Divergence of $\prod_{n=1}^{\infty} a\sin(n)$ for $a>1$ to $0$ or $\infty$

Question

Consider the product $$\displaystyle\prod_{n=1}^{\infty} a\sin(n).$$

When $a=1$, clearly this product diverges to $0$, as $|{\sin{(n)}}|\le1$, so the value of the partial products can only decrease or stay the same (and they will only decrease if $n\in \mathbb{N}$). However, the divergence to 0 or infinity of this product becomes much more complex if $a>1$.

The question of divergence to $0$ is certainly not intuitive, as when $\sin(n)\approx0,$ the partial product will drop significantly, but when $|\sin(n)|>{1\over{a}}$ the partial product will grow (and this happens frequently for large enough $a$).

What seems to happen is that, for $a\le2$ the partial sums chaotically grow, before eventually divergine to $0$. For example, $\displaystyle\prod_{n=1}^{307} 2\sin(n)\approx1402$, but $\displaystyle\prod_{n=1}^{5000} 2\sin(n)\approx 4.8\times10^{-13}$.

When I ask Mathematica about the infinite product when, it rather quickly claims that the product does not converge for any $a$ (I do not know whether this also means it does not diverge to 0). Numerical evidence seems to suggest divergence to $0$ otherwise for $a\le2$. For $a>2$, it seems to fairly quickly diverge to infinity.

$a=2$ is probably the most interesting case, as approximately $50$% of the time it should grow, since $0<|\sin(n)|<1$ and the partial product grows iff $|\sin(n)|>1/2$.

Is it known whether this product diverges to $0$ for even one example of $a$ where $1<a\le2$? And if $a=2$?

If it doesn't necessarily diverge to $0$, does it remain bounded? And can divergence to infinity be proven for any $a>2$?

Answer 1

As I wrote in another answer, we can use inequalities such as:

$$|\sin(x)\sin(x+1)|\le\sin^2(1)$$

to show that

$$0=\prod_{n=1}^\infty a\sin(n)$$

for all $|a|\le\csc(1)\approx1.188$. Similarly one can try to bound the product of 3 consecutive terms, and according to WolframAlpha:

$$|\sin(x)\sin(x+1)\sin(x+2)|\le0.292$$

which gives us $|a|\le1/\sqrt[3]{0.292}\approx1.507$. More results:

$$\prod_{n=0}^5|\sin(x+n)|\le0.0820\Rightarrow|a|\le1/\sqrt[6]{0.0820}\approx1.517$$

$$\prod_{n=0}^9|\sin(x+n)|\le0.00964\Rightarrow|a|\le1/\sqrt[10]{0.00964}\approx1.590$$

$$\prod_{n=0}^{12}|\sin(x+n)|\le0.00158\Rightarrow|a|\le1/\sqrt[10]{0.00158}\approx1.642$$

$$\prod_{n=0}^{21}|\sin(x+n)|\le0.000000544\Rightarrow|a|\le1/\sqrt[22]{0.000000544}\approx1.926$$

which seems to give good bounds near multiples of $\pi$ and worse bounds in between.

If this calculation is accurate, we have:

$$\prod_{n=0}^{222}|\sin(x+n)|\le238\cdot2^{-223}\Rightarrow|a|\le2/\sqrt[223]{238}\approx1.951$$

Unfortunately the next time $n$ is closer to a multiple of $\pi$ than $223$ is not so close, so one may wish to try and improve on this instead.

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Geometrically averaging $\sin(x)$ on $[0,\pi]$, it is interesting to note that we have

$$\frac12=\exp\left[\frac1\pi\int_0^\pi\ln(\sin(x))~\mathrm dx\right]$$

which gives divergence to $0$ for any $|a|<2$ and unbounded for $|a|>2$ by the equidistribution theorem and lower bounds to $|\sin(n)|$ given by the irrationality measure of $\pi$ being bounded.