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: 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.
A: 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 $\sum_{n = 1}^\infty \log(n\sin\frac{1}{n})$.
A: It is interesting to notice that, since:
$$ \sin z = z\prod_{n\geq 1}\left(1-\frac{z^2}{n^2 \pi^2}\right)\tag{1}$$
we have:
$$ \log\left(m\sin\frac{1}{m}\right)=\sum_{n\geq 1}\log\left(1-\frac{1}{\pi^2 n^2 m^2}\right)=-\sum_{n\geq 1}\sum_{k\geq 1}\frac{1}{k\pi^{2k}(nm)^{2k}}\tag{2}$$
and so, summing over $m$ too:
$$\sum_{m\geq 1}\log\left(m\sin\frac{1}{m}\right)=-\sum_{k\geq 1}\frac{1}{k\pi^{2k}}\sum_{s\geq 1}\frac{d(s)}{s^{2k}}=\color{red}{-\sum_{k\geq 1}\frac{\zeta(2k)^2}{k \pi^{2k}}}\tag{3}$$
that converges quite fast.
A: Hint: Compute $$\lim_{x\to0}\frac{\log(x^{-1}\sin x)}{x^2}.$$
A: Hint: Show that $\log ({\sin x \over x})$ has a Taylor expansion of the form $cx^2 + $ higher order terms, and use $x = {1 \over n}$.
