Every convergent sequence is bounded.
Previous Definitions:
- Absolute value definition and other properties
- Sequence definitions and convergence definitions
- Subsequences
- Show the limit template
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
$$
\begin{align*}
|a_n| \leq M \quad \text{for all $n \in \mathbb{N}$}.
\end{align*}
$$
\(\blacksquare\)