I am looking for a sequence in $\Bbb Q$ that converges to $0$ under the Euclidian metric and to $1$ under the 2-adic metric $d_2$. This metric is defined by the 2-adic absolute value as $d_2(x, y) = |x-y|_2 = 2^{-\text{ord}_2(x-y)}$ for $x ≠ y$. As $|3^{-n}|_2 = 2^{-0} = 1$ for all $n ∈ \Bbb N$, this sequence seems to "be the constant sequence $(1)_{n ∈ \Bbb N}$ w.r.t. the 2-adic metric", so my guess would be that the sequence $(3^{-n})_{n ∈ \Bbb N}$ will suffice. However, I run into a problem when I try to run the proof neatly. Indeed, what I want to show is the following:
$$(\forall \varepsilon ∈ \Bbb R_{>0})(\exists N ∈ \Bbb N)(\forall n ∈ \Bbb N)\left(n ≥ N \Rightarrow d_2(1, 3^{-n}) = 2^{-\text{ord}_2(3^{-n} - 1)} < \varepsilon\right), $$
but this suddenly isn't so clear.. The value of $\text{ord}_2(3^{-n} - 1)$ really seems to depend on $n$ in an non-trivial way. Am I even right that this sequence will suffice? If so, how do I prove it correctly?
You can take$$x_n=1-\frac{2^n}{2^n-1}.$$It clearly converges to $0$ with respect to the Euclidean distance. Besides, $d_2(x_n,1)=2^{-n}$ and so…