The Limit Superior and Limit Inferior of Sequences of Real Numbers

# The Limit Superior and Limit Inferior of Sequences of Real Numbers

We are now ready to define two important concepts for sequences of real numbers - the limit superior and limit inferior.

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

Recall from The Supremum and Infimum Properties for Subsets of Sets of Real Numbers page that if $A, B \subseteq \mathbb{R}$ are nonempty and if $A$ is bounded above and $B \subseteq A$ then:

(1)
\begin{align} \quad \sup B \leq \sup A \end{align}

Similarly, if $A$ is bounded below then:

(2)
\begin{align} \quad \inf B \geq \inf A \end{align}

Let $(a_n)_{n=1}^{\infty}$ be a sequence that is bounded above. Then the set $\{ a_n : n \in \mathbb{N} \}$ is bounded above and any subset of this set is bounded above. Therefore $\displaystyle{\sup_{k \geq n} \{ a_k \}}$ exists for every $n \in \mathbb{N}$ and furthermore, the sequence $\displaystyle{(\sup_{k \geq n} \{ a_k \})_{n=1}^{\infty}}$ is a decreasing sequence. Hence the limit of this sequence will be the infimum of this set, that is:

(3)
\begin{align} \quad \limsup_{n \to \infty} a_n = \lim_{n \to \infty} \sup_{k \geq n} \{ a_k \} = \inf_{n \geq 1} \left \{ \sup_{k \geq n} \{ a_k \} \right \} \end{align}

Similarly, if $(a_n)_{n=1}^{\infty}$ is a sequence that is bounded below, then the set $\{ a_n : n \in \mathbb{N} \}$ is bounded below and any subset of this set is bounded below. Therefore $\displaystyle{\inf_{k \geq n} \{ a_k \}}$ exists for every $n \in \mathbb{N}$ and furthermore, the sequence $\displaystyle{(\inf_{k \geq n} \{ a_k \})_{n=1}^{\infty}}$ is an increasing sequence. Hence the limit of this sequence will be the supremum of this set, that is:

(4)
\begin{align} \quad \liminf_{n \to \infty} a_n = \lim_{n \to \infty} \inf_{k \geq n} \{ a_k \} = \sup_{n \geq 1} \left \{ \inf_{k \geq n} \{ a_k \} \right \} \end{align}

We now have two different ways to express the limit superior and limit inferior of a sequence of real numbers. We now establish some important results regarding the general limit of $(a_n)_{n=1}^{\infty}$ and the limit superior and limit inferior.

 Lemma 1: Let $(a_n)_{n=1}^{\infty}$ be a sequence of real numbers. Then $\displaystyle{\liminf_{n \to \infty} a_n \leq \limsup_{n \to \infty} a_n}$.
• Proof: For all $n \in \mathbb{N}$ we have that:
(5)
\begin{align} \quad \inf_{k \geq n} \{ a_k \} \leq \sup_{k \geq n} \{ a_k \} \end{align}
• If we take the limit of both sides of the inequality above we get that $\displaystyle{\liminf_{n \to \infty} a_n \leq \limsup_{n \to \infty} a_n}$. $\blacksquare$.
 Theorem 2: Let $(a_n)_{n=1}^{\infty}$ be a sequence of real numbers. Then $\displaystyle{\lim_{n \to \infty} a_n = A}$ if and only if $\displaystyle{\liminf_{n \to \infty} a_n = A = \limsup_{n \to \infty} a_n}$.
• Proof: $\Rightarrow$ Suppose that $\displaystyle{\lim_{n \to \infty} a_n = A}$. Then for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then:
(6)
\begin{align} \quad \mid a_n - A \mid < \epsilon \quad \Leftrightarrow \quad A - \epsilon < a_n < A + \epsilon \end{align}
• So for $n \geq N$ we have that:
(7)
\begin{align} \quad A - \epsilon < \sup_{k \geq n} \{ a_n \} < A + \epsilon \quad \Leftrightarrow \quad \biggr \lvert \sup_{k \geq n} \{ a_n \} - A \biggr \rvert < \epsilon \quad (*) \end{align}
(8)
\begin{align} \quad A - \epsilon < \inf_{k \geq n} \{ a_n \} < A + \epsilon \quad \Leftrightarrow \quad \biggr \lvert \inf_{k \geq n} \{ a_n \} - A \biggr \rvert < \epsilon \quad (**) \end{align}
• Thus $\displaystyle{\limsup_{n \to \infty} a_n = A}$ and $\displaystyle{\liminf_{n \to \infty} a_n = A}$.
• $\Leftarrow$ Suppose that $\displaystyle{\limsup_{n \to \infty} a_n = A}$ and $\displaystyle{\liminf_{n \to \infty} a_n = A}$. Then for all $\epsilon > 0$ there exists $N_1, N_2 \in \mathbb{N}$ such that if $n \geq N_1$ we have that $(*)$ holds and if $n \geq N_2$ we have that $(**)$ holds. Let $N = \max \{ N_1, N_2 \}$. Then if $n \geq N$ both $(*)$ and $(**)$ holds so:
(9)
\begin{align} \quad A - \epsilon < \inf_{k \geq n} \{ a_n \} \leq a_n \leq \sup_{k \geq n} \{ a_n \} < A + \epsilon \quad \Leftrightarrow \quad \biggr \lvert a_n - A \biggr \rvert < \epsilon \end{align}
• So $\displaystyle{\lim_{n \to \infty} a_n = A}$. $\blacksquare$