Properties of the Limit Superior Inferior of a Sequence of Real Numbers

# Properties of the Limit Superior Inferior of a Sequence of Real Numbers

We will now establish some more very important properties of the limit superior and limit inferior of a sequence of real numbers.

 Theorem 1: Let $(a_n)_{n=1}^{\infty}$ and $(b_n)_{n=1}^{\infty}$ be sequences of real numbers. Then: a) $\displaystyle{\limsup_{n \to \infty} (a_n + b_n) \leq \limsup_{n \to \infty} a_n + \limsup_{n \to \infty} b_n}$. b) $\displaystyle{\liminf_{n \to \infty} (a_n + b_n) \geq \liminf_{n \to \infty} a_n + \liminf_{n \to \infty} b_n}$.
• Proof of a) Let $(a_n)_{n=1}^{\infty}$ and $(b_n)_{n=1}^{\infty}$ be sequences of real numbers. Let $\sup_{k \geq n} \{ a_k \} = A_n^*$ and let $\sup_{k \geq n} \{ b_k \} = B_n^*$. Then for all $k \geq n$ we see that $a_k \leq A_n^*$ and $b_k \leq B_n^*$. So then for all $k \geq n$ we have that:
(1)
\begin{align} \quad a_k + b_k \leq A_n^* + B_n^* \end{align}
• This shows that:
(2)
\begin{align} \quad \sup_{k \geq n} \{ a_k + b_k \} \leq A_n^* + B_n^* = \sup_{k \geq n} \{ a_k \} + \sup_{k \geq n} \{ b_n \} \end{align}
• Taking the limit as $n \to \infty$ shows us that $\displaystyle{\limsup_{n \to \infty} (a_n + b_n) \leq \limsup_{n \to \infty} a_n + \limsup_{n \to \infty} b_n}$. $\blacksquare$
• Proof of b) Once again, let $(a_n)_{n=1}^{\infty}$ and $(b_n)_{n=1}^{\infty}$ be sequences of real numbers. Let $\inf_{k \geq n} \{ a_k \} = A_{n*}$ and let $\inf_{k \geq n} \{ b_k \} = B_{n*}$. Then for all $k \geq N$ we see that $A_{n*} \leq a_k$ and $B_{n*} \leq b_k$. So then for all $k \geq n$ we have that:
(3)
\begin{align} \quad A_{n*} + B_{n*} \leq a_k + b_k \end{align}
• This shows that:
(4)
\begin{align} \quad \inf_{k \geq n} \{ a_k \} + \inf_{k \geq n} \{ b_k \} = A_{n*} + B_{n*} \leq \inf_{k \geq n} \{ a_k + b_k \} \end{align}
• Taking the limit as $n \to \infty$ shows us that $\displaystyle{\liminf_{n \to \infty} (a_n + b_n) \geq \liminf_{n \to \infty} a_n + \liminf_{n \to \infty} b_n}$. $\blacksquare$
 Theorem 2: Let $(a_n)_{n=1}^{\infty}$ and $(b_n)_{n=1}^{\infty}$ be two sequences of real numbers. a) If $a_n, b_n > 0$ for all $n \in \mathbb{N}$ and $\displaystyle{\limsup_{n \to \infty} a_n}$ and $\displaystyle{\limsup_{n \to \infty} b_n}$ are both finite then $\displaystyle{\limsup_{n \to \infty} (a_nb_n) \leq \left ( \limsup_{n \to \infty} a_n \right ) \left ( \limsup_{n \to \infty} b_n \right )}$. b) If $a_n, b_n > 0$ for all $n \in \mathbb{N}$ and $\displaystyle{\liminf_{n \to \infty} a_n}$ and $\displaystyle{\liminf_{n \to \infty} b_n}$ are both finite then $\displaystyle{\liminf_{n \to \infty} (a_nb_n) \geq \left ( \liminf_{n \to \infty} a_n \right ) \left ( \liminf_{n \to \infty} b_n \right )}$.
• Proof of a): Let $a_n, b_n > 0$ for all $n \in \mathbb{N}$. Let $A = \limsup_{n \to \infty} a_n$ and $B = \limsup_{n \to \infty} b_n$ where $A$ and $B$ are finite.
• Fix $n \in \mathbb{N}$ and let $\displaystyle{\sup_{k \geq n} \{ a_k \} = A_n}$ and $\displaystyle{\sup_{k \geq n} \{ b_k \} = B_n}$. Then for all $k \geq n$ we have that $0 < a_k \leq A_n$ and $0 < b_k \leq B_n$. Multiplying these together shows that $0 < a_kb_k < A_nB_n$ for all $k \geq n$. Therefore $A_nB_n$ is an upper bound to $\displaystyle{\{ a_kb_k \}_{k \geq n}}$ and so:
(5)
\begin{align} \quad \sup_{k \geq n} \{ a_kb_k \} \leq A_nB_n \end{align}
• Taking the limit as $n \to \infty$ gives us that:
(6)
\begin{align} \quad \limsup_{n \to \infty} a_nb_n & \leq \limsup_{n \to \infty} A_nB_n \\ \quad & \leq AB \\ \quad & \leq \left ( \limsup_{n \to \infty} a_n \right ) \left ( \limsup_{n \to \infty} b_n \right ) \quad \blacksquare \end{align}
• Proof of b) Let $a_n, b_n > 0$ for all $n \in \mathbb{N}$. Let $A = \liminf_{n \to \infty} a_n$ and $B = \liminf_{n \to \infty} b_n$ where $A$ and $B$ are finite.
• Fix $n \in \mathbb{N}$ and let $\displaystyle{\inf_{k \geq n} \{ a_k \} = A_n}$ and $\displaystyle{\inf_{k \geq n} \{ b_k \} = B_n}$. Then for all $k \geq n$ we have that $0 \leq A_n \leq a_k$ and $0 \leq B_n \leq b_k$. Multiplying these together shows that $0 \leq A_nB_n < a_kb_k$ for all $k \geq n$. Therefore $A_nB_n$ is a lower bound to $\displaystyle{\{ a_kb_k \}_{k \geq n}}$ and so:
(7)
\begin{align} \quad A_nB_n \leq \inf_{k \geq n} \{ a_k b_k \} \end{align}
• Taking the limit as $n \to \infty$ gives us that:
(8)
\begin{align} \quad \liminf_{n \to \infty} a_nb_n & \geq \lim_{n \to \infty} A_nB_n \\ \quad & \geq AB \\ \quad & \geq \left ( \liminf_{n \to \infty} a_n \right ) \left ( \liminf_{n \to \infty} b_n \right ) \quad \blacksquare \end{align}