A bounded sequence $(x_n)$ is a sequence which satisfies $|x_n|<M\ \forall n\in\mathbb{N}$ for some $M\in\mathbb{R^+}$.
So an unbounded sequence $(y_n)$ is a sequence such that $\forall N\in\mathbb{R^+}\ \exists n_k\in\mathbb{N}$ such that $|y_{n_k}|>N$.
So my question is this - provided I have correctly defined an unbounded sequence, is it necessary for such a sequence to have infinitely many value above any particular real number?
What about for example, $(x_n)=\tan \left(\dfrac{\pi}{2n}\right)$?
This is the reason for my query math.stackexchange.com/questions/401033
Note that for unboundedness you should have infinitely many such $n_k$'s, as per the discussion in the question you have linked. If not, then we could just pick a $B > |y_{n_k}|$ to be a sufficient bound.
Edit: For the sequence $(x_n)$ in your question, it is undefined for $n=1$ (since $\tan(\frac{\pi}{2})$ is undefined). If we start counting indices at $n=2$, then it is bounded. Just pick some $B > \tan(\frac{\pi}{4})=1$. This bound works because $\tan$ is increasing in the interval $[0, \frac{\pi}{4}]$, and because $\frac{\pi}{2n} \rightarrow 0$ as $n \rightarrow \infty$, so $(x_n)$ tends to $0$ as $n$ increases, without ever exceeding $B$.