For the following question:
For $n\in \mathbb{N}$ and $\alpha \in \mathbb{R}$, let $S_{n}(\alpha)=\sum_{k=1}^{n} (-1)^{\lfloor{k\alpha\rfloor}}.$ Prove that if $\alpha$ is irrational, then $S_{n}(\alpha)=0$ for infinitely many $n \in \mathbb{N}$
I don't think the sttatement is true for every irrational $\alpha$. I tried to do computations but I am not sure if I am interpreting $S_{n}(\alpha)$ correctly. As examples for $\alpha = \pi, e$, $n$ up to 4, if ${\lfloor{\alpha\rfloor}}$ is even, then $S_{n}(\alpha)\neq 0$ for every $n$ and $S_{n}(\alpha)=0$ for odd values of $n$.
$S_{n}(\alpha)$ for:
$n = 1$ and $\alpha = \pi$, $S_{1}=(\pi)=-1$
$n = 2$ and $\alpha = \pi$, $S_{2}=(\pi)= (-1)^{\lfloor{\pi\rfloor}} +(-1)^{\lfloor{2\pi\rfloor}} = (-1) + 1 = 0$
$n = 3$ and $\alpha = \pi$, $S_{3}=(\pi)= (-1)^{\lfloor{\pi\rfloor}} +(-1)^{\lfloor{2\pi\rfloor}} + (-1)^{\lfloor{3\pi\rfloor}} = (-1) + 1 + (-1) = -1 $
$n = 4$ and $\alpha = \pi$, $S_{4}=(\pi)= (-1)^{\lfloor{\pi\rfloor}} +(-1)^{\lfloor{2\pi\rfloor}} + (-1)^{\lfloor{3\pi\rfloor}} + (-1)^{\lfloor{4\pi\rfloor}} = (-1) + 1 + (-1) + 1 = 0 $
For:
$n = 1$ and $\alpha =e$, $S_{1}=(e)=1$
$n = 2$ and $\alpha =e$, $S_{2}=(e)= (-1)^{\lfloor{e\rfloor}} +(-1)^{\lfloor{2e\rfloor}} = 1 + (-1) = 0$
$n = 3$ and $\alpha =e$, $S_{3}=(e)= (-1)^{\lfloor{e\rfloor}} +(-1)^{\lfloor{2e\rfloor}} + (-1)^{\lfloor{3e\rfloor}} = 1 + (-1) + 1 = 1 $
$n = 4$ and $\alpha =e$, $S_{4}=(e)= (-1)^{\lfloor{e\rfloor}} +(-1)^{\lfloor{2e\rfloor}} + (-1)^{\lfloor{3e\rfloor}} + (-1)^{\lfloor{4e\rfloor}} = 1 + (-1) + 1 + 1 = 2$
Can someone in the community check to see if I am interpreting the question correctly with my computations or what I am not seeing. Thank you in advance.