9 Limit Theorems for Sequences
The theorems of this section allow complicated limits to be evaluated more easily.
Theorem.
Suppose (
x
n
) is a convergent sequence and suppose (
y
n
) is a convergent se
quence. Then the sequence (
x
n
+
y
n
) is convergent. Moreover, if
x
n
→
a
and
y
n
→
b
then
(
x
n
+
y
n
)
→
(
a
+
b
).
proof.
The basic idea is that if
x
n
is close to
a
, and
y
n
is close to
b
, then
x
n
+
y
n
ought
to be close to
a
+
b
.
Let
±>
0. Then there is
N
1
such that
n>N
1
implies

x
n

a

<±/
2. There is also
N
2
such that
n>N
2
implies

y
n

b

<±/
2. Let
N
= max
{
N
1
,N
2
}
. So if
n>N
, then
n>N
1
and
n>N
2
. It follows that

x
n

a

<±/
2 and

y
n

b

<±/
2. So if
n>N
, we have

(
x
n
+
y
n
)

(
a
+
b
)

=

x
n

a
+
y
n

b

≤
x
n

a

+

y
n

b
≤
±/
2+
±/
2=
±
so we are done.
We would like to prove the similar theorem concerning multiplication, but we need to know
about bounded sequences Frst.
DeFnition.
We say (
x
n
)is
bounded
if there is a number
M
such that

x
n
≤
M
for all
n
. We call
M
a
bound
for the sequence. If the sequence is not bounded, we say it is
unbounded
. That is, for any
M
, there is
n
such that

x
n

>M
.
Examples.
x
n
=
2
n
n
+1
is bounded (
M
= 2 works).
The sequence
x
n
=ln
n
is unbounded. ±or example, if we let
M
= 100, we can Fnd
n
such
that ln
n>M
, i.e., ln
n>
100. It turns out
n
=3
×
10
43
works.
Theorem.
If (
x
n
) is convergent, it must be bounded.
proof.
Suppose
x
n
→
L
. Let
±
= 1. There is
N
such that
n>N
implies

x
n

L

<±
=1.
So for
n>N
, we must have

x
n
≤
L

+ 1. (Use

x

y
 ≤ 
x

y

.) So if we let
M
=

L

+ 1, we have a bound for
x
n
, for
n>N
. But it’s possible that

x
k

>

L

+ 1 for
some
k
between 1 and
N
. So we let
M
= max
{
x
1

,

x
2
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 Spring '09
 WEISBART
 Limits, lim, Limit of a sequence, Xn

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