Prove: $\displaystyle\liminf_{n\to\infty}(-a_n)=-\limsup_{n\to\infty}(a_n)$
My general idea was: if $a$ is partial limit (PL) of $a_n$, then $-a$ is a PL of $-a_n$ so it follows that $s$ is the maximal PL iff $-s$ is the minimal PL.
But how do you show it rigorously ?
Let $n \in \mathbb{N}$. By properties of the supremum and infimum, we have that $$\sup\{−a_k : k \geq n\} = −\inf\{a_k : k \geq n\}.$$ Taking limits on both sides, we find that $$ \limsup_{n \to \infty}\{−a_k : k \geq n\} = \lim_{n \to \infty}(−\inf\{a_k : k \geq n\}) = - \liminf_{n \to \infty} \{a_k : k \geq n\}.$$ Therefore, $$ \limsup−a_n = −\liminf a_n.$$
Note that you can apply this argument to the sequence $(-a_n)_{n \in \mathbb{N}}$ to obtain a similar result, namely $$ \liminf -a_n = -\limsup a_n.$$
This follows easily from the fact that
$$ \sup( A) = - \inf (- A) $$