Proving Cauchy Sequence with Triangle Inequality

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Homework Help Overview

The problem involves a sequence {xn} in ℝn and requires demonstrating that if the sum of the norms || xn - xn+1 || for n ≥ 1 is finite, then the sequence is Cauchy. The discussion centers around the properties of sequences and the implications of the triangle inequality.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between the finiteness of the sum of distances and the convergence of the sequence. Some question how to derive an ε argument from the given conditions, while others reference the definition of Cauchy sequences and the properties of convergent sequences.

Discussion Status

The discussion is active, with participants providing insights into the definitions and properties of Cauchy sequences and convergence. There is a recognition of the need to connect the finiteness of the sum to the Cauchy condition, but no explicit consensus has been reached on the best approach to do so.

Contextual Notes

Some participants note the challenge of proving convergence from the given condition and question the implications of the finite sum on the behavior of the sequence. There is also mention of the need for clarity on the definition of a Cauchy sequence.

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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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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
 
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...
 
If the sum of the distances between adjacent points converges then the distances must form a monotonically decreasing sequence such that \lim \limits_{n\to\infty}||x_n - x_{n+1}|| = 0 since distances are always non-negative.

What is the definition of a Cauchy sequence?
 
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.
 
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 {p_n} such that for every \epsilon &gt; 0 there exists an integer N such that n,m&gt;N implies d(p_n,p_m) &lt; \epsilon

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

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

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

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