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 $\displaystyle{\lim_{n \to \infty} a_n = A}$. $\blacksquare$