How to construct a sequence that has a subsequence convergent to every $k\in \Bbb{N}$?

Question

How to construct a sequence $\{a_{n}\}^{\infty}_{n=1}$, such that for every $k\in \Bbb{N}$, $\{a_{n}\}^{\infty}_{n=1}$ has a subsequence convergent to $k$?

A subsequence is such as $2,4,6,...$ in $1,2,3,4,5...$ and we know that if a sequence converges to $L$ then every subsequence converges to $L$.

Perhaps I did not get the meaning of the question, but if a sequence is convergent, it has only one limit. How can we construct a sequence that converges to multiple limits?

Answer 1

$\{1,1,2,1,2,3,1,2,3,4,\dots\}$

Answer 2

Try $a_n=n-\lfloor\sqrt n\rfloor^2$.

Answer 3

The set of rational numbers $\mathbb{Q}$ is numerable, then there exists a bijective mapping $\varphi:\mathbb{N}\to\mathbb{Q}$. By defining $$a_n=\varphi(n),\qquad n\in\mathbb{N}$$ we get a sequence such that for every real number $L$ there exist a subsequence $\{a_{n_{m}}\}_{m\in\mathbb{N}}$ such that $a_{n_{m}}\longrightarrow L$.