I ##S_{n}(\alpha)=\sum_{k=1}^{n} (-1)^{\lfloor{k\alpha\rfloor}}##

[elias001]

TL;DR Summary

I would like to know how to show $S_{n}(\alpha)=\sum_{k=1}^{n} (-1)^{\lfloor{k\alpha\rfloor}}$.

For the following question, I am not sure how to go about solving it.

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 tried the following computational examples to check if the statement is true.

$S_{n}(\alpha)$ 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 $\alpha =e$, $S_{4}=(e)= (-1)^{\lfloor{e\rfloor}} +(-1)^{\lfloor{2e\rfloor}} + (-1)^{\lfloor{3e\rfloor}} + (-1)^{\lfloor{4e\rfloor}} = 1 + (-1) + 1 + 1 = 2$

Thank you in advance.

 

[anuttarasammyak]

I let EXCEL calcuate up to n=40 for approximate values of sqrt(2) and e.

Without losing generality we can make $|\alpha|<1$ by adding/deleting even numbers. For example $\pi-4$ instead of $\pi$, $\sqrt{2}-2$ instead of $\sqrt{2}$ work. $\alpha$ and -$\alpha$ are equivalent in the behavior with opposite siganure, so further we can reduce $0<\alpha<1$. Please find the graph showing for each k on x axis, +1 or -1. We find that + and - are almost same in number which shows sum =0 would take place infinite times though it is not proved yet.

 

[Office_Shredder]

Either you have some theorem that just surprisingly crushes this, or my guess is you're going to have to think about how you can approximate irrational numbers with rational numbers and use what happens to this with rational numbers. So the first thing I would do is think about what happens if $\alpha$ is rational actually.