Find a sequence such that ${d_{3}}(x,z)=0$

Question

For $z\in \mathbb{Z}$ find a non constant sequence which tend to $z$ in $(\mathbb{Z},d_{3})$ when $d$ is the 3-adic metric $$ d_{7}(x,y)= \begin{cases} 0, x=y\\ \frac{1}{3^{k(x,y)}}, x\neq y \end{cases} $$

when $k(x,y)=max\{i:3^{i}|x-y\}$

A sequence $x$ tend to $z$ if the metric $d(x,z)=0$ so in this case we have to find $x$ such that:

$$d_{3}(x,z)=0\iff \frac{1}{3^{k(x,z)}}\rightarrow 0\iff k(x,z)\rightarrow \infty$$

$$max\{i:3^i|x-z\}\rightarrow \infty\iff x-z=3k, k\in \mathbb{Z}\iff x=3k+z ,k\in \mathbb{Z}$$

So if we take $x=(-1)^{n}*3n+z$ it will be a correct answer?

Answer 1

I think the question contains a lot of misprints and one error ($max\{i:3^i|x-z\}\rightarrow \infty$ $\Leftarrow x-z=3k$), because we need $x_n-z$ be divisible by arbitrarily large powers of $3$), which can be fixed, for instance, by putting $x_n=z+3^n$ for each $n$.