Finding all cluster points of a sequence and proving it rigorously This is a problem from Kenneth Brown's Intro to Real Analysis:
Find all the cluster points of the sequence:
$$x_n = \cos\left(\frac{n\pi}{4}\right) + \sin\left(\frac{n\pi}{2}\right) + 2^{-n} + (-1)^n$$
I know that as n gets large, $2^{-n}$, goes to zero, but from here, how do you reconcile the different intervals for the trig functions and the oscillating negative one?
I need to find all cluster points and prove them. Thanks!

First, you can show that if $(v_n)_{n\in\mathbb{N}}$ is a sequence such that $$\lim\limits_{n\to+\infty}v_n=0,$$ then the cluster points of $(u_n+v_n)_{n\in\mathbb{N}}$ and $(u_n)_{n\in\mathbb{N}}$ are the same. I will give the proof of one inclusion: if $a\in \mathbb{R}$ is a cluster point for $(u_n)_{n\in\mathbb{N}}$, then by definition for all $\varepsilon'>0$ and for all $N'\in\mathbb{N}$, it exists $n\in\mathbb{N}$ such that $n\geq N'$ and $|u_{n}-a|<\varepsilon'$. Now, for $\varepsilon>0$ and $N\in\mathbb{N}$, take $N''\geq N$ such that $n\geq N''\implies|v_n|<\frac{\varepsilon}{2}$ (which holds by hypothesis on $(v_n)_{n\in\mathbb{N}}$), then take $n$ given by the previous property with $(\varepsilon',N')=(\frac{\varepsilon}{2},N'')$. Thus, you have a $n$ greater than $N$ such that
$$|u_n+v_n-a|\leq |u_n-a|+|v_n|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon,$$
which shows that $a$ is a cluster point of $(u_n+v_n)_{n\in\mathbb{N}}$.
So we only have to find the cluster points of the sequence defined by $y_n=\cos\left(\frac{n\pi}{4}\right) + \sin\left(\frac{n\pi}{2}\right) + (-1)^n$. But the value of $y_n$ only depends of the remainder of the division of $n$ by $8$, so this sequence is $8$-periodic, and the cluster points are $\{y_0,y_1,\dots,y_7\}$. You finally juste have to compute these eight values to get the cluster points.
