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Prove Cauchy sequence & find bounds on limit

  1. Sep 11, 2005 #1
    Here's the problem statement:

    Prove that [itex]x_1,x_2,x_3,...[/itex] is a Cauchy sequence if it has the property that [itex]|x_k-x_{k-1}|<10^{-k}[/itex] for all [itex]k=2,3,4,...[/itex]. If [itex]x_1=2[/itex], what are the bounds on the limit of the sequence?

    Someone suggested that I use the triangle inequality as follows:

    let [itex]n=m+l[/itex]
    [tex]|a_n-a_m|=|a_{m+l}-a_m|[/tex]
    [tex]|a_{m+l}-a_m|\leq |a_{m+l}-a_{m+l-1}|+|a_{m+l-1}-a_{m+l-2}|+...+|a_{m+1}-a_m|[/tex]

    Now by hypothesis, [itex]|a_k-a_{k-1}|<10^{-k}[/itex], so

    [tex]|a_{m+l}-a_m|<10^{-(m+l)}+10^{-(m+l-1)}+...+10^{-(m+1)}[/tex].

    It looks like we have an [itex]\epsilon[/itex] such that [itex]|a_n-a_m|<\epsilon[/itex]. Before we get to the bounds on the limit, is that correct? Is anything missing?
     
  2. jcsd
  3. Sep 12, 2005 #2

    Tom Mattson

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    You might take it a little further:

    [tex]|a_{m+l}-a_m|<\sum_{i=0}^l 10^{m+l-i}[/tex]

    [tex]|a_n-a_m|<\sum_{i=0}^{n-m} 10^{n-i}[/tex]

    I'll let a real mathematician help you the rest of the way.
     
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