Proving Cauchy Sequence with Triangle Inequality

In summary, we were given a sequence in n-dimensional real numbers that satisfies the condition that the sum of the distances between adjacent points is less than infinity. We need to show that this sequence is Cauchy, which means that the distances between the points get smaller and smaller. This can be proven by using the definition of a limit, where we can show that the sequence converges to a limit L. From there, we can use the fact that every convergent sequence in ℝn is also a Cauchy sequence, or we can use the definition of a Cauchy sequence directly. In either case, we can find natural numbers N and M such that the definition of a Cauchy sequence is satisfied for all n, m
  • #1
SpringPhysics
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0

Homework Statement


If a sequence {xn} in ℝn satisfies that sum || xn - xn+1 || for n ≥ 1 is less than infinity, then show that the sequence is Cauchy.


Homework Equations


The triangle inequality?


The Attempt at a Solution


|| xm - xn || ≤ || Ʃ (xi+1 - xi) from i=n to m-1||
Using the triangle inequality and the given condition, I only get that the norm is less than infinity. I do not know how to transform this into an ε argument. Is there a property of finite sums of telescoping norms that would help?
 
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  • #2
every convergent sequence in a metric space is a cauchy sequence
so you could either take that as a theorem or prove it (it's pretty much a one liner)
so you just want to show that your series converges
 
  • #3
I don't see how to show that the series converges if I am only given that the sum is finite. All I get from subtracting partial sums is that the norm of the difference is yet again finite...
 
  • #4
If the sum of the distances between adjacent points converges then the distances must form a monotonically decreasing sequence such that [itex]\lim \limits_{n\to\infty}||x_n - x_{n+1}|| = 0[/itex] since distances are always non-negative.

What is the definition of a Cauchy sequence?
 
  • #5
So by definition of a limit,
lim n→∞ xn = xn+1
which means that the sequence converges to some limit L.

From here:
- I can use either that any convergent sequence in ℝn must be Cauchy
- or that the above implies that there is some N, M (natural numbers) such that

|| xn - L || < ε/2 for all n > N
|| xm - L || < ε/2 for all m > M

so that the definition of a Cauchy sequence is satisfied for all n, m > max{N, M}.

Is that correct?

By the way...where do you find the code to format the limit? I couldn't find the latex for it.
 
  • #6
SpringPhysics said:
So by definition of a limit,
lim n→∞ xn = xn+1
which means that the sequence converges to some limit L.

From here:
- I can use either that any convergent sequence in ℝn must be Cauchy
- or that the above implies that there is some N, M (natural numbers) such that

|| xn - L || < ε/2 for all n > N
|| xm - L || < ε/2 for all m > M

so that the definition of a Cauchy sequence is satisfied for all n, m > max{N, M}.

Is that correct?

By the way...where do you find the code to format the limit? I couldn't find the latex for it.

A cauchy sequence is a sequence [itex]{p_n}[/itex] such that for every [itex]\epsilon > 0[/itex] there exists an integer [itex]N[/itex] such that [itex]n,m>N[/itex] implies [itex]d(p_n,p_m) < \epsilon[/itex]

This means that the difference between each neighbouring members of the squence get smaller and smaller.

In the reals, suppose [itex]{p_n} \rightarrow P[/itex] then there exists N such that [itex]n,m>N[/itex]
implies [itex]d(P,p_n) < \frac{\epsilon}{2}[/itex] and [itex]d(P,p_m) < \frac{\epsilon}{2}[/itex]
Therefore [itex]d(p_n,p_m) \leq d(P,p_n) + d(P,p_m) < \epsilon [/itex]
So every congergent sequence in R is a cauchy sequence

Can you see now what a cauchy sequence is?
 
  • #7
So that's basically what I said in my previous post, right?
 

1. How do you define a Cauchy sequence?

A Cauchy sequence is a sequence of real numbers where for any positive number, there exists a point in the sequence after which all subsequent elements are within that distance from each other. In other words, the elements of a Cauchy sequence get arbitrarily close to each other as the sequence progresses.

2. What is the importance of proving a sequence is Cauchy?

Proving that a sequence is Cauchy is important because it guarantees that the sequence converges to a limit. This is a fundamental concept in analysis and is used to prove the convergence of many important series and integrals.

3. How do you show a sequence is Cauchy?

To show that a sequence is Cauchy, we must prove that for any positive number, there exists a point in the sequence after which all subsequent elements are within that distance from each other. This can be done by using the definition of a Cauchy sequence and choosing a suitable point in the sequence to start from.

4. What are some common techniques for proving a sequence is Cauchy?

There are various techniques for proving a sequence is Cauchy, including using the triangle inequality, the Bolzano-Weierstrass theorem, and the Cauchy convergence criterion. Other techniques may involve using the limit of a subsequence of the sequence or using the Cauchy-Schwarz inequality.

5. Can a sequence be Cauchy but not convergent?

No, a sequence cannot be Cauchy but not convergent. If a sequence is Cauchy, it must converge to a limit. However, the converse is not always true - a sequence may converge to a limit without being Cauchy.

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