Proving that $\limsup a_n$ and $\liminf a_n$ are subsequential limits of $\{a_n\}$

Question

Prove that if $\{a_n\}$ is a sequence, then $\limsup a_n$ and $\liminf a_n$ are subsequential limits of $\{a_n\}$.

I don't know the case where $\limsup a_n = \infty$.

Answer 1

$$\limsup a_n = \inf_{k}\big( \sup_{n\ge k} a_n\big)$$ This is $+\infty$ iff $\forall k:\ \sup_{n\ge k} a_n=+\infty$. Then for each given previous index $k$ and $z\in\Bbb N$, there is a next index $n\ge{k+1}$ such that $a_n>z$. This selected subsequence is thus bigger than the sequence $1,2,3,4,5,..$, hence its limit is $+\infty$.

Answer 2

For the case you asking for let's take this sequence: $$(-1)^{n}n$$ which is: $$\{-1,2,-3,4,-5 \cdots\}$$ for this sequence is $\liminf = - \infty$ and $\limsup = \infty$. This are limits of subsequences of odd and even terms respectivly.