Limit of the product of sin(1/k) I'm having trouble with the following exercise:
$$ \lim_{n \to \infty} \prod_{k=1}^{n} \sin ( \frac{1}{k} )$$ 
As we already know by continuity $\lim_{n \to \infty} \sin \frac{1}{n} =0 $
So it seems like the limit can be expanded as:
$$ \lim_{n \to \infty} [  \sin(1) \sin (1/2)\sin(1/3)\cdots\sin(1/n)] =0 $$
but if such thing is correct, don't have any idea for the proof. 
Thanks in advance. 
 A: Since $x>0 \implies \sin(x) < x$ we get
$$0 < \prod_{k=1}^n \sin\left(\frac{1}{k}\right) < \prod_{k=1}^n \frac{1}{k}=\frac{1}{n!}$$
Then use squeeze theorem.
A: Another approach with less machinary is that since $0 < \frac{1}{n} \leq 1 < \frac{ \pi}{2}$, hence $ 0 < \sin \frac{1}{n} < \sin 1 $, so 
$$0 < \prod_{k=1}^n \sin\left(\frac{1}{k}\right) \leq  ( \sin 1 ) ^ n.$$
A: Amazing could be an approximation of the partial product.
$$a_n= \prod_{k=1}^n \sin\left(\frac{1}{k}\right)\implies \log(a_n)=\sum_{k=1}^n \log \left(\sin \left(\frac{1}{k}\right)\right)$$
Using Taylor expansions for large values of $k$
$$\log \left(\sin \left(\frac{1}{k}\right)\right)=-\log \left({k}\right)-\frac{1}{6 k^2}-\frac{1}{180
   k^4}+O\left(\frac{1}{k^6}\right)$$ giving
$$\log(a_n)\sim -\log(n!)-\frac{1}{6}H_n^{(2)}-\frac{1}{180}H_n^{(4)}$$ Using Stirling expansion and the asymptotics of the generalized harmonic numbers, this would give
$$\log(a_n)\sim n(1-\log(n))-\frac 12 \log(2\pi n)-\frac{\pi ^2 \left(450+\pi ^2\right)}{16200}+\frac 1 {12n}+O\left(\frac{1}{n^2}\right)$$
For illustration, just a few numbers for $\log(a_n)$
$$\left(
\begin{array}{ccc}
 n & \text{approximation} & \text{exact}  \\
 2 & -0.89031 & -0.90777 \\
 3 & -2.01647 & -2.02497 \\
 4 & -3.41660 & -3.42170 \\
 5 & -5.03435 & -5.03782 \\
 6 & -6.83165 & -6.83421 \\
 7 & -8.78153 & -8.78352 \\
 8 & -10.8639 & -10.8656 \\
 9 & -13.0635 & -13.0649 \\
 10 & -15.3679 & -15.3691
\end{array}
\right)$$
