Proof.
Let (
s
n
) be a convergent sequence, and let
s
be the limit, or equivalently
s
= lim
s
n
.
Apply the definition of convergence with
= 1, then we get some large number
N
∈
N
, such that
if
n > N
, then

s
n

s

<
1.
By the triangle inequality,

s
n

s

<
1 =
⇒

s
n
 ≤ 
s
+
s
n

s
 ≤ 
s

+

s
n

s
 ≤ 
s

+ 1
.
Define
M
= max
{
s
1

,

s
2

,
· · ·
,

s
N

,

s

+ 1
}
.
Then we have

s
n
 ≤
M
if
n
≤
N

s

+ 1
≤
M
if
n > N
.
Therefore

s
n
 ≤
M
for all
n
∈
N
, and so
{
s
n
}
is bounded.
Remark
9.3
.
In the proof above, we only apply the defining condition (6.1) for
a single
= 1. Actually, we can take
to be any finite positive number in the
proof, and this is left as an exercise.
Theorem 9.4.
If s sequence
(
s
n
)
converges to
s
, and
k
∈
R
, then the sequence
(
ks
n
)
converges to
ks
. That is,
lim(
ks
n
) =
k
lim
s
n
.
Proof.
If
k
= 0, the proof is trivial as
ks
n
≡
0, and lim(0
·
s
n
) = 0 = 0
·
lim
s
n
.
Assume
k
6
= 0 now. Let
>
0 be any positive real number, and we need to
show that

ks
n

ks

<
,
for large
n
∈
N
.
Note that the above inequality is equivalent to saying that

s
n

s

<

k

,
for large
n
∈
N
.
24
XIN ZHOU
Since lim
s
n
=
s
, for

k

>
0, by (6.1) there exists
N >
0, such that
if
n
≥
N
, then

s
n

s

<

k

.
Therefore for
n > N
, we have

ks
n

ks

<
.
Theorem 9.5.
If
(
s
n
)
converges to
s
and
(
t
n
)
converges to
t
, then
(
s
n
+
t
n
)
converges to
s
+
t
. That is to say,
lim(
s
n
+
t
n
) = lim
s
n
+ lim
t
n
.
Proof.
Let
>
0 be any positive real number, and we want to show that

(
s
n
+
t
n
)

(
s
+
t
)

<
,
for all large
n
∈
N
.
Note that

(
s
n
+
t
n
)

(
s
+
t
)
 ≤ 
s
n

s

+

t
n

t

.
(
Hint
: so we can try to show

s
n

s

<
/
2 and

t
n

t

<
/
2.)
Since lim
s
n
=
s
,
∃
N
1
∈
N
, such that
if
n
≥
N
1
, then

s
n

s

<
/
2
.
Similarly by lim
t
n
=
t
,
∃
N
2
∈
N
, such that
if
n
≥
N
2
, then

t
n

t

<
/
2
.
Let
N
= max
{
N
1
, N
2
}
.
Then if
n > N
, we have

(
s
n
+
t
n
)

(
s
+
t
)
 ≤ 
s
n

s

+

t
n

t

<
.
Theorem 9.6.
If
(
s
n
)
converges to
s
and
(
t
n
)
converges to
t
, then
(
s
n
·
t
n
)
converges to
s
·
t
. That is to say,
lim(
s
n
·
t
n
) = (lim
s
n
)
·
(lim
t
n
)
.
Discussion
:

s
n
t
n

st

=

s
n
t
n

s
n
t
+
s
n
t

st

≤ 
s
n

t
n

t

+

t

s
n

s

≤
M

t
n

t

+

t

s
n

s

.
•
In the above inequality,

t
n

t

and

s
n

s

can be very small for large
n
;
• 
s
n

is bounded by Theorem 9.2, i.e.

s
n
 ≤
M
.
MATH 117 FALL 2017 UCSB
25
Proof.
Let
>
0 be any positive real number. By Theorem 9.2,
∃
M >
0, such
that

s
n
 ≤
M
, for all
n
∈
N
.
Since lim
t
n
=
t
,
∃
N
1
∈
N
, such that
if
n
≥
N
1
, then

t
n

t

<
/
(2
M
)
.
Similarly by lim
s
n
=
s
,
∃
N
2
∈
N
, such that
if
n
≥
N
2
, then

s
n

s

<
2(

t

+ 1)
.
Let
N
= max
{
N
1
, N
2
}
.
Then if
n > N
, we have

s
n
t
n

st
 ≤ 
s
n

t
n

t

+

t

s
n

s

≤
M
·
2
M
+

t
 ·
2(

t

+ 1)
<
.
Lemma 9.7.
If
(
s
n
)
converges to
s
and if
s
n
6
= 0
for all
n
∈
N
, and if
s
6
= 0
,
then
lim
1
s
n
=
1
s
.
Discussion
:
1
s
n

1
s
=
s
n

s
s
n
s
≤

s
n

s

m

s

.
We know that

s
n

s

is very small for
n
very large. To show the right hand
side is small, we need the denominator

s
n
s

to be bounded away from 0. By
Example 7.5, we do have that
m
= inf
{
s
n

:
n
∈
N
}
>
0
.
Proof.
Let
>
0 be any positive real number, and
m
be defined as above.
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 Fall '08
 Akhmedov,A
 Math, Natural Numbers