Find limsup and liminf without considering subsequences I am trying to review some ideas form the calculus course and came across the much hated (back then) topic of $\limsup$, $\liminf$, $\sup$ and $\inf$. I tried to solve the following problem,
Find $\limsup$, $\liminf$, $\sup$ and $\inf$ of $a_n$ , where
$$
a_n=(-1)^n + \frac{1}{n} + 2\sin{\frac{n\pi}{2}}
$$
As far as I remember to solve such problems, one uses the fact that, if a subsequence $a_{n_k}$ of the sequence $a_n$ has an accumulation point, than this is also an accumulation point of $a_n$. It is trivial, how to find the subsequence and obtain that
$$
 \sup a_n=\frac{3}{2}, \inf a_n =-1, \liminf a_n = -1 \limsup a_n = 1 
$$ 
This method works fine, but according to my book the problem can be solved much faster using the following theorem, which does not require finding all the subsequences of $a_n$.
$$
\limsup a_n = \lim_{k\to \infty} \sup (a_k,a_{k+1},\dots)\\
\liminf a_n = \lim_{k\to \infty} \inf (a_k,a_{k+1},\dots)
$$ 
Is it possible to apply this theorem to the sequence $a_n$ as defined above and solve the problem without using the sub-sequences? 
Somehow, I am not able to understand what $\sup (a_k,a_{k+1},\dots)$ will be in my case.
 A: Your answer is partly wrong: $\liminf a_n=-3$, not $-1$. You can see that by taking $n=4k+3$ for $k\in\Bbb N$: $\sin\frac{n\pi}2=-1$ there. I did a numerical confirmation of my answer by taking $n$ out beyond $1000$.
That theorem of yours does have the advantage of not requiring you to find all the relevant subsequences of $a_n$. However, it has the great disadvantage of requiring you to find infinitely many suprema and infima. It is often easier to find the desired subsequences. Just consider how I got the right answer to $\liminf$ in your problem.
It is good to have a theorem to allow an alternate approach, but remember that there are reasons the primary approach is use. Use the right tool for the job!
Finding $\lim_{k\to \infty} \inf (a_k,a_{k+1},\dots)$ is not easy: it is, of course, $-3$. In fact, $\inf (a_k,a_{k+1},\dots)=-3$, so the limit is not needed. Proving that seems to require considering the subsequence $a_{4k+3}$ so we have not escaped subsequences. Your problem is definitely not easier with your theorem. The subsequence tool is much easier here.
Let me know if you really need a proof that $\inf (a_k,a_{k+1},\dots)=-3$.
