Does $\sum_{n=1}^{\infty}\ln(n\sin(\frac{1}{n}))$ converge?

Using Taylor series, $$\sin x\sim x-\frac{x^3}{3!}+o(x^3)$$$$\ln(1+x)\sim x-\frac{x^2}{2}+\frac{x^3}{3}+o(x^3)$$ When $n\rightarrow\infty$, $$n\sin\frac{1}{n}\sim n(\frac{1}{n}-\frac{1}{3!n^3}+o(\frac{1}{n^3}))=1-\frac{1}{6n^2}+o(\frac{1}{n^2})$$

Thus, $$\ln(n\sin\frac{1}{n})\sim -\frac{1}{6n^2}+o(\frac{1}{n^2})-\frac{\left(-\frac{1}{6n^2}+o(\frac{1}{n^2})\right)^2}{2}+o\left(-\frac{1}{6n^2}+o(\frac{1}{n^2}\right)=-\frac{1}{6n^2}+o(\frac{1}{n^2})$$

As $\sum\dfrac{1}{n^2}$ is convergent, so is $\sum\ln(n\sin\dfrac{1}{n})$.

Hope this can help you.