Convergent sequences are bounded

Statement

Let $(a_n)$ be a convergent sequence of real numbers. Then $(a_n)$ is bounded: there is a real number $K \ge 0$ such that $|a_n| \le K$ for all $n \in \mathbb{N}$.

Remarks

Let $L$ be the limit. Take $\varepsilon = 1$ in the epsilon-N definition to get $N$. Then $K = \max(|a_0|, \dots, |a_{N-1}|, |L| + 1)$ works. The converse is false. The sequence $(-1)^n$ is bounded and divergent. The statement holds in each metric space, where bounded means contained in some ball.

Explanations

L1 human

After some index, all terms are close to the limit, so they lie in a small interval. Only finitely many terms come before that index, and finitely many numbers are always bounded.