limit superior and limit inferior proof

Question

$$\limsup \left(\frac 1{a_n} \right)=\frac 1{\liminf(a_n )} $$

I know this is true base on the definition of $\limsup$ and $\liminf$, but I don't know how to prove it formally.

Answer 1

Assume that a subsequence $a_{n_k} \to a$. Of course we need some assumptions, and I'll assume that $a_n>0$ for any $n$. The $$\frac{1}{a_{n_k}}\to \frac{1}{a},$$ and since $a=\liminf_{n \to +\infty} a_n$ is the smallest possible choice, we deduce at once that the largest possible limit point of $1/a_n$ is $1/a$. Hence $$\frac{1}{a} = \limsup_{n \to +\infty} \frac{1}{a_n}.$$

Answer 2

$$\lim \sup \left(\frac {1}{a_n}\right) = \inf_{k=1}^{\infty}\sup_{n=k}^{\infty}\left(\frac {1}{a_n}\right) = \left(\frac {1}{\sup_{k=1}^{\infty}\inf_{n=k}^{\infty} a_n}\right)\dots$$