The Limit Sup/Inf of the Ratio of Terms in Pos. Seqs. of Real Numbers

# The Limit Superior/Inferior of the Ratio of Terms in Positive Sequences of Real Numbers

 Theorem 1: Let $(a_n)_{n=1}^{\infty}$ be a sequence of real numbers such that $a_n > 0$ for all $n \in \mathbb{N}$. Then we have the following chain of inequalities: $\displaystyle{\liminf_{n \to \infty} \frac{a_{n+1}}{a_n} \leq \liminf_{n \to \infty} (a_n)^{1/n} \leq \limsup_{n \to \infty} (a_n)^{1/n} \leq \limsup_{n \to \infty} \frac{a_{n+1}}{a_n}}$.
• Proof: We begin by establishing $\displaystyle{\liminf_{n \to \infty} \frac{a_{n+1}}{a_n} \leq \liminf_{n \to \infty} (a_n)^{1/n}}$.
• Since $a_n > 0$ for all $n \in \mathbb{N}$ we see that $\displaystyle{\liminf_{n \to \infty} \frac{a_{n+1}}{a_n} \geq 0}$. So there are two cases to consider.
• Case 1: Suppose that $\liminf_{n \to \infty} \frac{a_{n+1}}{a_n} = A$ for some $A \in \mathbb{R}$. So for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ we have that:
(1)
\begin{align} \quad \biggr \lvert \inf_{k \geq n} \left \{ \frac{a_{k+1}}{a_k} \right \} - A \biggr \rvert < \epsilon \quad \Leftrightarrow \quad A - \epsilon < \inf_{k \geq n} \left \{ \frac{a_{k+1}}{a_k} \right \} < A + \epsilon \end{align}
• Then for all $n \geq N$ we have that:
(2)
\begin{align} \quad A - \epsilon & \leq \frac{a_{n+1}}{a_n} \\ \quad a_n(A - \epsilon) & \leq a_{n+1} \end{align}
• So inductively we see that for any $n \geq N$:
(3)
\begin{align} \quad a_{N+1} & \geq a_N(A - \epsilon) \\ \quad a_{N+2} & \geq a_{N+1}(A - \epsilon) \leq a_N(A - \epsilon)^2 \\ \quad & \vdots \\ \quad a_{N+n} & \geq a_{N+n-1}(A - \epsilon) \leq a_N(A -\epsilon)^n \end{align}
• Taking the $n^{\mathrm{th}}$ root of both sides of above yields:
(4)
\begin{align} \quad (a_{N+n})^{1/n} \geq a_N^{1/n} (A - \epsilon) \end{align}
• And taking the limit inferior of both sides yields:
(5)
\begin{align} \quad \liminf_{n \to \infty} (a_n)^{1/n} = \liminf_{n \to \infty} (a_{N+n})^{1/n} \geq \liminf_{n \to \infty} a_N^{1/n} (A - \epsilon) = A - \epsilon \end{align}
• But $\epsilon > 0$ is arbitrary which shows that $\displaystyle{\liminf_{n \to \infty} \frac{a_{n+1}}{a_n} \leq \liminf_{n \to \infty} (a_n)^{1/n}}$.
• Case 2: Suppose that $\displaystyle{\liminf_{n \to \infty} \frac{a_{n+1}}{a_n} = \infty}$. Then there exists an $M \in \mathbb{R}$, $M > 0$ such that if $n \geq N$ then:
(6)
\begin{align} \quad \frac{a_{n+1}}{a_n} & \geq M \\ a_{n+1} & \geq Ma_n \end{align}
• So recursively we get that:
(7)
\begin{align} a_{N+1} & \geq Ma_N \\ a_{N+2} & \geq Ma_{N+1} \geq M^2a_N \\ \quad & \vdots \\ a_{N+n} & \geq Ma_{N+n-1} \geq ... \geq M^na_N \end{align}
• Taking the $n^{\mathrm{th}}$ root of both sides yields:
(8)
\begin{align} \quad (a_{N+n})^{1/n} \geq Ma_N^{1/n} \end{align}
• And taking the limit inferior of both sides gives us that $\displaystyle{\liminf_{n \to \infty} (a_n)^{1/n} = \liminf_{n \to \infty} (a_{N+n})^{1/n} \geq \liminf_{n \to \infty} Ma_N^{1/n} = M}$. But this holds for all $M > 0$ which shows that $\displaystyle{\liminf_{n \to \infty} (a_n)^{1/n} = \infty}$, so the inequality holds in either case.
• We now establish the inequality $\displaystyle{\limsup_{n \to \infty} (a_n)^{1/n} \leq \limsup_{n \to \infty} \frac{a_{n+1}}{a_n}}$. There are two cases to consider.
• Case 1: Suppose that $\limsup_{n \to \infty} \frac{a_{n+1}}{a_n} = A$ for some $A \in \mathbb{R}$. Then for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then:
(9)
\begin{align} \quad A - \epsilon \leq \frac{a_{n+1}}{a_n} \leq A + \epsilon \end{align}
• In particular, for $n \geq N$ we have that:
(10)
\begin{align} \quad \frac{a_{n+1}}{a_n} & \leq A + \epsilon \\ \quad a_{n+1} & \leq a_n(A + \epsilon) \end{align}
• So for $n \geq N$:
(11)
\begin{align} \quad a_{N+1} & \leq a_N(A + \epsilon) \\ \quad a_{N+2} & \leq a_{N+1}(A + \epsilon) \leq a_N(A + \epsilon)^2 \\ \quad & \vdots \\ \quad a_{N+n} & \leq a_{N+n-1}(A + \epsilon) \leq ... \leq a_N(A + \epsilon)^n \end{align}
• Taking the $n^{\mathrm{th}}$ root of both sides of the inequality above yields:
(12)
\begin{align} \quad (a_{N+n})^{1/n} \leq a_N^{1/n} (A + \epsilon) \end{align}
• Taking the limit superior of both sides of the inequality above gives us that $\displaystyle{\limsup_{n \to \infty} (a_n)^{1/n} = \limsup_{n \to \infty} (a_{N+n})^{1/n} \leq \limsup_{n \to \infty} a_N^{1/n}(A + \epsilon) = A + \epsilon}$. But $\epsilon > 0$ is arbitrary. This shows that $\displaystyle{\limsup_{n \to \infty} (a_n)^{1/n} \leq \limsup_{n \to \infty} \frac{a_{n+1}}{a_n}}$.
• Case 2: Suppose that $\displaystyle{\limsup_{n \to \infty} \frac{a_{n+1}}{a_n} = \infty}$. This it is trivially true that $\displaystyle{\limsup_{n \to \infty} (a_n)^{1/n} \leq \infty}$.
• So in total we conclude that:
(13)
\begin{align} \quad \displaystyle{\liminf_{n \to \infty} \frac{a_{n+1}}{a_n} \leq \liminf_{n \to \infty} (a_n)^{1/n} \leq \limsup_{n \to \infty} (a_n)^{1/n} \leq \limsup_{n \to \infty} \frac{a_{n+1}}{a_n}} \quad \blacksquare \end{align}