Convergence in p-adics and reals

Question

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.

Answer 1

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 $\Bbb R$?

Answer 2

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 $A=p_1p_2\cdots p_m$.

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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.

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Remark: By using linear combinations 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)