1
Dr. Bradley
Quoc Phung
230A
Theorem Limit of Sequences
Theorem:
Suppose sn and
lim k .sn k .s
lim( sn tn ) s t
lim( sn .tn ) s.t
lim
tn are real sequences with
lim sn s , lim tn t
x
x
x
x
x
x
1 1
sn s
Sn 0 and
Provide
S 0
Prove: lim( sn .tn ) s.t
x
Let > 0.
sntn st sntn snt snt st
sn tn t t sn s
Since lim tn t ,
x
N1 such that n > N1
tn t
2M
when | Sn | < M for all n.
Since lim sn s ,
x
So.
N2 such that n > N1
sntn st sn tn t t sn s
M
Therefore:
sn s
2
2M
lim( sn .tn ) s.t
x
2
t
2 t 1
Q.E.D.
2 t 1
2
Dr. Bradley
Quoc Phung
230A
Theorem 1:
If sequence { an }is convergent , then { an } is bounded.
Solution:
*
We need to prove
*
| a n| = | a n
Assume { an }is convergent to a
0,
| a n| < M ,
for all n N.
- a+ a| |an -a|+|a|
lim an a
n
N* such that
Let = 1, then N* such that
Therefore,
| a n|
n > N*
| an - a | < 1
|an
|an -a|+|a|
1
Let M = Max { a1 , a2 , … , a n-1 ,
Then | a n |
Therefore:
| an - a | <
=
=
n > N*
- a+ a|
+
|a|
1+|a|}
M
for n > N*
If sequence { an }is convergent , then { an } is bounded. Q.E.D.
3
Dr. Bradley
Quoc Phung
230A
Theorem 2:
If a sequence { an }is bounded, monotonic , then { an } is convergent.
Solution:
* We need to prove:
lim an L
n
0,
N* such that
n > N*
| an - L | <
Suppose a sequence { an }is bounded and monotone increasing.
Let A = { an | n N } be a nonempty bounded set.
Then by the Completeness axiom:
If every nonempty subset A of R is bounded above,
then subset A has a least upper bound.
Therefore
A = { an | n N } has a least upper bound Sup A = L.
Sup A - = L -
Sup A = L
aM
Since L= Sup A, then L- is not Sup A.
So there exist M such that L - aM L
A sequence { an }is monotone increasing
From ( 1 ) and ( 2 )
i e:
0,
that L - a n
M such that
L - aM
aM a n L
(1)
for n > M.
aM a n L L + ( 2)
L+
| an - L | <
n>M
| an - L | <
Therefore: If a sequence { an }is bounded, monotonic , then { an } is convergent Q.E.D.
4
Dr. Bradley
Quoc Phung
230A