Examples of Cauchy sequences in the rational numbers that do not converge in said set with respect to the p-adic topology

Question

I am working through a set of notes on Algebraic Number theory, and I find myself attempting to construct some examples of Cauchy sequences not converging in $\mathbb{Q}$, with respect to the p-adic topology.

To share my work thus far, [the work below does not offer the answer, left in for posterity's sake, see @Lubin 's response below]

I considered the following sequence, $\{a_n\}_{n=1}^\infty$, where $a_n = \sum_{k=1}^{n} p^k$, with respect to the p-adic topology. This is certainly Cauchy, as the difference between subsequent terms is $|a_n-a_{n-1}| = \frac{1}{p^n}$. As I understand it, the limit of the $a_n$'s is certainly not in $\mathbb{Q}$, namely, $\sum_{k=1}^{\infty} p^k$ does not lie in the rationals. Does this example hold up?

[Additionally, I have found one (related) example at the following link, also on Stack Exchange. Showing that Q is not complete with respect to the p-adic absolute value ]

To state my question: Is my initial example sound? Is there another sequence I should consider that more elegantly illustrates this concept?

Answer 1

The $p$-adic series $\sum_{k=1}^\infty p^k$ can be evaluated by high-school methods, to $p/(1-p)$, which of course is a rational number.

Indeed, you have the same situation as for decimal expansion of reals: if the $p$-adic expansion is (eventually) periodic, i.e. periodic in the tail, then it represents a rational number. Thus, to get an irrational element of $\Bbb Z_p$ or $\Bbb Q_p$, you need to have a nonperiodic expansion.

If you’re starting out, let me make a recommendation to you. Don’t think of a $7$-adic number as an expansion $a_0+a_1\cdot7+a_2\cdot7^2+\cdots$, but rather as an infinite $7$-ary (base-seven) expansion, running (potentially) infinitely to the left, rather than to the right. Now, once you have an addition and multiplication table for numbers in the $7$-ary notation, you can do additions, subtractions, and multiplications just as you learned in elementary school. Division, I confess, is set up differently, but you should be able to work that out.

Answer 2

This question already has an excellent answer, but I will provide another one which makes no reference to $\Bbb Q_p$. More precisely, I will prove that the sequence$$\left(\sum_{k=1}^mp^{k!}\right)_{m\in\Bbb N}\label{s}\tag1$$is a non-convergent Cauchy sequence of elements of $(\Bbb Q,d_p)$ working entirely within the rational numbers.

It is easy to prove that the sequence $\eqref{s}$ is a Cauchy sequence; it follows from the fact that if $m,n\in\Bbb N$ and $m>n$, then $d_p(\sum_{k=1}^mp^{k!},\sum_{k=1}^np^{k!})=p^{-(n+1)!}$.

On the other hand, if the sequence \eqref{s} converges in $(\Bbb Q,d_p)$, let $\frac ab$ be its limit, where $a\in\Bbb Z$ and $b\in\Bbb N$ are coprime numbers. For each $m\in\Bbb N$, $d_p(\sum_{k=1}^mp^{k!},0)=p^{-1}$ and therefore, since the set $\{q\in\Bbb Q\mid d_p(q,0)=\alpha\}$ is a closed set for every $\alpha>0$, $d_p(a/b,0)=p^{-1}$ too. But then $p\nmid b$; otherwise, we would have $p\nmid a$ and so $d_p(a/b,0)\geqslant p$.

It follows from the fact that$$\lim_{m\to\infty}\sum_{k=1}^mp^{k!}=\frac ab$$in $(\Bbb Q,d_p)$ that, for each $n\in\Bbb N$,$$\lim_{m\to\infty}\sum_{k=1}^mp^{(n+k)!}=\frac ab-\sum_{k=1}^np^{k!}.$$But, as above, since each sum $\sum_{k=1}^mp^{(n+k)!}$ belongs to the set $\left\{q\in\Bbb Q\mid d_p(q,0)=p^{-(n+1)!}\right\}$, we have that $a/b-\sum_{k=1}^np^{k!}$ belongs to that set too, and therefore$$\frac ab-\sum_{k=1}^np^{k!}\ne0\label{n}\tag2.$$Let $\nu_p$ be the $p$-adic valuation. Then, since $p\nmid b$,\begin{align}\nu_p\left(a-b\sum_{k=1}^np^{k!}\right)&=\nu_p\left(\frac{a-b\sum_{k=1}^np^{k!}}b\right)\\&=\nu_p\left(\frac ab-\sum_{k=1}^np^{k!}\right)\\&=(n+1)!.\end{align}In particular, $p^{(n+1)!}\mid a-b\sum_{k=1}^np^{k!}$.

On the other hand, for each $n\in\Bbb N$,\begin{align}\left\lvert a-b\sum_{k=1}^np^{k!}\right\rvert&\leqslant\lvert a\rvert+b\sum_{k=1}^np^{k!}\\&<\lvert a\rvert+b\sum_{k=0}^{n!}p^k\\&=\lvert a\rvert+b\frac{p^{n!+1}-1}{p-1}\\&<\lvert a\rvert+bp^{n!+1}\\&<p^{(n+1)!}\end{align}if $n\gg1$, since\begin{align}\lim_{n\to\infty}\frac{\lvert a\rvert+bp^{n!+1}}{p^{(n+1)!}}&=\lim_{n\to\infty}\left(\frac{\lvert a\rvert}{p^{(n+1)!}}+\frac{bp}{p^{n\times n!}}\right)\\&=0\\&<1.\end{align}But then, if $n\gg1$, the non-negative integer $\left\lvert a-b\sum_{k=1}^np^{k!}\right\rvert$ is both a multiple of $p^{(n+1)!}$ and smaller than it. Therefore, it must be equal to $0$, contradicting \eqref{n}. Or we can argue that $b\sum_{k=1}^Np^{k!}<b\sum_{k=1}^{N+1}p^{k!}$ and that therefore the numbers $b\sum_{k=1}^Np^{k!}$ and $b\sum_{k=1}^{N+1}p^{k!}$ cannot be both equal to $a$.