I have to prove that given a sequence {an}n, with lim sup an = infinity, prove that there is a sub-sequence ank of an that diverges to infinity.
I used the Theorem which states that for any sequence which has a limit (possibly infinite), all of it's sub-sequences have the same limit. As an clearly diverges to infinity, the limit of all sub-sequences is infinity, and thus all sub-sequences diverge to infinity.
Does this work, or is there something I am shortcutting passed something important?
Best Answer
No, it doesn't work; mainly because the problem statement does not require the sequence to have a limit $+\infty$.
The symbol $\limsup$ doesn't mean a limit, but rather an 'asymptotic upper bound'.
See Wikipedia Limit superior and limit inferior for definitions.
In your case the sequence is said to be unbound from above (it reaches larger and larger values), but not necessarily to converge to $+\infty$. As an example sequences $$a_n = \begin{cases} n & \text{if }\ 2\vert n\\0&\text{otherwise}\end{cases}$$ or $$a_n = n\cdot(-1)^n$$ both satisfy $$\limsup_{n\to\infty} a_n = +\infty$$ although none of them has $\lim a_n = \infty$.
Instead, the former one has $\liminf = 0$ whilst the latter has $\liminf = -\infty$.