Is $\sum_{n=2}^{\infty}\sin {\left(n\pi +\frac{1}{n^{\alpha}\log n}\right)}, \alpha<0$ divergent?

Question

I konw $\sum_{n=2}^{\infty}\sin {\left(n\pi +\frac{1}{n^{\alpha}\log n}\right)}=\sum_{n=2}^{\infty}(-1)^n\sin {\left( \frac{1}{n^{\alpha}\log n}\right)},$ if I could prove $\sin{\left( \frac{1}{n^{\alpha}\log n}\right)}\nrightarrow 0$(Maybe it can be simplified to $\sin{n^{\alpha}},\alpha>0)$ ,then this series is divergent. But I don't konw how to get it. Thanks for your time!

Answer 1

For $\alpha<0$ and $n\to\infty$, $$\frac{1}{n^{\alpha}\log n}~\to\infty$$ and $\lim_{z\to\infty}\sin(z)$ is undefined, so the series diverges as the general term doesn't tend to $0$.

Answer 2

For $\alpha>0$, the general term is alternating and convergent to 0, so the series converge.