Theorem
If \(\lim\limits_{n \rightarrow \infty} x_n = x\), there exists some \(M > 0\) such that \(|x_n| \leq M\) for all \(n \in \mathbb{N}\).

Strategy/Notes

First, one might say that the sequence converges to \(x\), isn’t that a bound? It’s not because the sequence might bounce back and forth really crazy before converging finally to \(x\). So maybe the earlier terms reach \(10x\) or \(-10x\) before finally settling down to \(x\).

Is the converse true? must a bounded sequence converge? No. Consider the sequence \(\{-1,1,-1,1,...\}\). This sequence is bounded but doesn’t converge.

Proof

Assume \(\lim\limits_{n \rightarrow \infty} x_n = x\). Consider \(\epsilon = 1\). Choose \(N \in \mathbb{N}\) such that when \(n \geq N\)

$$ \begin{align*} |x_n - x| &< 1. \end{align*} $$

Observe now that \(|x_n| = |(x_n - x) + x|\). By the triangle inequality then,

$$ \begin{align*} |x_n| &\leq |x_n - x| + |x| \\ |x_n| &< 1 + |x| \quad \text{(By our choice of $\epsilon$ above)}. \end{align*} $$

Set

$$ \begin{align*} M = \max{\{|x_1|, |x_2|, |x_3|, ....,|x_{N}|\}}. \end{align*} $$

Therefore, we have \(|x_n| \leq \max\{M, 1 + |x|\}\) for all \(n \in \mathbb{N}\) as required. \(\ \blacksquare\)

References