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

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The discussion revolves around demonstrating that the sequence s_n, defined as the sum of a specific series in the p-adic metric, is Cauchy while also showing that its equivalence class cannot be represented as a rational number. Participants suggest starting with the definition of a Cauchy sequence and exploring the differences between terms in the sequence. A proposed method involves bounding the difference between terms and utilizing summation techniques similar to those used in proving the Cauchy property for other series. The conversation emphasizes the importance of applying foundational concepts to tackle the problem effectively. Overall, the focus is on finding a structured approach to proving the properties of the sequence s_n.
arturo_026
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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 can't think of how to show that s_n is cauchy.

Thank you very much for any advice and hints.
 
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Well, when in doubt, just start with the definition and see if you can shove your particular case into it. Can you bound the difference between the j-th and k-th terms, for j<k, in terms of j? Maybe by considering the difference between the j-th and (j+1)-st terms, and summing them up, like you do to show the series 2^-n is Cauchy?
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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