Limit Sup/Inf Proper Divergence Criterion for 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}

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}

$\Rightarrow$ 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}

$\Rightarrow$ 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 $-\infty$ . $\blacksquare$

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