Does $\sum_{n=1}^{\infty}\ln(n\sin(\frac{1}{n}))$ converge? I must determine whether the following series converges: $$\sum_{n=1}^{\infty}\ln\left(n\sin\left(\frac{1}{n}\right)\right)$$
I know that in general, I must use the limit comparison test, but I cannot find an expression to which I can compare it. For instance, I have tried the usual process:
For $n$ large, we have that $\lim_{n\to \infty}n\sin\frac1n=1$, and so, $\ln(1)=0$. This fails the divergence test, but it cannot be concluded automatically that the series is convergent either. How may I proceed here? Any help would be appreciated.

Taylor series of $\sin x$ is 
$$\sin x = x - \frac{x^3}{3!} + O(x^5)$$
Hence
$$\sin \frac{1}{n} = \frac{1}{n}- \frac{1}{6n^3} + O(\frac{1}{n^5})$$
So
$$n\sin \frac{1}{n} = 1- \frac{1}{6n^2} + O(\frac{1}{n^4})$$
Taylor series of $\ln$ is 
$$\ln(1+x) = x + O(x^2)$$
For $x = -\frac{1}{6n^2} + O(\frac{1}{n^4})$, we get
$$\ln(1 - \frac{1}{6n^2} + O(\frac{1}{n^4})) = -\frac{1}{6n^2} + O(\frac{1}{n^4})$$
Then you'd get a sum of convergent series since all those sums have $p > 1$ according to the p-series test.
