If limit exists then sequence is bounded Suppose we have a sequence $\{a_n\}_n \subseteq \Bbb R: a_n\rightarrow l\in \Bbb R \implies \{a_n\}_n$ is bounded.
But this doesn't also imply that the sequence is monotone right? And as a counterexample can we take $\left\{\frac{(-1)^n}{n}\right\}_n$, which converges to $0$ but it has terms of alternating sign?
Thanks in advance.

Let us consider a sequence $(a_n)_{n\in\mathbb{N}}$ with limit $l$ in $\mathbb{R}$. If $\epsilon > 0$, one can find $N$ such that for all $n\geqslant N$, $a_{n}$ belongs to the interval $\left]l-\epsilon, l+\epsilon\right[$. If $m=\min_{n< N} a_n$ and $M=\max_{n< N} a_n$, then the terms of the sequence satisfy
\begin{equation}
\forall n \in \mathbb{N}, \qquad
\min\lbrace m, l-\epsilon\rbrace  \leqslant a_n \leqslant \max\lbrace M, l+\epsilon\rbrace \, .
\end{equation}
Therefore, the sequence is bounded.
The proposed counter-example $a_n = (-1)^n/n$, which vanishes at infinity and is alternating signs, shows that a convergent sequence is not necessarily monotonic.
