show that $\lim\limits_{n\to \infty}\sup (-1)^{n}n=\infty$

Question

How do i show formally that : $\lim\limits_{n\to \infty}\sup (-1)^{n}n=\infty$.
I know that if n is odd, then lim inf will be $-\infty$ and if n is even then lim sup will be $\infty$. However, i dont know how to prove it and show how i arrived to the conclusion of both of these statements. In general, i dont know how to prove what is lim inf or lim sup since it doesnt have an epsilon definition like normal limits.

Answer 1

Recall that $\limsup_{n \to \infty} (-1)^n n = \inf_{n \in \mathbf{N}} \sup_{m \geq n} (-1)^m m = \inf_{n \in \mathbf{N}} \{ +\infty \} = +\infty$.

To see why for all $n \in \mathbf{N}$, $\sup_{m \geq n} (-1)^m m = +\infty$, it suffices to show for all $n, T \in \mathbf{N}$, there exists $M \geq n$ such that $(-1)^M M \geq T$. Since then we have $\sup_{m \geq n} (-1)^n n \geq (-1)^M M \geq T$. This is true for any $T \in \mathbf{N}$ and thus $\sup_{m \geq n} (-1)^n n = +\infty$. (Otherwise we have contradiction by the Archimedean property of Real Number.)

It is left to show the sufficient condition: Let $n, T \in \mathbf{N}$ be given. Choose $M > \max\{ T, n \}$ with $M$ even, then $(-1)^M M = M \geq T$.

This concludes the proof.

Answer 2

Hint: Fix $n$ and compute $\sup_{m \geq n} (-1)^m m$. Use the definition of limit superior along with your findings.