Prove that $ \sum_{n=1}^\infty \ln\big(n\sin \frac{1}{n}\big)$ converges.

Prove that $\displaystyle \sum_{n=1}^\infty \;\ln\left(n\sin\frac{1}{n}\right)$ converges.

My Work: $$\left|\ln \left(n \sin \frac{1}{n}\right)\right| \leq\left|\ln \left(n \sin \frac{1}{n^{2}}\right)\right| \leq\left|\ln \left(\sin \frac{1}{n^{2}}\right)\right|$$ I was going to use comparison test. But now stuck. Please give me a hint.

Hint: try to use the fact that $$ \ln\left(n\sin \frac{1}{n}\right) = \ln\left(n \left( \frac{1}{n}+O\left(\frac{1}{n^3}\right) \right)\right) = \ln\left( 1+O\left(\frac{1}{n^2} \right)\right) = O\left(\frac{1}{n^2} \right) $$ and the comparison test.

Let's amply use Taylor series to give the leading orders.

Note that $\sin\left( \frac{1}{n} \right) \approx \frac{1}{n} - \frac{1}{6n^3}$. So $n \sin \left( \frac{1}{n} \right) \approx 1 - \frac{1}{6n^2}$.

Note also that $\ln(1 - x) \approx x + x^2 + \dots$, so that $$\ln\left( n \sin \frac{1}{n} \right) \approx \frac{1}{6n^2} + \frac{1}{36n^4} + \dots$$

This means that you are wondering about $$ \sum_{n \geq 1} \sum_{j \geq 1} \frac{1}{(6n^2)^j} \ll \sum_{n \geq 1} \frac{1}{n^2},$$ which converges.