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Determining if a sequence is convergent and/or a Cauchy sequence

  1. Nov 12, 2012 #1
    1. The problem statement, all variables and given/known data
    Let {pn}n[itex]\in[/itex]P be a sequence such that pn is the decimal expansion of [itex]\sqrt{2}[/itex] truncated after the nth decimal place.
    a) When we're working in the rationals is the sequence convergent and is it a Cauchy sequence?
    b) When we're working in the reals is the sequence convergent and is it a Cauchy sequence?


    2. Relevant equations
    A sequence {pn}n[itex]\in[/itex]P converges to p if ([itex]\forall\epsilon>0[/itex])([itex]\exists N \in P[/itex])([itex]\forall n\geq N[/itex])(| pn-p| < [itex]\epsilon[/itex]).
    It is a Cauchy sequence if ([itex]\forall\epsilon>0[/itex])([itex]\exists N \in P[/itex])([itex]\forall n,m\geq N[/itex])(|pn-pm|< [itex]\epsilon[/itex]).


    3. The attempt at a solution
    I haven't gotten very far with this. Obviously, the sequence converges to p=[itex]\sqrt{2}[/itex]. Thus when working in the rationals, it doesn't converge and when working in the reals it does converge (since [itex]\sqrt{2}[/itex] is a real number but not rational). However, I'm stuck trying to prove this using the mentioned definitions, and can't get anywhere with trying to prove if the sequence is Cauchy.
     
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  3. Nov 12, 2012 #2

    Dick

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    Pick a specific ε, say ε=1/10000. How large does N have to be?
     
  4. Nov 12, 2012 #3
    The thing is, though, I need to show that this is true for all ε, not just an arbitrary one. On the other hand, if I'm trying to prove that the series doesn't converge to a p, then I'd need to show that !(∀ϵ>0)(∃N∈P)(∀n≥N)(| pn-p| < ϵ), or (∃ϵ>0)(∀N∈P)(∃≥N)(|pn-p| [itex]\geq[/itex] ϵ). Just to clarify, P is the set of all positive integers.
     
  5. Nov 12, 2012 #4

    Dick

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    I know. I'm suggesting you think about the N corresponding to specific case ε=0.0001 to get you started. Working it out is really not much different from a general value of ε. And I'm talking about the Cauchy part of the proof here.
     
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