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.

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.
