If $\limsup x_n=\infty$, show that there is a divergent subsequence $\{x_{r_n}\}$.

Question

If $\limsup x_n=\infty$, show that there exists a properly divergent subsequence $\{x_{r_{n}}\}$ of $\{x_n\}$ such that $\lim{x_{r_n}}=\infty$.

I have no idea to prove this. Please solve this problem.

Answer 1

Recall the definition $\limsup_n x_n = \lim_{k \rightarrow \infty} \sup\{x_n: n \ge k \}$. Denote by $s_k$ that last $\sup$, and note that $k

So there must be some $k_1$ such that $s_{k_1} \ge 2$, for the $s_k$ to tend to $\infty$. This means we must have some $x_{n_1} > 1$, for some $n_1 \ge k_1$, or else $1$ would have been a smaller upperbound for $\{x_n: n \ge k_1 \}$, contradicting its sup is at least 2.

Now there must be some $k_2 \ge n_2$, such that $s_{k_2} > 4$, for the same reason as before and again we find some $n_2 \ge k_2$ with $x_{n_2} > 3$, similarly.

This way we construct a subsequence $x_{n_l}$ that tends to infinity as well.

Answer 2

By definition limsup is the supremum of the set of subsequential limits of $x_n$. Thus by definition exists a subsequence $x_{r_n}$ of $x_n$ which diverges to $\infty$ and we know that when a sequence diverges to infinity then every subsequence of it diverges to infinity also.(or when a sequence converges then every subsequence of it converges also to the same limit of the sequence)

Also note that every sequence is a subsequence of itself.