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.

A slightly different way than the two current answers, dealing mostly with intuition:
$$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} + \ldots$$
or alternately,
$$\sin\left(\frac 1x \right) = \frac{1}{x} - \frac{1}{3!x^3} + \frac{1}{5!x^5} + \ldots$$
Multiplying by $x$ gives that
$$x\sin\left( \frac 1x \right) = 1 - \frac{1}{3!x^2} + \ldots \approx 1 - \frac{1}{6x^2},$$
and where the approximation is extremely good for high $x$ (error $\ll \frac{1}{x^4}$).
You are now interested in how $\log \left( 1 - \frac{1}{6x^2}\right)$ behaves. Fortunately, we know that
$$ \log (1 - x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \ldots$$
and thus
$$ \log \left( 1 - \frac{1}{6x^2} \right) = \frac{1}{6x^2} + \ldots \approx \frac{1}{6x^2},$$
where the error is moderately good for large $x$ (no worse than $\frac{1}{x^2}$ for sufficiently large $x$).
So at long last, you are wondering about the sum
$$ \sum_{x \geq 1} \frac{1}{6x^2} = \frac{\pi^2}{36},$$
though the particular value doesn't matter, but only the fact that it converges.
