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Homework Help: Show a sequence is Cauchy

  1. Jan 30, 2012 #1
    1. The problem statement, all variables and given/known data
    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.


    2. Relevant equations
    The triangle inequality?


    3. 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?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 30, 2012 #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
     
  4. Jan 31, 2012 #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...
     
  5. Jan 31, 2012 #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?
     
  6. Jan 31, 2012 #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.
     
  7. Jan 31, 2012 #6
    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?
     
  8. Jan 31, 2012 #7
    So that's basically what I said in my previous post, right?
     
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