> [!NOTE] Lemma > Any [[Convergence|convergent real sequence]] is [[Bounded Sequence|bounded]]. **Proof**: Let $(a_{n})$ such that $a_{n}\to a$ Then, taking $\varepsilon=1$ in the definition of convergence, there exists $N$ such that for all $n\geq N,$ $|a_{n}-a|<1.$By [[Absolute Value Satisfies Triangle Inequality|triangle inequality]], for all $n\geq N,$ $|a_{n}| = |a_{n}-a+a| \leq |a_{n}-a|+|a| \leq 1+|a|$Let $A = \max(|a_{1}|,|a_{2}|,\dots,|a_{N-1}|,|a|+1)$then $|a_{n}|\leq A$ for all $n \in \mathbb{N}$.