Trying to understand $\sup$ and $\limsup$ of a sequence. The following is the sequence and a problem that I am working on.

$\{x_n\} = (-1)^n + \frac{1}{n} + 2\sin(\frac{n\pi}{2})$
Find the $\sup$, $\inf$, $\limsup$ and $\liminf$ of this sequence.

Writing out the sequence, I noticed that because of the trig part, there were 4 main subsequences that tells me that $\{1,-3\}$ are the limit points where three of the main subsequences all go to $1$.
The first term gave me the largest term $2$, so that is the $\sup$.
It seemed like there was no inf because the decreasing subsequence goes to $-3$ but there was not term less than that.
So this is my summary:
$$\sup x_n = 2, \inf x_n = \varnothing, \limsup x_n = 1, \liminf x_n = -3 $$
I am a dilettante in analysis and I feel like I'm starting to get it... but I need to confirm if I'm doing this right.
Any comments?
Note: the sequence was edited so the problem is different.
 A: Note that
$$\sin\frac{n\pi}2=\begin{cases}\;\;0&,\;\;n=0,2\bmod 4\\{}\\\;\;1&,\;\;n=1\bmod 4\\{}\\\!\!-1&,\;\;n=3\bmod 4\end{cases}$$
and from here
$$(-1)^n+1+2\sin\frac{n\pi}2=\begin{cases}\;\;2&,\;\;n=0,1,2\bmod 4\\{}\\\!\!-2&,\;\;n=3\bmod 4\end{cases}$$
Since the above cover all the possibilities for an index $\,n\,$ , the set of partial limits simply is $\;\{-2\,,\,2\}\;$
A: Your analysis is fine. We say precisely the same thing below.
Look first at the sum of the first and third terms. We start at $n=1$. The first terms go $-1,1,-1, 1,\dots$, cycling with period $2$. The third terms go $2,0,-2,0,2,\dots$, cycling with period $4$.
So their sum goes $1,1,-3,1, 1,1,-3,1,\dots,$ cycling with period $4$. 
When we add the middle term, we get
$$1+1,1+\frac{1}{2}, -3+\frac{1}{3}, 1+\frac{1}{4}, 1+\frac{1}{5}, 1+\frac{1}{6}, -3+\frac{1}{7}, \dots.$$
The big guy is easy to pick out, it is the first term $2$. There is no littlest.
The biggest number $a$ which has an infinite subsequence that has $a$ as a limit is $1$, and the smallest number $b$ which has an infinite subsequence that has $b$ as a limit is $-3$. 
