lim inferior and Lim superior

Question

Let $(S_n) ⊂ (0, +∞)$ be a sequence of real numbers.

Prove that $$\liminf\limits_{n \to \infty} \frac{1}{S_n} = \frac{1}{\limsup\limits_{n \to \infty } S_n} $$.

How can i tackle this problem? can anybody help me out

Answer 1

To obtain desired equality we need to assume, that $\lim \sup S_n > 0$. Now assuming, that $E$ is set of limit points for $S_n$, we have $\sup E = \lim \sup S_n $. For simplicity let's take case, when $+\infty \notin E$. It's obvious, that if $\alpha \in E$, then $\frac{1}{\alpha}$ will be limit point for $\frac{1}{S_n}$, so if we denote by $F$ set of limit points for $\frac{1}{S_n}$, then all we need to prove is $$(\sup E)^{-1} = (\sup \{ \alpha \})^{-1}= \inf F = \inf \left\{ \frac{1}{\alpha} \right\}$$

1. $\forall \alpha \in E$ we have $ \alpha \leqslant \sup E \Rightarrow \frac{1}{\alpha} \geqslant \frac{1}{\sup E}$

2. Let's take $\forall \epsilon_1 >0$ and consider $\frac{1}{\sup E} + \epsilon_1 $. It is possible to find such $\epsilon_2 >0$, that $\frac{1}{\sup E} + \epsilon_1 = \frac{1}{\sup E - \epsilon_2}$. From definition of $\sup E$ we found some $\alpha_0 \in E$, that $\alpha_0 > \sup E - \epsilon_2 \Rightarrow \frac{1}{\alpha_0} <\frac{1}{\sup E - \epsilon_2} = \frac{1}{\sup E} + \epsilon_1$, which finish proof, that $\frac{1}{\sup E} = \inf F$