Augustin Cauchy was not the first to come up with the criteria. Leonhard Euler, for example, used a similiar criteria. It may be that the significance of the criteria became appreciated in the context of Cauchy's great work on the foundations of calculus.
Definition 1: Cauchy Sequence
A sequence si is Cauchy Sequence if and only if given any positive number ε, there exists an integer N such that if m,n are greater than N, then absolute(sm - sn) is less than ε
In other words, elements of the sequence get arbitrarily close to one another.
I will need a few properties of absolute inequalities:
Lemma 1: absolute(a - b) ≤ absolute(a) + absolute(b)
Proof:
(1) Case I: a - b is nonnegative
So abs(a-b) = a - b
If b ≥ 0, then a - b ≤ a ≤ abs(a) + abs(b)
If b is less than 0, then a - b = a + abs(b) ≤ abs(a) + abs(b)
(2) Case II: a - b is negative
So abs(a-b) = -(a-b) = b - a = abs(b - a)
Using step #1, we know that abs(b - a) ≤ abs(b) + abs(a) = abs(a) + abs(b)
So that:
abs(a - b) ≤ abs(a) + abs(b)
QED
Lemma 2: abs(a) - abs(b) ≤ abs(a - b)
Proof:
(1) Case I: a - b is nonnegative so that abs(a - b) = a - b
If a ≥ 0, then abs(a - b) = a - b ≥ abs(a) - abs(b)
NOTE: It is = except for the case where b is negative.
If a is less than 0, then abs(b) is greater than abs(a) and abs(a) - abs(b) must be a negative number.
(2) Case II: a - b is negative and abs(a - b) = -(a - b) = b - a
If a is ≥ 0, then abs(b) is greater than a and abs(a) - abs(b) is a negative number.
If a is less than 0 and b is less than 0, then b - a = b + abs(a) = abs(a) - abs(b) so that abs(a - b) = abs(a) - abs(b)
If a is less than 0 and b ≥ 0, then b - a = b + abs(a) = abs(a) + abs(b) ≥ abs(a) - abs(b).
QED
Here are some properties of Cauchy Sequences which I will use below:
Lemma 3: Any convergent sequences is a Cauchy Sequence
Proof:
(1) Let ai be a convergent sequence (that is, as ai gets larger, it approaches a limit) so that it's limit = L. [See definition 7, here for definition of a convergent sequence]
(2) So from the above definition, we know for any positive number ε, there exists a positive number N such that:
if n is greater than N, then absolute(an - L) is less than ε
(3) So, for a value (1/2)ε, there exists an integer N such that if n is greater than N, absolute(an - L) is less than (1/2)ε
(3) So let's assume that we have two integers m,n both greater than N.
(4) This means that in both cases absolute(am - L) is less than (1/2)ε and absolute(an - L) is less than (1/2)ε
(5) This gives us:
absolute(am - an) = absolute([am - L] - [an - L]) ≤ absolute(am - L) + absolute(an - L) [See Lemma 1 above]
(6) Finally,
absolute(am - L) - absolute(an - L) is less than (1/2)ε + (1/2)ε = ε
QED
Lemma 4: A Cauchy Sequence has a bound
Proof:
(1) Let (ai) be a Cauchy Sequence.
(2) Then, for any positive number ε greater than 0, there is an integer N such that:
for any integer m,n ≥ N, abs(an - am) is less than ε [See Definition of a Cauchy Sequence above]
(3) So that if we ε = 1 (since ε can be any positive number), we have:
abs(an) - abs(am) ≤ abs(an - am) less than 1 for all n,m ≥ N. [See Lemma 2 above]
(4) Let m = N (since m can be any integer ≥ N), then we have:
abs(an) - abs(aN) is less than 1 which means that:
abs(an) is less than abs(aN) + 1 for n ≥ N.
(5) Now, let n = N (since n can be any integer ≥ N), then we have:
abs(aN) - abs(am) is less than 1 which means that:
abs(am) is greater than abs(aN) - 1 for all n ≥ N.
(6) So, for all n, we have:
abs(an) is less than max { abs(a1), ..., abs(aN-1), abs(aN) + 1 }
and
abs(an) is greater than min { abs(an1), ..., abs(aN-1), abs(aN) - 1 }
(7) This shows that for all finite subsets of the sequence, there exists a bound for ai where upper bound = max { abs(a1), ..., abs(aN-1), abs(aN) + 1 } and a lower bound = min { abs(a1, ..., abs(aN-1), abs(aN)-1 }
QED
Lemma 5: If a Cauchy Sequence has a subsequence convergent to b, then the Cauchy sequence itself converges to b.
Proof:
(1) Let an be a Cauchy sequence with the subsequence ain convergent to b.
(2) By the definition of convergence, we know that for a positive number ε/2, there exists an integer M such that for all n ≥ M abs(ain - b) ≤ ε/2.
(3) By the definition of a Cauchy sequence, we know that there exists an integer n0 such that for all m,n ≥ n0, abs(an - am) ≤ ε/2.
(4) Now, if iM (this is the start of the subsequence that converges) is less than n0, we can always find a M' which is greater than M such that iM' ≥ n0.
We can assume this since we are assuming an infinite subsequence.
(5) So for all n ≥ n0, we have:
abs(an - b) = abs(an - aiM' + aiM' - b) ≤ abs(an - aiM') + abs(aiM' - b) [See Lemma 1 above]
(6) Now, abs(an - aiM') + abs(aiM' - b) is less than ε/2 + ε/2 = ε
We know that abs(an - aiM') is less than ε/2 from the definition of a Cauchy sequence.
We know that abs(aiM') is less than ε/2 from the definition of the convergent sequence.
(7) Now putting this all together gives us that:
abs(an - b) ≤ ε
Which by definition (see Definition 7, here) means that lim(an) = b.
QED
Lemma 6: Every real Cauchy sequence is convergent.
Proof:
(1) By Lemma 4 above, every Cauchy sequence is bounded.
(2) So, by the Bolzano-Weierstrass Theorem (see Theorem, here), every Cauchy sequence has a convergent subsequence.
(3) So, by Lemma 5 above, every Cauchy sequence is convergent.
QED
Lemma 7: A sequence of reals converges if and only if it is a Cauchy sequence
Proof:
(1) By Lemma 6 above, we know that a Cauchy sequence is convergent.
(2) By Lemma 3 above, we know that a convergent sequence is a Cauchy sequence.
QED
Here is the Criterion:
Theorem: Cauchy's Criterion
A series ai is convergent (that is, has a finite limit) if and only if for every positive number ε, there exists a positive integer N such that:
for all n greater than N and p ≥ 1:
absolute(an+1 + an+2 + ... + an+p) is less than ε
Proof:
(1) Let sn = ∑ ai
The assumption here is that i ranges from 0 to n.
(2) sn converges if and only if it is a Cauchy Sequence. [See Lemma 7 above]
(3) Assume than sn is a Cauchy Sequence.
(4) Then, for every positive number ε, there exists a number N such that for all integers n,m greater than N, absolute(sm - sn) is less than ε [See Definition 1 above]
(5) Let's assume that m is greater than n.
At this point, we've made no assumption about m or n and this is consistent with our assumption in step #3.
(6) We know that there exists an integer p ≥ 1 such that m = n + p
(7) Based on the definition of sn, we can see that:
absolute(sm - sn) = absolute(sn+p - sn) = absolute(an+1 + an+2 + ... + an+p)
(8) Now, using step #6, we can see that ∑ ai is convergent if and only if the conditions of the given apply.
QED
References
- Somsack Chaitesipaseut, Cauchy's Criterion for Convergence
- Cauchy's Criterion for Convergence, PlanetMath.org
- Cauchy Sequences
- Cauchy Sequences and the Completeness of the Reals, Mathology
10 comments :
Are Cauchy sequences bounded for all numbers or just for real numbers?
Great question. Cauchy sequences are applicable to any metric space.
Wikipedia has a very good article on the subject here.
-Larry
adnan
is every cauchy sequence is convergent?
Hi Adnan,
Yes, every Cauchy sequence, by definition, converges.
If a sequence doesn't converge, then it isn't a Cauchy sequence.
-Larry
Thank you, thank you, thank you. This post is so much easier to understand then my text book.
Finally! Thorough, clear explanations. Thank you so, so much.
is there a condition like if
abs(Xn+1 - Xn)=c^n
and abs(Xn+2 -Xn+1)=c*abs(Xn=1 -Xn)
where c<1 then its a cauchy sequence.
if at all its true then i request you to publish a rigorous proof of this!
Thank you for the post - enjoyed reading it. There's a little typo in the following sentence:
"Here are some properties of Caucy Sequences which I will use below"
(Cauchy's name misspelled).
-Zeljko
Hi Zeljko,
Thanks for noticing. I've updated the post.
-Larry
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