Proof about infinite limsups and subsequences

Question

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: $$limsup \space s_n = lim_{N \to \infty} \space sup(s_n:n>N)$$

Answer 1

As the sequence $\;\sup\limits_{k>n} a_k$ is non-increasing, the hypothesis means each $\;\sup\limits_{k>n} a_k=\infty$. In other words, for any $A$ and any $n$, there exist $k>n$ such that $a_k>A$.

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

By construction, this is an increasing subsequence wich tends to $+\infty$.

Answer 2

If there is no subsequence $a_{n_k}$ s.t. $a_{n_k} \to \infty$, then $a_{n_k} < \infty$ for all $k$. This implies $\limsup a_n < \infty$. 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.

Answer 3

We need to show that given any $M\in\mathbb{N}$, there is a subsequence $(a_{M_k})$ such that $a_{M_k} > M \forall k$.

Fix $M$. By definition of lim sup and hypothesis, there is an $N_M\in\mathbb{N}$ such that $n\geq N_M \Rightarrow \sup(s_m:m>n) > M+1$

Take $n=N_M$. Then there must be a $m_1 > N_M$ such that $a_{m_1} > M$ by definition of sup.

But again by definition of limit, $m_1>N_M \Rightarrow \sup(s_m: m > m_1) > M+1$ so there must be a $m_2 > m_1$ such that $a_{m_2} > M$.

And so on. Hence, $(a_{m_n})$ is a subsequence such that $a_{m_n}>M\forall n$