Determine the convergence of $\sum_{n=2}^{\infty} \sin \left(n \pi+\frac{1}{\ln n}\right)$.
$\sum(-1)^{n}\sin(\frac{1}{\ln(n)})$ converges by Leibniz Test .
But $\sum\sin(\frac{1}{\ln(n)})$ is divergent by comparing with $\frac{1}{\ln(n)}$.
That is $$\lim_{n\to\infty}\frac{\sin(\frac{1}{\ln(n)})}{\frac{1}{\ln(n)}}=1$$ .
And $\sum\frac{1}{\ln(n)}$ diverges by Cauchy Condensation test.(Or just by simple comparison with $\frac{1}{n}$ as mentioned in the comments.)