How do I show convergence in the 2-adics?

Question

How do I show convergence in the 2-adics in general, but particularly for the series given below?

If, to discriminate between convergence in different $p$-adics I define $\infty_p=\lim_{n\to\infty} p^n$ then it would seem reasonable to state $\infty_2=0_2$ in the 2-adics. I'm unsure of however, whether other series are convergent in the 2-adics such as whether $\infty_3=0_2$

I have an infinite sum which is divergent in the integers but in the 2-adics it converges to $\infty_2=0$.

The case I have in mind is; show that for every $x_0\in3\mathbb{N}_{>0}$:

$$f(x_0)=\lim_{n\to\infty} \left(3^n\left(x_m-2^{v(x_0)}\right)+\sum_{k=0}^n3^{n-k}2^{v(x_{k})} \right)=0_2$$

where $v(\cdot)$ is the (additive) 2-adic valuation $v:\Bbb Q_2^\times\to\Bbb Z$ and $x_{m+1}=2x_m+v(x_m$).

It actually converges to $2^r$ for some $r$ in finite steps for all $x_m$ and lands on every alternate power of $2$ thereafter.

How might one typically identify convergence in such a series? Straight off the bat I can see that it's sufficient to show its value converges to the inverse of its 2-adic norm and therefore if $x$ converges to $0_2$, then $\frac{x}{2^{v(x)}}=1$

I can also see that this problem is equivalent to stating that for any given $x_0$, for sufficiently large $r$, there exists a series $x_0+x_{1}+\ldots$ in which $x_{m+1}=2x'_m+2^{v(x)}$ which sums to either $2^r$ or $2^{r+1}$ where $x'_m$ denotes the partial sum to $x_m$. I mention this partly to highlight the alternating nature of the series in that for sufficiently high $2^r$ it lands on alternate powers of $2$ and not on every power of $2$.

I'm not necessarily asking you to solve this problem (although that would be welcome!) as it's clearly a challenging one but guidance as to a typical method would be appreciated. There is much I am unsure of.

Answer 1

This belongs in a comment, but it will be too long.

You raise in your comment to @Wojowu the issue of the $p$-adic limit of $(1+p)^n$, as $n\to\infty$, and this is something I know somewhat much about.

It’s easiest to see what’s going on in the case $p\ge3$, for there, the integral powers of $1+p$ form a dense subgroup of the multiplicative group $1+p\Bbb Z_p$, the group of “principal units” of the $p$-adic ring $\Bbb Z_p$. This means that in (perhaps) the strongest sense possible, $\lim_{n\to\infty}(1+p)^n$ does not exist. The question that I would have asked you to pose instead would be the nature of $\lim(1+p)^{p^n}$, and maybe I should challenge you to show that the value of this limit is $1$. Granting the truth of this claim, you can show that $(1+p)^z$ makes good sense for no matter what $p$-adic integer $z$. For instance, $3$-adically, $10$ is a square, so that $4^{\sqrt{10}}$ makes good sense as an element of $\Bbb Z_3\subset\Bbb Q_3$. You can even use this to exhibit an isomorphism between the additive group $\Bbb Z_p$ and the multiplicative group $1+\Bbb Z_p$, by $z\mapsto(1+p)^z$, though verifying that it is an isomorphism requires some work.

In the case $p=2$, the situation is just a little more complicated, because $1+2\Bbb Z_2$ is not isomorphic to $\Bbb Z_2$, rather the picture is $\{\pm1\}\times\Bbb Z_2$, and the subgroup of $1+2\Bbb Z_2$ that’s isomorphic to $\Bbb Z_2$ is topologically generated by $5$. That is, the $\Bbb Z$-powers of $5$ form a dense subgroup of $1+4\Bbb Z_2$. The upshot again is that the powers of $3$ again have no limit, and but rather form a subgroup of the principal units whose closure is of index two in $1+2\Bbb Z_2$.

Answer 2

I have tried to understand what you wrote, but please forgive me that I could not understand very much of it.

First, what is $x_m$? is it a single positive integer divisible by $3$? Or a sequence of positive integers divisible by $3$ indexed by $m$? If it is a sequence you should better write $(x_m)_{m=1}^{\infty}$ if the index $m$ starts with $1$, $(x_m)_{m=0}^{\infty}$ if it starts with $0$, etc. So $(m^2)_{m=1}^{\infty}$ is the sequence of squares $1$, $4$, $9$, $16$, $\ldots$. You may also write $(x_m)$ if it is unimportant where the index $m$ starts.

Second, I am very confused about your notation $\infty_p$. There is to my knowledge no standard notation for this, but if you want to indicate that a sequence $(x_n)$ converges $p$-adically to a limit $b$, say, you may write this as $$\lim_{n\to\infty} x_n=b\text{ } (p).$$ Roughly speaking this means that the $p$-adic absolute value $|x_n-b|_p$ gets smaller and smaller and eventually tends to $0$ if $n$ tends to infinity, more precisely, for every $N>0$ there is $M>0$ such that$$|x_n-b|_p\leq p^{-N} \text{ for every }n\geq M.$$From this definition it is trivial that$$\lim_{n\to\infty} p^n=0 \text{ } (p).$$Indeed, take any $N>0$ and let $M=N$. Then for every $n\geq M$ we have$$|p^n|_p=p^{-n}\leq p^{-M}=p^{-N}.$$There is a nice criterion with which you can check if a sequence in $\mathbb{Q}_p$ converges to a limit or not. This is important to show that in fact the sequence $((p+1)^n)$ does not converge in $\mathbb{Q}_p$. I assume you have understood the construction of the field of $p$-adic numbers $\mathbb{Q}_p$ from some source, which is essential here. (This is of course not easy.) The $p$-adic numbers are in fact some abstract constructions, which have no relation whatsoever with the real numbers. In fact, the field of rational numbers $\mathbb{Q}$ can be "completed" to the field $\mathbb{R}$ of real numbers, but the field of $2$-adic numbers $\mathbb{Q}_2$, the fields of $3$-adic numbers $\mathbb{Q}_3$, etc. are all other completions of $\mathbb{Q}$, which have no relation to each other. $p$-adic numbers can be represented as sums with powers of $p$, with coefficients from $\{0,1,\ldots,p-1\}$, starting with a power of $p$ with exponent positive or negative or $0$ and with increasing powers of $p$. For instance, a $5$-adic number is $$3(5^{-2})+2(5^{-1})+4(5^0)+2(5^1)+0(5^2)+3(5^3)+\ldots$$

We may write this numbers as $32.4203\ldots$ $(5)$ with infinitely many digits from $\{0,\ldots,4\}$ on the right. You can add and multiply $p$-adic numbers represented in this way similarly as real numbers, with carry, except that you have to work from left to right instead of right to left.

The only thing you can say, that the field of rational numbers $\mathbb{Q}$ is contained in $\mathbb{R}$, $\mathbb{Q}_2$, $\mathbb{Q}_3$, etc. You should not try to find any connection of $p$-adic numbers with "real life". It does in general not make sense to ask whether a sequence of $3$-adic numbers converges in $\mathbb{Q}_2$. You may only ask whether sequences of rational numbers converge in $\mathbb{R}$, $\mathbb{Q}_2$, $\mathbb{Q}_3$, etc.

The criterion states that $(x_m)$ converges to a limit in $\mathbb{Q}_p$ if and only if the sequence of differences of consecutive terms $(x_{m+1}-x_m)$ converges to $0$ in $\mathbb{Q}_p$. There is no analogue of this criterion for sequences of real numbers.

Theorem. Let $(x_m)$ be a sequence in $\mathbb{Q}_p$. Then:

(1) Then $(x_m)$ converges in $\mathbb{Q}_p$ if and only if$\ldots$

(2) $\ldots\lim_{m\to\infty} (x_{m+1}-x_m)=0 \text{ }(p)$, that is, for every $N>0$ there is $M>0$ such that $$|x_{m+1}-x_m|_p\leq p^{-N} \text{ for every }m\geq M.$$ Proof. First assume (1), that is, $(x_m)$ converges in $\mathbb{Q}_p$ to a limit, say $b$. This means that for every $N>0$ there is $M>0$ such that $$|x_m-b|_p\leq p^{-N} \text{ for every }n\geq M.$$We now use the ultrametric inequality or strong triangle inequality for $p$-adic numbers: if $c$, $d$ are any two numbers in $\mathbb{Q}_p$, then$$|c+d|_p\leq \max(|c|_p,|d|_p).$$ Applying this with $c=x_{m+1}-b$, $d=x_m-b$ we get $$|x_{m+1}-x_m|_p\leq \max (|x_{m+1}-b|_p,|b-x_m|_p)\leq p^{-N} \text{ for }m\geq N.$$ This implies (2). So (1) implies (2).

Next assume (2). Let $N>0$, and choose $M>0$ such that $$|x_{m+1}-x_m|_p\leq p^{-N} \text{ for every }m\geq M.$$Then for every pair of indices $(m,n)$ with $n>m\geq M$ we have by repeatedly applying the ultrametric inequality, $$\begin{align*} |x_n-x_m|_p & = |(x_n-x_{n-1})+(x_{n-1}-x_{n-2})+ \ldots +(x_{m+1}-x_m)|_p \\ & \leq \text{max}(|x_n-x_{n-1}|_p,|x_{n-1}-x_{n-2}|_p,\ldots,|x_{m+1}-x_m|_p) \\ & \leq p^{-N}. \end{align*}$$But this means that $(x_m)$ is a Cauchy sequence, and an important property of $\mathbb{Q}_p$ is that such a sequence does indeed converge to a limit in $\mathbb{Q}$_p. This proves (1). So (2) implies (1), and (1) and (2) are equivalent.$$\tag*{$\square$}$$Example: $((p+1)^m)$ does not converge in $\mathbb{Q}_p$ to any limit. For consider the sequence of consecutive terms $((p+1)^{m+1}-(p+1)^m)$. Taking the $p$-adic absolute value we get $$|(p+1)^{m+1}-(p+1)^m|_p=|p(p+1)^m|_p=|p|_p.|p+1|^m=p^{-1}.$$Hence assertion (2) from the above theorem, that for any $N>0$ there should be an $M>0$ such that $$|(p+1)^{m+1}-(p+1)^m|_p\leq p^{-N} \text{ for all }m\geq M$$ is not satisfied.