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Cauchy sequence

  1. Aug 18, 2005 #1
    Question

    Consider the sequence [itex]\{p^n\}_{n\in\mathbb{N}}[/itex]. Prove that this sequence is Cauchy with respect to the p-adic metric on [itex]\mathbb{Q}[/itex]. What is the limit of the sequence?
     
  2. jcsd
  3. Aug 18, 2005 #2
    Solution

    Let [itex]p_n = 1 + p + p^2 + \dots + p^{n-1}[/itex]. Then we have

    [tex]|p_{n+k}-p_n|_p = \left|p^n + p^{n+1} + \dots + p^{n+k-1}\right|_p[/tex]
    [tex]= \left|p^n(1+p+p^2 + \dots + p^{k-1})\right|_p[/tex]
    [tex]= \frac{1}{p^n}[/tex]

    So for any [itex]\epsilon > 0[/itex], we can choose an [itex]N\in\mathbb{N}[/itex] such that [itex]p^N \geq \frac{1}{\epsilon}[/itex], so if [itex]n > N[/itex] we have

    [tex]|p_{n+k} - p_n|_p < \frac{1}{p^N} \leq \epsilon[/tex]

    Therefore [itex]\{p^n\}_{n\in\mathbb{N}}[/itex] is Cauchy.
     
  4. Aug 18, 2005 #3
    ii) Since

    [tex]|p^n|_p = \frac{1}{p^n} \rightarrow 0 \quad \mbox{as} \quad n\rightarrow \infty[/tex]

    the limit

    [tex]\lim_{n\rightarrow\infty}^p p^n = 0[/tex]

    Hence this sequence is actually a null sequence with respect to the p-adic norm.
     
  5. Aug 19, 2005 #4

    shmoe

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    Science Advisor
    Homework Helper

    You've shown the corresponding sequence of partial sums is Cauchy, not the sequence itself. In your other p-adic thread you showed the terms, not the partial sums, go to zero. You seem to be mixing the sequence itself with the sequence of partial sums (or you swapped your replies).
     
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