The Connection Between the Limit Sup/Inf of a Seq. of Real Numbers

The Connection Between the Limit Superior Inferior of a Sequence of Real Numbers

Recall from 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 defined to be:

(1)
\begin{align} \quad \limsup_{n \to \infty} = \lim_{n \to \infty} \left ( \sup_{k \geq n} \{ a_n \} \right ) \end{align}

Similarly, the limit inferior of $(a_n)_{n=1}^{\infty}$ is defined to be:

(2)
\begin{align} \quad \liminf_{n \to \infty} = \lim_{n \to \infty} \left ( \inf_{k \geq n} \{ a_n \} \right ) \end{align}

In the following theorem we will state a connection between the limit superior and limit inferior of a sequence of real numbers.

 Theorem 1: Let $(a_n)_{n=1}^{\infty}$ be a sequence of real numbers. Then $\displaystyle{\limsup_{n \to \infty} (-a_n) = - \liminf_{n \to \infty} a_n}$.
• Proof: For each $n \in \mathbb{N}$ we note that:
(3)
\begin{align} \quad \sup_{k \geq n} \{ -a_k \} = - \inf_{k \geq n} \{ a_k \} \end{align}
(4)
\begin{align} \quad \inf_{k \geq n} \{ a_k \} \leq a_k \end{align}
• Multiplying both sides by $-1$ shows us that:
(5)
\begin{align} \quad -a_k \leq -\inf_{k \geq n} \{ a_k \} \end{align}
• Therefore $\displaystyle{ -\inf_{k \geq n} \{ a_k \}}$ is an upper bound to the set $\displaystyle{\{ -a_k \}_{k \geq n}}$, so:
(6)
\begin{align} \quad \sup_{k \geq n} \{-a_k \} \leq -\inf_{k \geq n} \{ a_k \} \quad (*) \end{align}
• Now let $v$ be any upper bound to the set $\displaystyle{\{ -a_k \}_{k \geq n}}$. Then for all $k \geq n$ we have that $-a_k \leq v$. So $-v \leq a_k$ for all $k \geq n$. Therefore $-v$ is a lower bound to the set $\displaystyle{\{ a_k \}_{k \geq n}}$ and thus $\displaystyle{-v \leq \inf_{k \geq n} \{a_k \}}$, so $v \geq -\inf_{k \geq n} \{a_k \}$. But $v$ is an upper bound to the set $\displaystyle{\{ -a_k \}_{k \geq n}}$ which shows that:
(7)
\begin{align} \quad -\inf_{k \geq n} \{ a_k \} \leq \sup_{k \geq n} \{ -a_k \} \quad (**) \end{align}
• Combining $(*)$ and $(**)$ shows us that:
(8)
\begin{align} \quad \sup_{k \geq n} \{ - a_k \} = -\inf_{k \geq n} \{ a_k \} \end{align}
• Taking the limit as $n \to \infty$ shows us that $\displaystyle{\limsup_{n \to \infty} (-a_n) = - \liminf_{n \to \infty} a_n}$. $\blacksquare$