# $(a_n)_{n \geq 1}=\mathbb{Q}_+$ and $\sqrt[n]{a_n}$ is convergent

Is there any sequence $(a_n)_{n \geq 1}$ such that it contains all positive rational numbers without repetition, and $\sqrt[n]{a_n}$ is convergent?

My first guess is that there is no such sequence. I tried to build $a_n$ just like the sequence in the proof that $\mathbb{Q}$ is countable: $$1,2,1/2,1/3,3,4,3/2,2/3,1/4,1/5,2/4$$ and so on. I'm not sure if this works, however

## 2 Answers

Actually the standard one

$$(a_n) = \left( \frac 11, \frac 21, \frac 12, \frac 31, \frac 13, \frac 41, \frac 32, \frac 23, \frac 14, \cdots\right)$$

works. The observation is that for the members $a_n$ in the $i$-th layer:

$$\frac i1, \frac{i-1}{2}, \cdots, \frac{2}{i-1}, \frac 1i,$$

we have

$$i \ge a_n \ge i^{-1}\Rightarrow \sqrt[n]{i} \ge \sqrt[n]{a_n} \ge (\sqrt[n]{i})^{-1}.$$

But clearly $n\ge i$, so

$$\sqrt[n]{n} \ge \sqrt[n]{a_n} \ge (\sqrt[n]{n})^{-1}.$$

So $\sqrt[n]{a_n} \to 1$ as $\sqrt[n]{n} \to 1$.

Pick an enumeration of $\mathbb{Q}_{+}$, $(q_i)_{i \geq 1}$.

We will define $a_n$ to be a reordering of $q_i$ such that $\frac{1}{2n}<a_n<2n$. Any such sequence will satisfy $\sqrt[n]{a_n}\rightarrow 1$.

Define $k_n= \min\{k| q_k \in (\frac{1}{2n},2n), k\notin \bigcup_{i=1}^{n-1}I_i \}$. We can verify that $\bigcup_{i=1}^{\infty}I_i =\mathbb{N}$, and by construction $I_i\neq I_j$ when $i\neq j$.

Finally, the sequence $a_n=q_{k_n}$ does the job.

Remark: If $\sqrt[n]{a_n}$ converges then the limit must be $1.$