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I have been trying to solve three closely related problems about convergence of sequences in the p-adic numbers. I managed to solve two, but got stuck on the last one. Apart from trying to find a solution for this last one, I'd also like to doublecheck my answers for the first two, just to make sure I'm not misunderstanding the theory.

Sidenote: I've been seeing different notations for the p-adic valuation, I am used to $\text{ord}_p(x)$.

Problem 1: Can you construct a sequence converging to $0$ in $\mathbb{Q}_2$, but to $1$ in $\mathbb{R}$?

The answer I came up with was $a_n = \frac{2^n}{2^n-1}$, convergence in the reals is easy to verify, while in $\mathbb{Q}_2$ you can note that $\text{ord}_2(a_n) = \text{ord}_2(2^n) - \text{ord}_2(2^n - 1) = n \to \infty$.

Problem 2: Can you construct a sequence converging to $0$ in $\mathbb{Q}_2$, but to $1$ in $\mathbb{Q}_3$?

The answer I came up with was $a_n = \frac{2^n}{2^n-3^n}$, where you can note that $a_n - 1 = \frac{3^n}{2^n-3^n}$. Consequently $\text{ord}_2(a_n) = n \to \infty$ and $\text{ord}_3(a_n - 1) = n \to \infty$.

Problem 3: Can you construct a sequence converging to $0$ in $\mathbb{Q}_2$, but to $1$ in both $\mathbb{Q}_3$ and $\mathbb{R}$?

First I thought simply trying to take the product of my above two sequences, until I realized that my second sequence actually converges to zero over the reals. I'm not entirely convinced it is even possible to find such a sequence, but I wouldn't know where to start on proving that.

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If you have $|a|>1$ in $K=\Bbb Q_p$ or $\Bbb R$ and also $|a|<1$ in $L=\Bbb Q_p$ or $\Bbb R$ then $1/(1-a^n)\to0$ in $K$ but $1/(1-a^n)\to1$ in $L$. So, can you find $a$ with $|a|>1$ in $\Bbb Q_2$ and $|a|<1$ in $\Bbb Q_3$ and $\Bbb R$?

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  • $\begingroup$ So I assume the most straightforward choice would be $a = 3 / 4$? I'll try to write the answer out fully to see if I come to the right conclusion. $\endgroup$ – Zeno Jun 16 at 12:09
  • $\begingroup$ Yes so I can see indeed that this converges correctly as it should. One small thing that leaves me wondering is how exactly you came up with that result you are using. Is this a well-known result, or at least tied to a well-known property? $\endgroup$ – Zeno Jun 16 at 12:27
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    $\begingroup$ @Zeno At least in hindsight, it is not that hard. To achieve $x_n\to 0$, it seems natural to look for $x_n=a^n$ with $|a|<1$ in the desired places. However, this will have $|a|=1$ for almost all other places, and perhaps $|a|>1$ for finitely many others. For the latter, we will thus have $|a^n|\to \infty$. Then applying a Möbius transformation to move the limits $0$ and $\infty$ to $1$ and $0$, respectively, is perhaps not too far-fetched. $\endgroup$ – Hagen von Eitzen Jun 16 at 13:08
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Given finitely many primes $p_1,\ldots, p_m$, there do exist sequences $(x_n)$ in $\Bbb Q$ such that $x_n\to 0$ in $\Bbb Q_{p_i}$ for $1\le i\le m$, but $x_n\to 1$ in $\Bbb R$. Indeed, just let $$x_n=\frac{A^n}{A^n-1}$$ where $A=p_1p_2\cdots p_m$.


Likewise, there do exist sequences $(x_n)$ in $\Bbb Q$ such that $x_n\to 0$ in $\Bbb Q_{p_i}$ for $2\le i\le n$ and $x_n\to0$ in $\Bbb R$, but $x_n\to 1$ in $\Bbb Q_{p_1}$: Let $B=p_2p_3\cdots p_m$. Find $k\in\Bbb N$ with $B^k>p_1$and let $$y_n=\frac{B^{kn}}{B^{kn}-p_1^n}.$$ Then automatically $\operatorname{ord}_{p_i}(y_n)=kn\to \infty$ for $2\le i\le m$ and $\operatorname{ord}_{p_1}(y_n-1)=n\to \infty$. But how can we fix the limit in $\Bbb R$? For each integer $N$, there exist infinitely many primes that are $\equiv 1\pmod N$, hence we can pick $q_n$ such that $q_n$ is prime and $q_n\equiv 1\pmod{p_1^nB^{kn}}$. Then automatically $q_n>B^{kn}$. Now let $$ x_n=\frac{y_n}{q_n}.$$ As $q_n\to 1$ in $\Bbb Q_{p_i}$ for $1\le i\le n$, the limits of $(x_n)$ are the same as the limits of $(y_n)$ in the $\Bbb Q_{p_i}$, $1\le i\le m$. And as $0<x_n<\frac1{B^{kn}-p_1^n}=\frac1{(B^k-p_1)(B^{k(n-1)}+\ldots+p_1^{n-1})}<\frac1{B^{k(n-1)}}$, we also have $x_n\to 0$ in $\Bbb R$ as desired.


Remark: By using linear combinatiosn of the above, we conclude that it is possible to prescribe a (rational) limit for any finite collection of places (i.e., $p$-adic valuations and/or the reals)

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  • $\begingroup$ I see. I just solved it using the other remark that was posted, but this is an exceedingly helpful result as well, thank you. I would not have guessed the same thing would be possible for an arbitrary set of limits. $\endgroup$ – Zeno Jun 16 at 12:37

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