Does $x=\dfrac{1}{2^n-3}$ converge in both directions in $\Bbb{Z}_2$ for $n$ is integer?
Solving $x=\dfrac{1}{2^n-3}$ for $x$ where $n$ is integer.
We get that the only the solutions in $\Bbb{Z}$ are $x\in\{-1,1\}$ and of course there is a countable set of solutions in $\Bbb{Q}$ given by:
$x\in\left\{\ldots,-\frac{4}{11},-\frac{2}{5},-\frac{1}{2},-1,1,\frac{1}{5},\frac{1}{13},\ldots\right\}$
Does this sequence $x_n$ converge in both directions in $\Bbb{Z}_2$?
To the right (i.e. $n\to\infty$) we have $x\to \dfrac{1}{2^{m}-3}$
To the left (i.e. $n\to-\infty$) we have $x\to \dfrac{-2^m}{2^m+2^{m-1}-1}=2^{2m}\cdot\dfrac{-1}{1+2^{2m-1}-2^m}$
I have no doubt that to the right it converges to $-\frac{1}{3}$. Is that correct?
To the left I'm not so sure. I think all but the $1$ in the denominator converges to $0$ leaving me with $0\cdot \frac{-1}{1}=0$. Is that correct?
To put this into a bit of perspective, I figure I will post this as an answer to a broader case. In general, if we look at the expression $\dfrac{1}{x^{-1} + a^{-1}}$ it tends towards $a$ as $x$ gets larger in magnitude and tends towards $0$ as $x$ gets smaller in magnitude, regardless of what field you're in, real or p-adic.
Specifically here we can let x play the role of $p^n$ and $a$ play the role of $\dfrac{-1}{3}$.