For $x$ irrational, is $a_{n} =\sum_{k=1}^{n}(-1)^{⌊kx⌋}$ unbounded? For $x$ irrational, define $a_{n} :=\sum_{k=1}^{n}(-1)^{⌊kx⌋}$. Can you prove that $\left\{a_n\right\}$ is unbounded?
I feel that it is not easy to treat every irrational $x$.
I have asked in S.E. and it seems that it is an advanced question. Thus I go for help here.
 A: As Ville Salo already wrote in his answer, your question can be phrased in terms of the difference between the number of elements of the sequence $y, 2y, \dots, ky$ which are contained in $[0,1/2]$, and $k$ times the length of $[0,1/2]$. Here $y = x/2$. In the language of discrepancy theory, you are asking whether the interval $[0,1/2]$ is a so-called bounded remainder set of the sequence $(n y~\text{mod}~1)_{n \geq 1}$. However, it is known that it is not: the only intervals with bounded remainder are those whose length is in $\mathbb{Z} + y \mathbb{Z}$, and since $y$ is irrational in your question the length of $[0,1/2]$ is not of such type. Bounded remainder sets were classified in this paper: H. Kesten, On a conjecture of Erdös and Szüsz related to uniform distribution mod 1, Acta Arith. 12(1966), 193–212.
A: It's indeed unbounded for every irrational $x$.
Let me identify points of $\mathbb{R}/\mathbb{Z}$ with their representatives on $[0,1)$, and order it by the usual order $<$ of $\mathbb{R}$ applied to the representatives.
Replace $x$ by $y = x/2$, and the question becomes whether for all $m$ there exists $k$ such that the orbit of $0$ in the irrational rotation on $\mathbb{R}/\mathbb{Z}$ by $y$ is in $[0,1/2)$ at least $m$ more times than in $[1/2,1)$, in the first $k$ time steps $y,2y,3y,...,ky$; or that this happens with $[0,1/2)$ and $[1/2,1)$ interchanged.
Since the irrational rotation by $2y$ is topologically transitive, we can find odd $k$ with $ky > 0$ arbitrarily small. For odd $k$ the set $Y_k = \{y,2y,...,ky\}$ has to intersect either $[0,1/2)$ or $[1/2, 1)$ more times than the other. Let's suppose the first case happens for infinitely many $k$. (The other case is symmetric, and one happens by the pigeonhole principle.)
Let now $P_m$ be the statement that there are $ky > 0$ arbitrarily small such that $[0,1/2)$ contains $m$ more elements of $Y_k$ than $[1/2,1)$ does. We have that $P_1$ holds. Observe that if $P_m$ holds and $k$ is as in the definition, then $[0, 1/2)$ also contains $m$ more elements of $Y_{k,a} = \{a+y,a+2y,...,a+ky\}$ than $[1/2,1)$ whenever $a$ is small enough, because $Y_k$ is disjoint from $\{0,1/2\}$.
Let now $m \geq 1$ be any integer such that $P_m$ holds. Let $\epsilon > 0$ be arbitrary and pick $k$ such that $0 < ky < \epsilon/2$ and $[0,1/2)$ contains at least $m$ more elements of $Y_k$ than $[1/2,1)$ does. Let $a$ be as in the previous paragraph.
Using $P_m$ again, take $0 < k'y < \min(\epsilon/2, a)$ such that $[0, 1/2)$ contains $m$ more elements of $Y_{k'} = \{y,2y,...,k'y\}$ than $[1/2,1)$ does. Then $[0,1/2)$ contains $2m$ more elements of $Y_{k'+k}$ than $[1/2,1)$. We have $0 < (k+k')y < \epsilon$, so $P_{2m}$ holds.
Thus $P_m$ holds for all $m$, and the claim follows.
A: We have
Theorem. Let $\psi(x)$ and $\varphi(x)$ be positive increasing functions such that 
$$\int_1^\infty \frac{dx}{\psi(x)}=+\infty,\qquad \int_1^\infty \frac{dx}{\varphi(x)}<+\infty.$$
Then for almost all $\alpha\in(0,1)$ we have 
$$\Omega(\log N\cdot \psi(\log\log N))\le\sup_{n\le N}\sum_{j=1}^n(-1)^{\lfloor j\alpha\rfloor}=O(\log N\cdot \varphi(\log\log N)).$$
This is proved in my paper with Jan van de Lune On Some oscillating sums, Uniform Distribution Theory 3 (2008) 35--72.
In this paper it is contained an algorithm to compute the sums. We obtain for example
$$S_{\sqrt{2}}(10^{1000})=-10,\quad S_{\sqrt{2}}(10^{10000})=166,\quad S_{\pi}(10^{10000})=11726.$$
With $S_\alpha(N)=\sum_{j=1}^n(-1)^{\lfloor j\alpha\rfloor}$.
A: Here is an argument essentially due to  fedja I learned about thirteen years ago on artofproblemsolving.com. 
Proposition: if $f$ is $2$-periodic Riemann integrable such that $\sup_{n \geq 1} \left|\sum_{k=1}^{n}f(kx)\right|<C<\infty$  for some irrational $x \in \mathbb{R} \setminus \mathbb{Q}$ then necessarily 
$$
\sum_{m\neq 0}\,  \left| \frac{\hat{f}(m)}{e^{i \pi m x}-1}\right|^{2}<2C^{2}. \quad (1)
$$.
Notice that the proposition solves the question. Indeed,  $f(t)=(-1)^{\lfloor{t}\rfloor}$ is 2-periodic with Fourier coefficients 
$\hat{f}(m) = \left| \frac{1}{2}\int_{0}^{2}f(t)e^{-i\pi m x}dx \right| \approx \frac{1}{m}$ for $m$ odd, and $\hat{f}(m)=0$ for even $m$. There are infinitely many odd numbers $m$ such that $\mathrm{dist}\left(mx, 2\mathbb{Z}\right) < \frac{C'}{m}$ (it does not follow directly from Dirichlet's rational approximation theorem, however, I think it is not difficult to adapt it here or use Minkowski's theorem on product of two linear forms). Therefore, in the left hand side of (1) there are infinitely many terms comparable to 1 so we get a contradiction. 
Proof of the proposition:
Let me first assume that $f$ is continuous and then I will explain later what to do in the arbitrary case. 
Let $S_{m} = \sum_{k=1}^{m-1}f(kx)$. Then the boundedness of $|S_{M+n}-S_{M}|$ implies 
$$
|\sum_{k=0}^{n-1} f(Mx+kx)|<2C \quad \text{for all} \quad M>1 \quad (2)
$$
and all $n\geq 1$. By density of the sequence $\{Mx\,  \mathrm{mod}\, (2)\}_{M>1}$ in $[0,2)$ we conclude  $$
|\sum_{k=0}^{n-1} f(t+kx)|\leq 2C \quad \text{for all} \quad t \in [0,2). \quad (3)
$$ 
Let us convolve $f$ with Fejer Kernel so that the new function $F$ is now a trigonometric polynomial with almost the same Fourier coefficients, and clearly it also satisfies the inequality (3). After expanding $F$ into its Fourier series  $F(s) = \sum_{m} \hat{F}(m) e^{i\pi m s}$ (finite sum), and using (3) for $F$ we obtain 
$$
\left| \sum_{m} \hat{F}(m)e^{i \pi m t} \, \frac{e^{i \pi m n x} -1}{e^{i \pi m x} -1} \right| <2C
$$ 
In particular its $L^{2}$ norm is bounded, i.e., 
$$
\sum_{m\neq 0} \left|\hat{F}(m)\,   \frac{e^{i \pi m n x} -1}{e^{i \pi m x} -1}\right|^{2} <4C^{2}\quad (4)
$$
Invoking the density of the sequence $\{n x\,  \mathrm{mod}\, (2)\}_{n \geq 1}$ again we can replace $nx$ by an arbitrary $s \in [0,2)$, after that  we integrate (4) with respect to $s$ over the interval $[0,2]$, and we use the identity $\int_{0}^{2} |e^{i \pi m s} -1|^{2}ds=4$, hence,  we conclude  (1) for $F$. Finally it remains to remove the convolution with Fejer kernel to conclude (1) for $f$ (we have nonnegative terms in the sum on the left hand side of (1), so we can just cut the sum and take the limit $\hat{F}(m) \to \hat{f}(m)$)
In general, if $f$ is not continuous, let $K_{N}$ be the Fejer kernel. Split $Mx=M_{1}x+M_{2}x$ in (2), multiply (2) by $K_{N}(M_{1}x)$ and take the average in $M_{1}$ and use the Riemann integral criteria for equidistribution of $\{M_{1} x\, \mathrm{mod}\, (2)\}_{M_{1}\geq 1}$ to conclude $\left| \sum_{k=0}^{n-1} F(M_{2}x + kx)\right|<2C$ and now due to continuity of $F$ and density of $\{M_{2} x\}$ we conclude (3) for $F$ and the rest of the argument proceeds in the same way. $\square$
