Does $\sum_{n=2}^{\infty}\frac{\sin n}{n \ln n}$ converge or diverge? 

We use Dirichlet's test (https://en.wikipedia.org/wiki/Dirichlet%27s_test) to show that this infinite series converges. Since $\frac{1}{n \ln n}$ is decreasing for ($n\geq 2$) and bounded, it is enough to show that 
\[S_n=\sum_{k=2}^{n}\sin k\]
is bounded. Notice that 
\[S_n=\text{Real}(\sum_{k=2}^{n}e^{ik})=\text{Real}(\frac{e^{2i}-e^{(n+1)i}}{1-e^{i}}).\]
Hence
\[S_n \leq |\frac{e^{2i-e^{(n+1)i}}}{1-e^{i}}|=|e^{2i}\frac{1-e^{(n-1)i}}{1-e^{i}}|\leq \frac{2}{|1-e^{i}|}<\infty.\]
Thus it follows from the Dirichlet's test that 
\[\sum_{n=2}^{\infty}\frac{\sin n}{n\ln n}\]
is convergenet.