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.

In the same spirit as mixedmath's answer, let us massively use Taylor expansions for large values of $n$. Then $$\sin(\frac 1n)=\frac{1}{n}-\frac{1}{6 n^3}+\frac{1}{120 n^5}-\frac{1}{5040
   n^7}+O\left(\left(\frac{1}{n}\right)^8\right)$$ $$n\sin(\frac 1n)=1-\frac{1}{6 n^2}+\frac{1}{120 n^4}-\frac{1}{5040
   n^6}+O\left(\left(\frac{1}{n}\right)^8\right)$$ $$\log \left(n \sin \left(\frac{1}{n}\right)\right)=-\frac{1}{6 n^2}-\frac{1}{180 n^4}-\frac{1}{2835
   n^6}+O\left(\left(\frac{1}{n}\right)^8\right)$$ Summing from $n=1$ to $\infty$, then $$S=\sum_{n=1}^\infty \ln\big(n\sin \frac{1}{n}\big)\approx -\frac 16 \sum_{n=1}^\infty\frac{1}{n^2}-\frac 1{180} \sum_{n=1}^\infty\frac{1}{n^4} -\frac 1{2835} \sum_{n=1}^\infty\frac{1}{n^6}$$ and the value of the successive sums are $\frac{\pi ^2}{6}$, $\frac{\pi ^4}{90}$, $\frac{\pi ^6}{945}$ which make $$S \approx -\frac{\pi ^2 \left(595350+1323 \pi ^2+8 \pi ^4\right)}{21432600}\approx -0.280527$$ while the numerical evaluation of the original sum is $-0.280556$.
Adding an extra term to the expansion would lead to $S \approx -0.280554$.
