Fix real numbers $\alpha, \beta \geq 0$ with $\alpha$ irrational, and let $N$ be a positive integer. For $x \in \mathbb{R}$, let $\{x\} := x - \lfloor x \rfloor$ denote the fractional part of $x$. Are there any estimates known for the following sums \begin{equation} S_1 := \sum_{n=1}^N \left(\{\alpha n + \beta\} - \dfrac{1}{2}\right) \end{equation} and \begin{equation} S_2 := \sum_{n=1}^N \left(\{\log(\alpha n + \beta)\} - \dfrac{1}{2}\right). \end{equation} Is it true that $S_1 = o(N)$ and $S_2 = o(N)$ as well? What are methods that may be useful for estimating such sums? Thanks a lot.

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    $\begingroup$ I would expect $S_1=\frac{N}{2}+o(N)$, though. It may be the case that $S_2=\frac{N}{2}+o(N)$ as well. $\endgroup$ – Batominovski Jan 22 '17 at 2:09
  • $\begingroup$ Can your guess be proven? $\endgroup$ – user152169 Jan 22 '17 at 2:20
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    $\begingroup$ Well, a heuristic argument is that $\big\{\alpha n +\beta\big\}$ and $\big\{\log(\alpha n+\beta)\big\}$ are (probably) uniformly distributed on $[0,1)$. If this is true, then we may even have the estimate $\frac{N}{2}+\mathcal{O}\big(\sqrt{N}\big)$ for both $S_1$ and $S_2$. $\endgroup$ – Batominovski Jan 22 '17 at 2:24
  • $\begingroup$ Makes a lot of sense. But not true until proven :) $\endgroup$ – user152169 Jan 22 '17 at 2:30
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    $\begingroup$ After a trial with $\alpha=\sqrt{2}$ and $\beta=0$, $\big\{\log(\alpha n+\beta)\big\}$ does not seem uniformly random. But it may still hold that, now with the new definitiosn, $S_2=o(N)$ (although I wouldn't expect $S_2=\mathcal{O}\big(\sqrt{N}\big)$ anymore). I'm too tired to think about $S_1$ now, but it should not be difficult to show $S_1=\mathcal{O}\big(\sqrt{N}\big)$. $\endgroup$ – Batominovski Jan 22 '17 at 2:48

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