Convergence of $\sum_{n=1}^{\infty} \log~ ( n ~\sin \frac {1 }{ n })$ Convergence of $$\sum_{n=1}^{\infty} \log~ ( n ~\sin \dfrac {1 }{  n  })$$
Attempt: 
Initial Check : $\lim_{n \rightarrow \infty } \log~ ( n ~\sin \dfrac {1 }{  n  }) = 0$
$\log~ ( n ~\sin \dfrac {1 }{  n  }) < n ~\sin \dfrac {1 }{  n  }$
$\implies \sum_{n=1}^{\infty} \log~ ( n ~\sin \dfrac {1 }{  n  }) < \sum_{n=1}^{\infty} n ~\sin \dfrac {1 }{  n  }$
But, $\sum_{n=1}^{\infty} n ~\sin \dfrac {1 }{  n  }$ is itself a divergent sequence. Hence, I don't think the above step is of any particular use.
Could anyone give me a direction on how to move ahead.
EDIT: I did think of trying to use the power series of $\sin$. Here's what I attempted : 
$\sin \dfrac {1}{n} = \dfrac {1}{n}-\dfrac {1}{n^3.3!}+\dfrac {1}{n^5.5!}-\cdots$
$\implies n \sin \dfrac {1}{n} = 1- \dfrac {1}{n^2.3!}+\dfrac {1}{n^4.5!}-\cdots$
I couldn't proceed further due to the $\log$.
Thank you for your help.

Since $$\log\left(n\sin \frac{1}{n}\right) = \log\left(n\left(\frac{1}{n} + O\left(\frac{1}{n^3}\right)\right)\right) = \log\left(1 + O\left(\frac{1}{n^2}\right)\right) = O\left(\frac{1}{n^2}\right)$$
and $\sum_{n = 1}^\infty \frac{1}{n^2}$ converges, so does $\sum_{n = 1}^\infty \log(n\sin\frac{1}{n})$.
