---
tag: 0S4JF619V7
kind: statement
role: proposition
name: Convergent sequences are bounded
msc: [40A05]
uses: [0AZ3R52593, 05FQ3VGAR9, 09G6KHS6GY]
sources:
  - url: https://en.wikipedia.org/wiki/Limit_of_a_sequence
    locator: Properties
---

## Statement

Let $(a_n)$ be a [[convergent sequence|0AZ3R52593]] of real numbers.
Then $(a_n)$ is [[bounded|05FQ3VGAR9]]: 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|09G6KHS6GY]] 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.