I know $\sum_{k=1}^{n} \sin(k)$ is bounded by a constant and $\sum_{k=1}^{n} \sin(k)$ is [not bounded][1] by a constant. Then, what about $\sum_{k=1}^{n} (|\sin(k)|-2/\pi)$?

From numerical calculation, $\max_{n=0...10^8}(\sum_{k=1}^{n} (|\sin(k)|-2/\pi))=0.0900478$ which is much small compared to $\max_{n=0...10^8}(\sum_{k=1}^{n} (|\sin(k-^2)|-2/\pi))=1882.86$.

So, I suppose $\sum_{k=1}^{n} (|\sin(k)|-2/\pi)$ can be bounded by a constant, but I don't know how to prove it. [1]: Is $\sum_{k=1}^{n} \sin(k^2)$ bounded by a constant $M$?