Proof that Convergent Sequences are Bounded

We are now going to look at an important theorem - one that states that if a sequence is convergent, then the sequence is also bounded.

Theorem: If $\{a_n \}$ is a convergent sequence, that is $\lim_{n \to \infty} a_n = L$ for some $L \in \mathbb{R}$, then $\{a_n \}$ is also bounded, that is for some $M > 0$, $\mid a_n \mid ≤ M$.

• Proof of Theorem: We first want to choose $N \in \mathbb{N}$ where $n ≥ N$ such that $\mid a_n - L \mid < \epsilon$. Choose $\epsilon$ to be some positive number for epsilon, let's say $\epsilon = 1$. Therefore $\mid a_n - L \mid < 1$ and by the triangle inequality ($\mid A \mid - \mid B \mid ≤ \mid A - B \mid$):

(1)

\begin{align} \quad \mid a_n \mid - \mid L \mid ≤ \mid a_n - L \mid < 1 \\ \mid a_n \mid - \mid L \mid < 1 \\ \mid a_n \mid < 1 + \mid L \mid \end{align}

• So if $n ≥ N$, then $\mid a_n \mid < 1 + \mid L \mid$.

• Now consider where $n < N$. This is a finite set so there exists a maximum value, call it $\mid a_p \mid$, that is $\mathrm{max} \{ \mid a_1 \mid, \mid a_2 \mid, ..., \mid a_p \mid, ..., \mid a_{N-1} \mid \} = \mid a_p \mid$.

• We will now summarize that if $n < N$, then the maximum value the sequence $\{ a_n \}$ takes on is $\mid a_p \mid$. If $n ≥ N$, then the maximum value the sequence $\{ a_n \}$ takes on is $1 + \mid L \mid$. Therefore, let $M = \mathrm{max} \{ \mid a_p \mid, 1 + \mid L \mid \}$. Therefore, $\forall n$, $\mid a_n \mid ≤ M$, so we have shown that $\{ a_n \}$ is bounded. $\blacksquare$