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Consider the series $$ \sum_{n=1}^{\infty} ( \{ n \xi \} - \frac{1}{2}) $$ where $\{ \}$ denotes the fractional part and $\xi$ is some irrational number. It clearly does not converge but can we show that it does not tend to $\infty$ or $-\infty$? Would it also be true that it bounded?

The problem seems to be related to equidistributed sequences. Since ${(\{ n \xi \} )}_{n \geq 1}$ is equidistributed we have that

$$ \frac{1}{N}\sum_{n=1}^{N}(\{n\xi\}- \frac{1}{2}) \to 0 $$ as $N \to \infty$. However, just the equidistribution property cannot be enough. Consider for example a sequence $(a_n)$ given by $a_n = 1$ if $n$ is odd or a power of $2$ and $a_n= -1$ otherwise. Then $(a_n)$ is equidistributed on $ \{ -1,1 \}$ but

$$ \sum_{n=0}^{\infty} a_n = \infty. $$

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This is studied in detail in Sums of Fractional Parts of Integer Multiples of an Irrational
The sum $$C(N)=\sum_{n=1}^{N} \bigl( \{ n \xi \} - \tfrac{1}{2}\bigr)$$ satisfies $|C(N)|>c\log N$ for infinitely many $N$. The coefficient $c\geq 1/720$.

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    $\begingroup$ Thank you! That is a great reference. This shows that the series cannot be bounded. However, I can't immediately see in the reference whether or not it is possible for $C(N) \to \infty$ as $N \to \infty$. $\endgroup$
    – Hampus Nyberg
    Commented Jul 10, 2023 at 16:06
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    $\begingroup$ The reference shows that $C(N) > c \log N$ and $C(N) < -c\log N$ both occur infinitely often; since $C(N)$ is Lipschitz in $N$, this already implies that $|C(N)|$ cannot diverge to infinity (there are infinitely many sign changes) $\endgroup$
    – Terry Tao
    Commented Jul 10, 2023 at 22:40
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    $\begingroup$ If I'm not mistaken the reference only shows that $C(N) < - c \log N$ for infinitely many $N$ under certain conditions. In fact, Fact 5 in the reference gives an example of an irrational number where $C(N)$ is bounded below. Also, what do you mean by $C(N)$ being Lipschitz in $N$ and how is a Lipschitz condition relevant here? $\endgroup$
    – Hampus Nyberg
    Commented Jul 11, 2023 at 10:04
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Another way to see unboundedness of the sum is to notice that for 2-periodic $f(t) = \{t/2\}-1/2$ we have $|\hat{f}(m)| \gtrsim \frac{1}{|m|}$ for $m\neq 0$. Since the series $$\sum_{m \neq 0} \left| \frac{\hat{f}(m)}{e^{i\pi m x}-1}\right|^{2} \gtrsim \sum_{m \neq 0, \mathrm{dist}(mx, 2\mathbb{Z})<C/m} 1 = \infty $$ whenever $x$ is irrational, the claim follows from this proposition

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