Every convergent sequence is bounded.

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Proof

Let \((a_n)\) be a sequence that convergences to some number \(L\). By definition (2.2.3), we know that for any \(\epsilon > 0\), there exists an \(N \in \mathbb{N}\) such that if \(n \geq N\), then

$$ \begin{align*} |a_n - L| < \epsilon \end{align*} $$

Suppose we let \(\epsilon = 1\), then we know when \(n \geq N\), \(a_n\) is will be in interval \((L - 1, L + 1)\). Precisely,

$$ \begin{align*} |a_n - L| < 1 \quad \text{whenever $n \geq N$}. \end{align*} $$

We want to bound the terms \(a_n\) using \(L\). Since we don’t know if \(L\) if positive or negative, then we want to bound these terms using \(|L|\). To do that, we can use the triangle inequality to see that

$$ \begin{align*} |a_n| &= |a_n - L + L| \leq |a_n - L| + |L| < 1 + |L| \quad \text{whenever $n \geq N$}. \end{align*} $$

This bounds all the terms that come after \(N\). For the remaining finite terms, we can let \(M\) be

$$ \begin{align*} M = \max\{|a_1|,|a_2|,...,|a_{N}|, |L|+1\} \end{align*} $$

So now every term in the sequence is bounded by \(M\). Therefore

$$ \begin{align*} |a_n| \leq M \quad \text{for all $n \in \mathbb{N}$}. \end{align*} $$

\(\blacksquare\)


References