CAUCHY SEQUENCE
DR.HAMED AL-SULAMI
Definition 0.1. A sequence {x
n
} of real numbers is said to be Cauchy sequence if for every ε > 0
there exists N ∈ N such that if n, m > N ⇒ |x
n
− x
m
| < ε.
A sequence is Cauchy if the terms eventually get arbitrarily close to each other.
Example 0.1. The sequence {
1
n
} is Cauchy. To see this let ε > 0 be given. Choose N ∈ N such that
1
N
<
ε
2
. Now, if n, m > N ⇒ |
1
n
−
1
m
| ≤
1
n
+
1
m
<
ε
2
+
ε
2
= ε.
Example 0.2. The sequence {
n
n+1
} is Cauchy. To see this let ε > 0 be given. Choose N ∈ N such that
1
N
<
ε
2
. Now, if n, m > N ⇒ |
n
n+1
−
m
m+1
| = |
m+1−n−1
(n+1)(m+1)
| ≤ |
m−n
nm
| <
1
n
+
1
m
<
ε
2
+
ε
2
= ε.
Lemma 1. Let sequence {x
n
} be a Cauchy sequence of real numbers. Then {x
n
} is bounded.
Proof. Since {x
n
} is a Cauchy sequence, then there exists N ∈ N such that if n, m > N ⇒ |x
n
− x
m
| < 3.
if n, m > N ⇒ |x
n
− x
m
| < 3
let m = N + 1, if n > N ⇒ |x
n
− x
N +1
| < 3 Note: |x
n
| − |x
N +1
| ≤ |x
n
− x
N +1
|
⇒ |x
n
| − |x
N +1
| ≤ |x
n
− x
N +1
| < 3
if n > N ⇒ |x
n
| < 3 + |x
N +1
|.
Let M = max{|x
1
|, |x
2
|, · · · |x
N
|, |x
N +1
| + 3}
Now, if n > N ⇒ |x
n
| < 3 + |x
N +1
| ≤ M
Now, if n ≤ N ⇒ |x
n
| < max{|x
1
|, |x
2
|, · · · |x
N
|} ≤ M
Thus ∀ n ∈ N, |x
n
| ≤ M.
¤
Date: April 1, 2006.
1
Cauchy Sequence Dr.Hamed Al-Sulami
Theorem 0.1. [Cauchy Convergence Criterion]
A sequence of real numbers is convergent if and only if it is a Cauchy sequence.
Proof. Let {x
n
} be a sequence of real numbers.
(⇒) Suppose that lim
n→∞
x
n
= x ∈ R. We want to show that {x
n
} is Cauchy sequence.
Let ε > 0 be given. Since lim
x→∞
x
n
= x ∴ ∃ N ∈ N 3
if n > N ⇒ |x
n
− x| <
ε
2
Also, if m > N ⇒ |x
m
− x| <
ε
2
.
Now, if n, m > N ⇒ |x
n
− x
m
| = |x
n
− x + x − x
m
| ≤ |x
n
− x| + |x
m
− x| <
ε
2
+
ε
2
= ε.
Thus {x
n
} is a Cauchy sequence .
(⇐) Suppose that {x
n
} is a Cauchy sequence.We want to show that {x
n
} is convergent.
Let ε > 0 be given. Since {x
n
} is a Cauchy sequence, then by Lemma 1 it is bounded.
Hence {x
n
} has a converge subsequence {x
n
k
}. Suppose lim
k→∞
x
n
k
= x ∈ R.
There exist N
1
, N
2
∈ N 3 if n, m > N
1
⇒ |x
n
− x
m
| <
ε
2
and, if k > N
2
⇒ |x
n
k
− x| <
ε
2
.
Now, fix k > N
2
such that n
k
> N
1
and, if n > N
1
⇒ |x
n
− x
n
k
| <
ε
2
and |x
n
k
− x| <
ε
2
.
Now, if n > N
1
⇒ |x
n
− x| = |x
n
− x
n
k
+ x
n
k
− x| ≤ |x
n
− x
n
k
| + |x
n
k
− x| <
ε
2
+
ε
2
= ε.
Thus {x
n
} converges . ¤
April 1, 2006 2
c
° Dr.Hamed Al-Sulami
Cauchy Sequence Dr.Hamed Al-Sulami
Example 0.3. Prove that any sequence of real numbers {x
n
} which satisfies
|x
n
− x
n+1
| =
1
5
n
, ∀ n ∈ N is convergent.
Proof.
If m > n ⇒ |x
n
− x
m
| = |x
n
− x
n+1
+ x
n+1
+ x
n+2
+ . . . + x
m−1
− x
m
|
≤ |x
n
− x
n+1
| + |x
n+1
+ x
n+2
| + . . . + |x
m−1
− x
m
|
=
1
5
n
+
1
5
n+1
+ . . . +
1
5
m−1
=
1
5
n−1
µ
1
5
+
1
5
2
+ . . . +
1
5
m−n
=
1
5
n−1
m−n
X
k=1
1
5
k
=
1
5
n−1
µ
1 −
1
5
m−n
Note that:
µ
1 −
1
5
m−n
< 1
<
1
5
n−1
.
Let ε > 0 be given, choose N ∈ N such that
1
5
n−1
< ε.
Now, if n, m > N ⇒ |x
n
− x
m
| <
1
5
n−1
< ε.
Thus {x
n
} is convergent. ¤
April 1, 2006 3
c
° Dr.Hamed Al-Sulami