How to prove you found ALL limit points: $x_n = (-1)^n + 1/n + 2\sin(\frac{n\pi}{2})$ How does one find all limit points? How do you prove you're not missing a limit point? For example, if $x_1,x_2,\cdots$ converges to $x$ and $y_1,y_2,\cdots$ converges to $y$, and we consider the shuffled sequence, obvious limit points are $x$ and $y$ but how can we prove those are the only two limit points?
For my problem, we have $$x_n = (-1)^n + 1/n + 2\sin\left(\frac{n\pi}{2}\right)$$ An observation is that for $n$ even, the third term is 0, so if we remove all the odd $n$, the resulting subsequence is defined such that $x_j = 1 + 1/2j$ which converges to 1. So one limiting point is 1. Something to note about this subsequence, moreover, is that its $\sup$ is $3/2$. If we don't find a larger $\sup$ in the subsequence we removed from the original sequence, then $3/2$ is the $\sup$ for the entire sequence. But we realize that $x_1 = -1 + 1 + 2 = 2$, which is the $\sup$ for the whole sequence. Now, let's look at the subsequence we removed. First, observe the first term is always -1, the second term converges to 0, and the third term bounces back and forth between 2 and -2. Hence, we have two limit points: $-1 + 0 + 2 = 1$ and $-1 + 0 - 2 = -3$. The $\inf$, then, is $-3$, as well as the $\liminf$, and the $\limsup = 1$.
How can I prove that I pointed out all limit points?
 A: HINT: Consider the sequences $x_{4n+r}$ for $r=0,1,2,3$. All of these sequences will converge to some limit $l_r$ which you can determine. To show that the set $L = \{l_r: r=0,1,2,3\}$ is exactly  the set of limit points of $x_n$ you need to consider an arbitrary convergent subsequence $x_{n_k}$ of $x_n$ and argue that it should have infinitely many terms in common with some $x_{4n+r}$ and thus converges to $l_r$.
A: We have that, starting from $n=4k$

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*$x_{4k} = (-1)^{4k} + \frac1{4k} + 2\sin\left(\frac{4k\pi}{2}\right)=1+\frac1{4k}+0=1+\frac1{4k}$


*$x_{4k+1} = (-1)^{4k+1} + \frac1{4k+1} + 2\sin\left(\frac{4k\pi}{2}+\frac \pi 2\right)=-1+\frac1{4k+1}+2=1+\frac1{4k+1}$


*$x_{4k+2} = (-1)^{4k+2} + \frac1{4k+2} + 2\sin\left(\frac{4k\pi}{2}+ \pi \right)=1+\frac1{4k+2}+0=1+\frac1{4k+2}$


*$x_{4k+3} = (-1)^{4k+3} + \frac1{4k+3} + 2\sin\left(\frac{4k\pi}{2}+\frac {3\pi} 2\right)=-1+\frac1{4k+3}-2=-3+\frac1{4k+3}$
and the same pattern repeats, indeed $4k+4=4(k+1)$, $4k+5=4(k+1)+1$, $4k+6=4(k+1)+2$, $4k+7=4(k+1)+3$ and so on, therefore we can take $k\to \infty$ and determine the limit points.
