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Let $S_n$ be a sequence comprising the sum of two sequences, one of which converges in $\Bbb Z_2$ and the other diverges in $\Bbb Q_2$ but converges in $\Bbb Q$.

An example is $x_{n+1}=4x_n+5\cdot2^{\nu_2(x_n)-2}$ where $2^{\nu_2(x_n)}$ measures the highest power of $2$ that divides $x$.

When $\nu_2(x_0)$ is even, this is the sum of a sequence converging to $-\frac13$ in $\Bbb Z_2$ and another converging to $\frac13$ in $\Bbb Q$, the sum of which converges to $0$.

When $\nu_2(x_0)$ is odd, this is the sum of a sequence converging to $-\frac23$ in $\Bbb Z_2$ and another converging to $\frac23$ in $\Bbb Q$

EDIT FOR CLARITY: let $S_n=a_n+b_n$ and let $a_0=1$ and let $b_0=0$

Then let $a_n$ be the binary number to the left of point and $b_n$ the part to the right.

Then proceed: $a_n=1,5,21,\ldots$ and $b_n=\frac12,\frac58,\frac{21}{64}\ldots$

The example given above does not converge in $\Bbb Q_2$ due to the component $\frac12,\frac58,\frac{21}{64}\ldots$ that converges in $\Bbb Q$. However if instead of $2^{\nu_2(x)}$ we take the power of $2$ that divides $x$ and then take either of the obvious morphisms $\phi:\Bbb Z\to\Bbb N$ and use the distance:

$d(x,y)=2^{-\phi(\nu_2(x))}$

Then the sequences above converge in the completed (ring?) of $\overline {\Bbb Z},d$ which is isomorphic to $\Bbb Z_2$.

It would seem therefore that $\Bbb Q_2$ embeds in this space, but in such a way that some sequences (e.g. the example given above) will converge where they do not converge in $\Bbb Q_2$. Is this correct?

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    $\begingroup$ Is "a sequence comprising the sum of two sequences" deliberately confusing terminology? Because when I hear "sum of two sequences", I think of some $c_n = a_n+b_n$ with explicitly given sequences $a_n, b_n$, and then one has the usual limit laws etc. But in your example, a sequence is given recursively, and it just so happens that the recursion formula is a sum of two terms. I doubt that anything can be inferred from that. E.g. look at $x_{n+1} =x_n +17$. Do you think one can make non-trivial statements about this from knowing the constant sequences $x_{n+1}=x_n$ and $x_{n+1}=17$? $\endgroup$ – Torsten Schoeneberg Oct 9 at 18:07
  • $\begingroup$ Besides, I don't even see what "summand" of your sequence is supposed to converge to $-1/3$ resp. $1/3$ in any metric, and what it would mean to converge "in $\mathbb Z_2$" resp. "in $\mathbb Q$" (both $\pm 1/3$ are in both sets), so I stopped reading there and vote to close. $\endgroup$ – Torsten Schoeneberg Oct 9 at 18:12
  • $\begingroup$ @TorstenSchoeneberg I meant it in the meaning you first thought of. I would never be deliberately confusing. If it's unclear how I've partitioned the sequence into a sum of two, I've separated them at the binary point into a sum of the part to the left of the point plus the part to the right. I hoped that would be clear where I indicate the component in $\Bbb Q$ is $\frac12,\frac58,\frac{21}{64}\ldots$ $\endgroup$ – samerivertwice Oct 9 at 18:15
  • $\begingroup$ @TorstenSchoeneberg with respect to $\frac13$ being in $\Bbb Q_2$ you literally clarified for me in the question I asked yesterday (or this morning depending on time zone) that $\frac12,\frac58,\frac{21}{64}\ldots$ diverges in $\Bbb Q_2$ despite the fact that $\frac13$ is in the set. $\endgroup$ – samerivertwice Oct 9 at 18:16
  • $\begingroup$ @TorstenSchoeneberg I've now written out the sequences $s_n=a_n+b_n$ which hopefully helps. $\endgroup$ – samerivertwice Oct 9 at 18:29

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