Limit Sup/Inf Proper Divergence Criterion for Seqs. of Real Numbers

# Limit Superior Inferior Proper Divergence Criterion for 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}

On the Limit Superior/Inferior Convergence Criterion for Sequences of Real Numbers page we also saw that if $(a_n)_{n=1}^{\infty}$ is a sequence of real numbers then:

(3)
\begin{align} \quad \liminf_{n \to \infty} a_n \leq \limsup_{n \to \infty} a_n \end{align}

Furthermore we saw that $(a_n)_{n=1}^{\infty}$ converges to a finite $A \in \mathbb{R}$ if and only if:

(4)
\begin{align} \quad \liminf_{n \to \infty} a_n = A = \limsup_{n \to \infty} a_n \end{align}

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

 Theorem 1: A sequence of real numbers $(a_n)_{n=1}^{\infty}$ diverges to $\infty$ if and only if $\displaystyle{\liminf_{n \to \infty} a_n = \infty = \limsup_{n \to \infty} a_n}$.
• Proof: $\Rightarrow$ Suppose that $(a_n)_{n=1}^{\infty}$ diverges to $\infty$. Then for all $M \in \mathbb{R}$, $M > 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then $a_n \geq M$.
• So for all $n \geq N$, $M$ is a lower bound to the set $\displaystyle{\{ a_k \}_{k \geq n}}$. So for all $n \geq N$:
(5)
\begin{align} \quad M \leq \inf_{k \geq n} \{ a_k \} \leq \sup_{k \geq n} \{ a_k \} \end{align}
• Therefore $\liminf_{n \to \infty} a_n = \infty = \limsup_{n \to \infty} a_n$.
• $\Rightarrow$ Suppose that $\displaystyle{\liminf_{n \to \infty} a_n = \infty = \limsup_{n \to \infty} a_n}$. Then for all $M \in \mathbb{R}$, $M > 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then:
(6)
\begin{align} \quad M \leq \inf_{k \geq n} \{ a_k \} \leq \sup_{k \geq n} \{ a_k \} \end{align}
• But for all $n \geq N$ we have that then $M \leq a_n$. So for all $M \in \mathbb{R}$, $M > 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then $M \leq a_n$. Thus $(a_n)_{n=1}^{\infty}$ diverges to $\infty$. $\blacksquare$
 Theorem 2: A sequence of real numbers $(a_n)_{n=1}^{\infty}$ diverges to $-\infty$ if and only if $\displaystyle{\liminf_{n \to \infty} a_n = -\infty = \limsup_{n \to \infty} a_n}$.
• Proof: $\Rightarrow$ Suppose that $(a_n)_{n=1}^{\infty}$ diverges to $-\infty$. Then for all $M \in \mathbb{R}$, $M < 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then $a_n \leq M$.
• So for all $n \geq N$, $M$ is an upper bound to the set $\displaystyle{\{ a_k \}_{k \geq n}}$. So for all $n \geq N$:
(7)
\begin{align} \quad \inf_{k \geq n} \{ a_k \} \leq \sup_{k \geq n} \{ a_k \} \leq M \end{align}
• Therefore $\liminf_{n \to \infty} a_n = -\infty = \limsup_{n \to \infty} a_n$.
• $\Rightarrow$ Suppose that $\displaystyle{\liminf_{n \to \infty} a_n = -\infty = \limsup_{n \to \infty} a_n}$. Then for all $M \in \mathbb{R}$, $M < 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then:
(8)
\begin{align} \quad \inf_{k \geq n} \{ a_k \} \leq \sup_{k \geq n} \{ a_k \} \leq M \end{align}
• But for all $n \geq N$ we have that then $a_n \leq M$. So for all $M \in \mathbb{R}$, $M < 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then $a_n \leq M$. Thus $(a_n)_{n=1}^{\infty}$ diverges to $-\infty$. $\blacksquare$