The question I'm having trouble with is the following:
Let $a_n$ be defined for all $n \in \mathbb{N}$ by $a_n = \frac{1}{n^3}$ if $n$ is even, and $a_n = \frac{-1}{n^2}$ if $n$ is odd. Define $A = \{a_n | n \in \mathbb{N} \}$ and $\sup_{n \geq k} a_n = \sup \{a_n | n \geq k \}$.
I need to prove that $$\sup_{k \geq 1 }\inf_{n \geq k} a_n = \inf_{k \geq 1} \sup_{n \geq k} a_n.$$
I'm incredibly confused as to how I'm supposed to take the supremum of an infimum or vice versa. I'd like to say that $\inf_{n \geq k} a_n = a_k$ if $k$ is odd, or $a_{k+1}$ if $k$ is even, with the opposite true for the supremum. However, I am unsure about how to proceed with finding $\sup_{k \geq 1 }\inf_{n \geq k} a_n$, much less proving the inequality.
Any help would be appreciated.
It is well known limit inferior and limit superior for sequences $$\limsup\limits_{n \to \infty}a_n = \inf\limits_{n} \sup\limits_{k \geqslant n}a_k$$ $$\liminf\limits_{n \to \infty}a_n = \sup\limits_{n} \inf\limits_{k \geqslant n}a_k$$ In another words they are $\sup$ and $\inf$ of limit points for given sequences. As your sequence is converged and $\lim\limits_{n \to \infty}a_n=0$, then it have only one limit point, which equals as to limit superior, so limit inferior.