Consider the series $$S_N = \sum_{n = 1}^{N} \frac{\sin n}{n}$$
Let $$S^{+}_{N} := \sum_{n=1}^{N} \max \left(\frac{\sin{n}}{n}, 0 \right) $$ $$S^{-}_{N} := \sum_{n=1}^{N} \min \left(\frac{\sin{n}}{n}, 0 \right) $$ i.e. $S^{+}_{N}, S^{-}_N$ are the partial sums of positive and negative elements of the sequence $\{ \frac{\sin n}{n} \}_{n \geq 1}$.
I want to study the following limit $$\lim_{N \to +\infty}{\frac{S^{+}_N}{S^{-}_N}}$$
Intuitively it seems that the limit equals to $-1$, as the cardinality of the set of integers such that $\sin n > 0$ is "approximately equal" to the cardinality of the set of integers for which $\sin n < 0$, thus in the limit both parts give "the same" contribution.
Are there any hints to make the statement rigorous? Is there any way to utilize the Weyl equidistribution theorem?
Because $$\lim_{N \rightarrow +\infty} S_N^{+}= +\infty$$ and $$\lim_{N \rightarrow +\infty} S_N$$ converges to a finite constant. Hence, $$\lim_{N \rightarrow +\infty} \frac{ S_N^{+}}{S_N^{-}}=\lim_{N \rightarrow +\infty} \frac{ S_N^{+}}{S_N- S_N^{+}} =\lim_{N \rightarrow +\infty} \frac{ 1}{ \frac{S_N}{S_N^+ }- 1 } = \frac{1}{0-1}=-1$$