$\limsup a_n=-\infty$ if $\{a_n\}$ does not have a subsequence which is bounded below I am trying to prove the second part of the following question stated as 1. If $\\{a_n\\}$ has a subsequence which is bounded below by $\alpha$ then $\limsup a_n\geq \alpha$. 2. If $\\{a_n\\}$ does not have a subsequence which is bounded below, then $\limsup a_n=-\infty$. Here is my solution for the first question; > Let $\\{a_{{n}_{k}}\\}\subset\\{a_n\\}$ be bounded below by $\alpha$. Then, $\exists\;\alpha\in \Bbb{R}:\alpha\leq \\{a_{{n}_{k}}\\},\;k\geq1$ $$\implies \alpha\leq a_{{n}_{k}}\leq \sup_{j \ge n_k}\\{a_{j}\\},\forall\;k\geq1$$ $$\implies \alpha\leq\lim_{k\to \infty} \sup_{j \ge n_k}\\{a_{j}\\}\leq \lim_{n\to \infty}\sup\\{a_{{n}}\\}$$ $$\implies \alpha\leq \lim_{n\to \infty}\sup\\{a_{{n}}\\}$$ Please, can anyone help me with the second question?

If $a_n$ has no subsequence which is bounded below, then $$\forall\alpha\in\mathbf R\space \forall N \space \exists n > N(a_n<\alpha) $$ for if not, the sequence $a_N,a_{N+1},\dots$ would be bounded below by $\alpha$. So for every $\alpha$ we can construct a subsequence whose sup is less than $\alpha$, and from there you should have plain sailing.