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

# Comparison Theorems for the Limit Superior/Inferior of Sequences of Real Numbers

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 $\displaystyle{\liminf_{n \to \infty} a_n \leq \liminf_{n \to \infty} b_n}$. $\blacksquare$
• Proof of b) If $a_n \leq b_n$ for all $n \in \mathbb{N}$ then we have that:
(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 $\displaystyle{\limsup_{n \to \infty} a_n \leq \limsup_{n \to \infty} b_n}$. $\blacksquare$