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.
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 $-3$.