Comparison Theorems for the Limit Sup/Inf of Seqs. of Real Numbers

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_k \} \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_k \} \right ) \end{align}

We will now look at some nice comparison theorems regarding the limit superior/inferior of sequences 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_k \} \leq \sup_{k \geq n} \{ a_k \} \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: Let

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

and

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

be sequences of real numbers such that

a_n \leq b_n

for all

n \in \mathbb{N}

. Then:
a)

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

.
b)

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

Proof of a) Suppose that $a_n \leq b_n$ for all $n \in \mathbb{N}$ . Then for all $n \in \mathbb{N}$ we have that:

(4)

\begin{align} \quad \inf_{k \geq n} \{ a_k \} \leq \inf_{k \geq n} \{ b_k\} \end{align}

Taking the limit as $n \to \infty$ shows us that $\displaystyle{\liminf_{n \to \infty} a_n \leq \liminf_{n \to \infty} b_n}$ . $\blacksquare$

(5)

\begin{align} \quad \sup_{k \geq n} \{ a_k \} \leq \sup_{k \geq n} \{ b_k \} \end{align}

Taking the limit as $n \to \infty$ shows us that $\displaystyle{\limsup_{n \to \infty} a_n \leq \limsup_{n \to \infty} b_n}$ . $\blacksquare$

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