If the subsequence converge, then will the sequence has at least upper bound or lower bound By the Bolzano-Weierstrass theorem, we know that any bounded sequence $\{x_n\}$ has a convergent subsequence. Also, if there exists a converge subsequence, it's not always true that the sequence $\{x_n\}$ is bounded($\{x_n\}$ doesn't always have both upper bound and lower bound), but is it possible to conclude that $\{x_n\}$ at least has one of them(have at least upper bound or lower bound) if there exists a converge subsequence $\{x_{n_k}\}$?
 A: No, consider the sequence
$$x_n=
\begin{cases}
1 & n\equiv 1\pmod{2}\\
n & n\equiv 0\pmod{4}\\
-n & n\equiv 2\pmod{4}
\end{cases}$$
So the sequence would look like $1,-2,1,4,1,-6,1,8,1,-10,...$, so all the odd terms are $1$, and all the even terms are either there positive or negative versions (alternating at each even number). It is clear that the subsequence of all the odd terms converges, so $x_n$ has a convergent subsequence, but the sequence $x_n$ is not bounded above nor below.
A: For the sequence :
$$x_n = n \sin \dfrac{2 n \pi}{3}$$
We have :

*

* $x_{3 n} = 3 n \sin 2 n \pi = 0 \to 0$ Converge.

* $x_{3 n + 1} = (3 n + 1) \sin \left(2 n \pi + \dfrac{2 \pi}{3}\right) = (3 n + 1) \sin \left(\dfrac{2 \pi}{3}\right) = \sqrt{3} \dfrac{2 n + 1}{2} \to +\infty$ so $(x_n)$ is not upper bounded.

* $x_{3 n + 2} = (3 n + 2) \sin \left(2 n \pi + \dfrac{4 \pi}{3}\right) = (3 n + 1) \sin \left(\dfrac{- \pi}{3}\right) = - \sqrt{3} \dfrac{2 n + 1}{2} \to -\infty$ so $(x_n)$ is not lower bounded.

 
