Prove that $\mathbb{Q}_+$ can be enumerated as $(q_n)$ such that $\lim\sqrt[n]{q_n}$ exists.

Question

Prove that the set of positive rationals can be enumerated as $(q_n),$ such that $\lim\sqrt[n]{q_n}$ exists.

Comment. I don't know if I should be looking for a certain "formula" on $q_n's$, or a way, a method to enumerate the positive rationals, in order to get the desired result.

Thanks for the help!

Answer 1

Let $\mathbb{Q}_{+} = \bigcup_{k = 1}^{\infty} Q_k$ with disjoint finite nonempty $Q_k$ such that $Q_k \subseteq [1/k, k]$; $$\text{say, take }Q_k=\{a/b : a, b \in \mathbb{Z}_{+}, a + b = k + 1, \gcd(a, b) = 1\}.$$ Now if, say, $Q_k=\{q_{n_k}, \ldots, q_{n_{k + 1} - 1}\}$ with $n_1 = 1$, then

• $n \mapsto q_n$ is an enumeration of $\mathbb{Q}_{+}$;
• $n_{k} \geqslant k$ (because $Q_k$ is nonempty for all $k$);
• $(1/k)^{1/k} \leqslant q_n^{1/n} \leqslant k^{1/k}$ when $n_k \leqslant n < n_{k + 1}$.

As $\displaystyle\lim_{k\to\infty} k^{1/k} = 1$, the last inequality implies $\displaystyle\lim_{n\to\infty} q_n^{1/n} = 1$.