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


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 $1.$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.