Basic Theorems Regard. the Limit Sup/Inf of a Seq. of Real Numbers 1

Table of Contents

Recall from the The Limit Superior and Limit Inferior of a Sequence of Real Numbers page that if $(a_n)_{n=1}^{\infty}$ is a sequence of real numbers then the limit superior of $(a_n)_{n=1}^{\infty}$ is:

(1)

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

The limit inferior of $(a_n)_{n=1}^{\infty}$ is:

(2)

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

We will now look some basic theorems regarding the limit superior and inferior of a sequence of real numbers.

Theorem 1: If

(a_n)_{n=1}^{\infty}

is a sequence of real numbers then

\displaystyle{\liminf_{n \to \infty} a_n \leq \limsup_{n \to \infty} a_n}

Proof: Let $(a_n)_{n=1}^{\infty}$ be a sequence of real numbers. Then by the definition of the infimum and supremum of a set, we have that for all $n \in \mathbb{N}$ that:

(3)

\begin{align} \quad \inf_{k \geq n} \{ a_n \} \leq \sup_{k \geq n} \{ a_n \} \end{align}

Taking the limit as $n \to \infty$ and applying the squeeze theorem gives us $\displaystyle{\liminf_{n \to \infty} a_n \leq \limsup_{n \to \infty} a_n}$ . $\blacksquare$

Theorem 2: A sequence

(a_n)_{n=1}^{\infty}

converges to a finite

A \in \mathbb{R}

if and only if

\displaystyle{\liminf_{n \to \infty} a_n = A = \limsup_{n \to \infty} a_n}

Proof: $\Rightarrow$ Suppose that $(a_n)_{n=1}^{\infty} = A$ . Then for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then:

(4)

\begin{align} \quad A - \epsilon < a_n < A + \epsilon \end{align}

So for all $n \geq N$ we have that $\displaystyle{A - \epsilon < \sup_{k \geq n} < A + \epsilon}$ , i.e., $\displaystyle{\mid \sup_{k \geq n} - A \mid < \epsilon}$ . So $\displaystyle{\limsup_{n \to \infty} a_n = A}$ .

Similarly we have that $\displaystyle{A - \epsilon < \inf_{k \geq n} < A + \epsilon}$ , i.e., $\displaystyle{\mid \inf_{k \geq n} - A \mid < \epsilon}$ . So $\displaystyle{\liminf_{n \to \infty} a_n = A}$ .

$\Leftarrow$ Suppose that $\displaystyle{\liminf_{n \to \infty} a_n = A = \limsup_{n \to \infty} a_n}$ for a finite $$A$$ . Since $\displaystyle{\liminf_{n \to \infty} a_n = A}$ we have that for a given $\epsilon > 0$ there exists an $N_1 \in \mathbb{N}$ such that if $n \geq N_1$ then:

(5)

\begin{align} \quad A - \epsilon < \inf_{k \geq n} a_n < A + \epsilon \quad (*) \end{align}

So we have that $A - \epsilon_1$ is an lower bound for $\displaystyle{\{ a_n \}_{k \geq n}}$ .

Similarly, since $\displaystyle{\limsup_{n \to \infty} a_n = A}$ we have that for $\epsilon >0$ there exists an $N_2 \in \mathbb{N}$ such that if $n \geq N_2$ then:

(6)

\begin{align} \quad \quad A - \epsilon < \sup_{k \geq n} a_n < A + \epsilon \quad (**) \end{align}

Let $N = \max \{ N_1, N_2 \}$ . So if $n \geq N$ then $$(*)$$ and $$(**)$$ hold, so for all $n \geq N$ :

(7)

\begin{align} \quad A - \epsilon < a_n < A + \epsilon \\ \quad \mid a_n - A \mid < \epsilon \end{align}

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