The Limit Superior and Limit Inferior of a Sequence of Real Numbers

Definition: Let $(a_n)_{n=1}^{\infty}$ be a sequence of real numbers. Then Limit Superior of $(a_n)_{n=1}^{\infty}$ denoted $\displaystyle{\limsup_{n \to \infty} a_n}$ is defined to be $\displaystyle{\limsup_{n \to \infty} a_n = \lim_{n \to \infty} \left ( \sup_{k \geq n} \{ a_k \} \right )}$. The Limit Interior of $(a_n)_{n=1}^{\infty}$ denoted $\displaystyle{\liminf_{n \to \infty} a_n}$ is defined to be $\displaystyle{\liminf_{n \to \infty} a_n = \lim_{n \to \infty} \left ( \inf_{k \geq n} \{ a_k \} \right )}$.

We will now establish some important notions regarding the limit superior and limit inferior of a sequence of real numbers.

Lemma 1: If $A$ and $B$ are subsets of real numbers such that $A \subseteq B$ then: a) $\sup B \leq \sup A$. b) $\inf A \leq \inf B$.

• Proof of a) Let $A \subseteq B$. If $M = \sup A$ then $M$ is an upper bound to the set $A$, so, for all $x \in A$ we have that $x \leq M$. But $A \subseteq B$, so for all $x \in B$ we also have that $x \leq M$. Therefore $M$ is an upper bound to $B$ and hence $\sup B \leq M = \sup A$. $\blacksquare$

• Proof of b) Let $A \subseteq B$. If $m = \inf A$ then $m$ is a lower bound to the set $A$, so, for all $x \in A$ we have that $m \leq x$. But $A \subseteq B$, so for all $x \in B$ we also have that $m \leq x$. Therefore $m$ is a lower bound to $B$ and hence $m = \inf A \leq \inf B$. $\blacksquare$

Notice that for any sequence $(a_n)_{n=1}^{\infty}$ of real numbers and for each $k \in \{ 1, 2, ... \}$ we have that:

(1)

\begin{align} \quad \{a_k, a_{k+1}, a_{k+2}, ... \} \supseteq \{ a_{k+1}, a_{k+2}, a_{k+3}, ... \} \end{align}

Define the sequences $(b_n)_{n=1}^{\infty}$ and $(c_n)_{n=1}^{\infty}$ as follows:

(2)

\begin{align} \quad (b_n)_{n=1}^{\infty} = \left ( \sup_{k \geq n} a_k \right )_{n=1}^{\infty} \end{align}

(3)

\begin{align} \quad (c_n)_{n=1}^{\infty} = \left ( \inf_{k \geq n} a_k \right )_{n=1}^{\infty} \end{align}

By Lemma 1 we see that $(b_n)_{n=1}^{\infty}$ is a decreasing sequence of real numbers while $(c_n)_{n=1}^{\infty}$ is an increasing sequence of real numbers. If $(a_n)_{n=1}^{\infty}$ is a bounded sequence of real numbers then the set $\{ a_n : n \in \mathbb{N} \}$ is bounded in $\mathbb{R}$ then any subset of $\{ a_n : n \in \mathbb{N} \}$ is also bounded and by the completeness property of the real numbers we have that the supremum and infimum of all of these sets exist. Moreover, the sequences $(b_n)_{n=1}^{\infty}$ and $(c_n)_{n=1}^{\infty}$ will both converge in $\mathbb{R}$ since $(b_n)_{n=1}^{\infty}$ is a decreasing sequence that is bounded below and $(c_n)_{n=1}^{\infty}$ is an increasing sequence that is bounded above.

If $(a_n)_{n=1}^{\infty}$ is bounded below but not bounded above then we say that:

(4)

\begin{align} \quad \limsup_{n \to \infty} a_n = \infty \end{align}

Similarly, if $(a_n)_{n=1}^{\infty}$ is bounded above but not bounded below then we say that:

(5)

\begin{align} \quad \liminf_{n \to \infty} a_n = -\infty \end{align}