Let $f(x)=\log|2\sin(x/2)|$ (the normalizing factor $2$ is chosen to have the average over the period equal to $0$). Does there exist $a>0$ such that all sums $\sum\_{k=1}^n f(ak)\ge 0$? The computations (run up to the values of $n$ where I could not rely on the floating point precision any more) show that it may be the case even for $a=2\pi\sqrt 2$ but I'll be happy with any $a$ (or with a proof that no such $a$ exists).


This question came up in our joint study of Leja interpolation points with Volodymyr Andriyevskyy but the connection is a bit too convoluted to be explained here :-)



Can all partial sums $\sum_{k=1}^n f(ka)$ where $f(x)=\log|2\sin(x/2)|$ be non-negative?
The answer to this question is yes, which is proved in the following [paper](https://arxiv.org/abs/2111.12974) that appeared on arXiv today. Specifically, part $(ii)$ of Theorem 2 of this paper, in the case $b = 1$, says that for $\beta = \frac{\sqrt{5}-1}{2}$ the smallest term of the sequence
$$P\_N(\beta) = \prod\limits\_{r = 1}^N 2|\sin(\pi r\beta)|$$
is the first term $P\_1(\beta)$ (this is the same as the sequence in OP up to denoting $a = 2\pi \beta$ and taking the exponent). Since
$$P\_1(\beta) = 2|\sin(\pi \beta)| = 1,86\ldots > 1,$$
the sequence in the OP for $a = 2\pi \beta$ is strictly positive and even has a uniform poisitve lower bound $\log(1,86\ldots) = 0,62\ldots$ .


Note also that the case $b = 2$ of the same Theorem gives us $\beta = \sqrt{2}-1$. By periodicity of sine, it's the same as $\beta' = \sqrt{2}$, which translates to $a = 2\pi \sqrt{2}$ from the OP. So, for this $a$ infimum is also the first term, which is also strictly positive.


