The infinite product of the sines of all positive integers is zero I tried to prove that
$\prod\limits_{k = 1}^\infty  {\sin k = 0}$
Define
$$P(m)=\prod\limits_{k = 1}^m  {\left|\sin k\right|}$$ 
then I take the logarithm of both sides
$$\log P(m)=\log \prod\limits_{k = 1}^m  {\left|\sin k\right|}$$ 
which can be written as
$$a_m=\log P(m)=\sum\limits_{k = 1}^m  {\log\,\left|\sin k\right|}$$
for any $k\in\mathbb{N}$ we have $0<\left|\sin k\right|<1$, so all terms of the sum above is strictly negative.
I say that the sequence diverges: $\left\{a_m\right\}_{m\in\mathbb{N}}\to\,-\infty$
This is a delicate point on which I have some doubts.
I know that 
$$\int\limits_0^\pi  {\log \left|\sin x\right|{\text{ d}}x}  =  - \pi \log 2$$
and that the function $\log \left|\sin x\right|$ is periodic with period $\pi$, 
so the integral from 1 to $\infty$ diverges to $- \infty$.
Then the series diverges as well for the integral test for convergence.
As $\log P(m)\to\, -\infty$ we can say that $P(m)\to\,0$ as $m\to\,\infty$
Therefore
$$\mathop {\lim }\limits_{m \to \infty } \prod\limits_{k = 1}^m {\left| {\sin k} \right| = \mathop {\lim }\limits_{m \to \infty } \prod\limits_{k = 1}^m {\sin k = \prod\limits_{k = 1}^\infty  {\sin k = } 0} }.\quad \blacksquare$$
 A: Since $$|\sin(2k-1)\sin(2k)| = \left|\frac{\cos 1-\cos(4k-1)}{2}\right| \le \frac{1+\cos 1}{2} < 1$$
we have
$$\begin{align}
& 0 \le \left|\; \prod_{k=1}^{n} \sin k\;\right| 
    \le \left|\; \prod_{k=1}^{\lfloor\frac{n}{2}\rfloor} \sin(2k-1)\sin(2k)\;\right|
\le \left(\frac{1+\cos 1}{2}\right)^{\lfloor\frac{n}{2}\rfloor} \\
\implies & 0 
\le \liminf_{n\to\infty} \left|\; \prod_{k=1}^{n} \sin k\;\right| 
\le \limsup_{n\to\infty} \left|\; \prod_{k=1}^{n} \sin k\;\right| 
\le \lim_{n\to\infty} \left(\frac{1+\cos 1}{2}\right)^{\lfloor\frac{n}{2}\rfloor} = 0\\
\implies &\lim_{n\to\infty} \prod_{k=1}^{n} \sin k = 0
\end{align}$$
A: One can prove using irrationality of $\pi$ that there are infinitely many natural numbers $\{n_k,k\geqslant 1\}$ such that $|\sin(n_k)|<1/2$. Then 
$$\left|\prod_{l=1}^{n_k}\sin l\right|\leqslant \left|\prod_{j=1}^k\sin n_j\right|\leqslant \frac 1{2^k}.$$
Conclude taking the limit $k\to +\infty$.
A: To complete your idea which is almost correct: note that the integral test for convergence is applicable only for monotone decreasing function so in our example you can't use this test, however we know that the sequence $(\log|\sin k|)$ does not converge to $0$ since the sequence $(\sin k)$ has not even a limit so the series 
$$\sum_{k=1}^\infty \log|\sin k|$$
is divergent, moreover its general term $\log|\sin k|$ has a negative sign hence we have
$$\sum_{k=1}^\infty \log|\sin k|=-\infty$$
which allows us to conclude
A: We have $\forall n\in \Bbb N :0<\left|\sin(n)\right|<1$. Therefore $P(m)$ is nonnegative and decreasing. Thus it converges.
Now, show that $0$ is a limit point of $\left|\sin(\Bbb N)\right|$. If you already know that $\sin(\Bbb N)$ is dense in $[-1,1]$, you can use that. Otherwise, check one of the many proofs on this site. The first step is usually to show that $0$ is a limit point.
It now follows that no matter what $\epsilon > 0$ we choose, the limit of $P(m)$ is smaller. Conclude that the limit is $0$.
A: $$\prod_{k=1}^n \left \vert \sin(k) \right \vert = \left(\prod_{k=1}^n \left(1-\cos^2(k)\right) \right)^{1/2}$$
And since $\displaystyle \sum_{k=1}^n \cos^2(k)$ diverges we can conclude from here, that the sequence converges to $0$.
A: Actually if you find a real number a with $0<a<1$ and any strictly increasing sub-sequence of integers, say $p(k)$ with the property that $|\sin p(k)|<a,k=1,2,\cdots$,
$$b_m=\prod_{k=1}^m|\sin p(k)|<a^m\to 0$$
You are done because you can decompose the natural numbers $N$ as $A \bigcap (N-A)$ where $A=\{ p(1),p(2),\cdots k=1,2,...\}$ and you have the product over A converging to zero and the product over $N-A$ bounded by 1.
Q.E.D.
