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Show a certain sequence in Q, with p-adict metric is cauchy

  1. Mar 7, 2012 #1
    I left this following question from my excercises last, hoping that solving the others will give me an insight onto how to proceed. But I still don't have a plan on how to start it:

    Consider the sequence s_n = Ʃ (k=0 to n) (t_k * p^k) in Q(rationals) with the p-adic metric (p is prime); where t_n is the sequence 1,2,1,1,2,1,1,1,2,... Show that s_n is Cauchy, but [s_n] (the equivalence class of s_n) cannot be expressed by a rational number.

    As far as what I know: I'm familiar with the p-adic metric, and what a cauchy sequence is. I just cant think of how to show that s_n is cauchy.

    Thank you very much for any advice and hints.
     
  2. jcsd
  3. Mar 8, 2012 #2

    morphism

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    Look at |s_n - s_m|. What power of p will divide this?
     
  4. Mar 8, 2012 #3
    Will it be the p-adic absolute value of the partial series from m+1 to n, so that way if I choose N large enough so p^-N is larger or equal to such series (for any n), and p^-N≥ε , then s_n will satisfy the cauchy criterion.
     
    Last edited: Mar 8, 2012
  5. Mar 8, 2012 #4

    morphism

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    I think you have the right idea.
     
  6. Mar 8, 2012 #5
    Great! thank you very much. I'll work on cleaning it up.
     
  7. Mar 8, 2012 #6

    morphism

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    No problem. By the way, in general, a series ##\sum a_n## converges in ##\mathbb Q_p## iff ##a_n \to 0## in the p-adic metric (compare to the case in ##\mathbb R## or ##\mathbb C##!). The proof of this general statement should be similar to the proof you're writing up.
     
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