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Proving a sequence is Cauchy?

  1. Jul 8, 2008 #1
    Hello all. I'm having trouble on the following homework problem. It seems like it should be easy, but I'm just now sure how to approach it

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

    Let [tex](s_n)[/tex] be a sequence st [tex]|s_{n+1} - s_n | < 2^{-n}, \forall n \in \mathbb{N}[/tex]

    show that [tex](s_n)[/tex] converges


    3. The attempt at a solution

    well I thought the easiest way to prove it would be to show it's a Cauchy sequence and therefore convergent, but perhaps that's the wrong approach. What's a good starting point for a problem like this?
     
    Last edited: Jul 8, 2008
  2. jcsd
  3. Jul 8, 2008 #2

    Dick

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    A good starting point is to show it's Cauchy. But you already knew that. Just do it. Cauchy means |s_n-s_m|<epsilon for n and m greater than N. The maximum difference between s_n and s_m involves summing a lot of large powers of 1/2. Use the triangle inequality.
     
  4. Jul 8, 2008 #3

    HallsofIvy

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    To show that a sequence is Cauchy you must show that [itex]|a_m- a_n|[/itex] goes to 0 as m and n go to infinity independently (in particular, you cannot assume that m= n+1).

    But [itex]|a_{n+2}- a_n|\le |a_{n+2}- a{n+1}|+ |a{n+1}- a_n|[/itex], [itex]|a_{n+3}- a_n|\le |a_{n+3}- a_{n+2}|+ |a_{n+2}- a{n+1}|+ |a{n+1}- a_n|[/itex], etc.

    You can use the given property on each of those and use induction to show the general case.
     
  5. Jul 8, 2008 #4

    HallsofIvy

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    asaaaa
     
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