Bounded Sequences of Real Numbers

Definition: A sequence $(a_n)$ of real numbers is said to be Bounded Above if there exists a real number $M \in \mathbb{R}$ such that $a_n \leq M$ for every $n \in \mathbb{N}$. A sequence $(a_n)$ is said to be Bounded Below if there exists a real number $m \in \mathbb{R}$ such that $m \leq a_n$ for every $n \in \mathbb{N}$. A sequence $(a_n)$ is said to be Bounded if it is both bounded above and bounded below.

Observe that if a sequence $(a_n)$ is bounded above and below then there exists $m, M \in \mathbb{R}$ such that $m \leq a_n \leq M$ for every $n \in \mathbb{N}$. Let $M_0 = \max \{ |m|, |M| \}$. Then $|a_n| \leq M_0$. Conversely, if $|a_n| \leq M_0$ for every $n \in \mathbb{N}$ then $-M_0 \leq a_n \leq M_0$ for all $n \in \mathbb{N}$ and so $(a_n)$ is bounded.

Thus, a sequence $(a_n)$ is bounded if and only if there exists an $M \in \mathbb{R}$ such that $|a_n| \leq M$ for every $n \in \mathbb{N}$. We will use this a lot in coming proofs

There are many examples of bounded sequences. The sequence $(-1)^n$ is bounded since $|(-1)^n| \leq 1$ for every $n \in \mathbb{N}$. This is an example of a bounded sequence that is divergent.

Also, the sequence $\left ( \frac{1}{n} \right )$ is bounded since we already know that for every $n \in \mathbb{N}$, $n \geq 1$ and so:

(1)

\begin{align} \quad \biggr \lvert \frac{1}{n} \biggr \rvert = \frac{1}{n} \leq 1 \end{align}

This is an example of a bounded sequence that is convergent. Thus we have established that convergence and boundedness are not equivalent properties. However, on The Boundedness of Convergent Sequences Theorem page we will see that if a sequence of real numbers is convergent then it is guaranteed to be bounded.