence
. If
C
is a con
stant, let
s
1
=
C, s
2
=
C, s
3
=
C,
· · ·
.
It is clear that this sequence has
limit
C
too, since for any
ǫ >
0,

s
n
−
C

= 0
< ǫ
∀
n
≥
1.
•
Another simple example from calculus is the sequence (
1
n
). This converges
to 0 since given
ǫ >
0, choose
N >
1
ǫ
, then
n
≥
N
⇒ 
1
n
−
0

=
1
n
≤
1
N
< ǫ
.
•
A sequence (
s
n
) is called
bounded
if the set
{
s
n
:
n
∈
N
}
is a bounded set.
That is, there are numbers
m
and
M
such that
m
≤
s
n
≤
M
for all
n
∈
N
.
This is the same as saying that
{
s
n
:
n
∈
N
} ⊂
[
m, M
]. It is easy to see
that this is equivalent to: there exists a number
K
≥
0 such that

s
n
 ≤
K
for all
n
∈
N
. (See the first lines of the last Section.)
Fact 1.
Any convergent sequence is bounded.
Proof:
Suppose that
s
n
→
s
as
n
→ ∞
.
Taking
ǫ
= 1 in the definition of
convergence gives that there exists a number
N
∈
N
such that

s
n
−
s

<
1 whenever
n
≥
N
. Thus

s
n

=

s
n
−
s
+
s
 ≤ 
s
n
−
s

+

s

<
1 +

s

whenever
n
≥
N
. Now let
M
= max
{
s
1

,

s
2

,
· · ·
,

s
N

,
1+

s
}
. We have

s
n
 ≤
M
if
n
= 1
,
2
,
· · ·
, N
, and

s
n
 ≤
M
if
n
≥
N
. So (
s
n
) is bounded.
•
A sequence (
a
n
) is called
nonnegative
if
a
n
≥
0 for all
n
∈
N
. To say that
a nonnegative sequence converges to zero is simply to say that:
∀
ǫ >
0
,
∃
Ns.t. a
n
< ǫ
∀
n
≥
N.
This is because

a
n
−
s

=

a
n
−
0

=

a
n

=
a
n
in this case.
•
The reason we introduce nonnegative sequences, is that the convergence
of a general sequence can be restated as another nonnegative sequence
converging to 0. In fact:
34
MATH 3333–INTERMEDIATE ANALYSIS–BLECHER NOTES
Fact 2.
If (
s
n
) is a general sequence then:
lim
n
s
n
=
s
⇐⇒
lim
n
(
s
n
−
s
) = 0
⇐⇒
lim
n

s
n
−
s

= 0
.
That is, the sequence (
s
n
) converges to
s
if and only if the nonnegative sequence
(

s
n
−
s

) converges to 0.
To prove this, simply look at the
ǫ
definition of lim
n
s
n
=
s
, and compare it
with the
ǫ
definition of lim
n

s
n
−
s

= 0.
•
Next we will prove some facts for nonnegative sequences, which we will
use later, together with Fact 2, to prove many other facts about general
sequences.
Fact 3.
If (
a
n
) and (
b
n
) are nonnegative sequences, with lim
n
a
n
= 0 and
lim
n
b
n
= 0, and if
C
≥
0, then
lim
n
a
n
+
b
n
= lim
n
Ca
n
= 0
.
Also, if lim
n
a
n
= 0 and if (
b
n
) is any bounded sequence, then lim
n
a
n
b
n
= 0.
Proof:
Let
ǫ >
0 be given. Then
∃
N
1
such that
a
n
< ǫ
whenever
n
≥
N
1
. In
fact, replacing
ǫ
by
ǫ/
2, we can say that
∃
N
1
such that
a
n
< ǫ/
2 whenever
n
≥
N
1
.
Similarly,
∃
N
2
such that
b
n
< ǫ/
2 whenever
n
≥
N
2
. Let
N
= max
{
N
1
, N
2
}
, so
that if
n
≥
N
then
n
≥
N
1
and
n
≥
N
2
. Thus if
n
≥
N
then
a
n
+
b
n
< ǫ/
2+
ǫ/
2 =
ǫ
.
We have proved that lim
n
a
n
+
b
n
= 0.
To prove lim
n
Ca
n
= 0, we can assume that
C >
0, because if
C
= 0 this
result is obvious. If
ǫ >
0 is given, then replacing
ǫ
by
ǫ/C
, we can say that
∃
N
3
such that
a
n
< ǫ/C
whenever
n
≥
N
3
. Thus
Ca
n
< ǫ
whenever
n
≥
N
3
. Hence
lim
n
Ca
n
= 0
.