Proof about infinite limsups and subsequences I am working on a final exam study guide, and came across this question:

Suppose limsup$(a_n)$ = $\infty$. Prove: There must exist a sub-sequence ${a_n}_k$ such that ${a_n}_k \to \infty$.

My initial thought was to show that since the limsup $=\infty$, then $a_n \to \infty$ as well. So here, I want to say that if I remove the first term, or even the second, no matter how many head terms I remove from $a_n$, I get that $a_k \to \infty$, and by extension, ${a_n}_k \to \infty$ as well. Is this a proper way of going about this?

I would just like to note and add, as it has caused some confusion (my fault for not noting) From my notes in class: $$limsup \space s_n = lim_{N \to \infty} \space sup(s_n:n&gt;N)$$
As the sequence $\;\sup\limits_{k&gt;n} a_k$ is non-increasing, the hypothesis means each $\;\sup\limits_{k&gt;n} a_k=\infty$. In other words, for any $A$ and any $n$, there exist $k&gt;n$ such that $a_k&gt;A$.

So let's start with $A=1$, and let $n_1$ the smallest $k&gt;0$ such  $a_k&gt;1$. Then let $n_2$ the smallest $k&gt;n_1$ such that $a_k&gt;\max(2, a_{n_1})$. More generally, supposing
$a_{n_1},\dots,a_{n_r}$ have been defined, $a_{n_{r+1}}$ will be defined as the smallest $k&gt;n_r$ such that $a_k&gt;\max(r+1,a_{n_r})$.

By construction, this is an increasing subsequence wich tends to $+\infty$.
If there is no subsequence $a_{n_k}$ s.t. $a_{n_k} \to \infty$, then $a_{n_k} &lt; \infty$ for all $k$. This implies $\limsup a_n &lt; \infty$.
It turns out that the implication above is not true. How about the following arguments?

Since $\limsup a_n = \infty$ and $\limsup a_n$ is a decreasing sequence, therefore the sequence $(a_n)$ is unbounded.