I am trying to solve the following problem.
Let $(a_n)$ be a positive sequence. Prove that
$$\dfrac {1}{\lim \sup a_n} = \lim \inf \left( \dfrac {1}{a_n} \right)$$ in the sense that if one side lies in $(0, \infty)$, then so does the other and the two sides are equal.
I am not really sure how to approach the last sentence.
Assuming everything exists, I believe that this problem comes down to showing that if $A$ is a set of positive numbers, then $$\inf \dfrac {1}{A} = \dfrac {1}{\sup A}$$
where $\frac{1}{A} = \left\{\frac{1}{a} \mid a \in A \right\}$.
I can prove that $\dfrac {1}{\sup A}$ is a lower bound for $1/A$, because for each $a \in A$ we have $a \le \sup A$, so $\frac 1a \ge \frac {1}{ \sup A}$. But I am having trouble proving that $\frac {1}{\sup A}$ is least among the upper bounds. I know that for $\epsilon > 0$, we have an $a \in A$ such that $\sup A - \epsilon < a \le \sup A$, but this turns into $$\dfrac {1}{\sup A - \epsilon} > a \ge \dfrac {1}{\sup A}$$ It makes sense that I can choose $\epsilon$ smaller and smaller and squeeze $a$ from above, but I don't know how to show it formally.
Any help is greatly appreciated.