Does this sequence $\sin(1)\sin(2)\cdots\sin(n)$ converge to 0? Or does it converge at all?
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1It converges to 0 because $n$ will be arbitrarily close to a multiple of $\pi$ infinitely often – Maximilian Janisch Dec 17 '19 at 18:24
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Welcome to math.se. See this formatting guide. – J.G. Dec 17 '19 at 18:31
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@MaximilianJanisch $\pi$ being irrational, does any of the $sin(n)$ has $n$ an exact multiple of $\pi$ (even though it goes from $1$ to $\infty$)? – Déjà vu Dec 17 '19 at 18:36
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3@RingØ No - for the reason that $\pi$ is irrational. Hence "arbitrarily close to" rather than "is" – Henry Dec 17 '19 at 18:39
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It converges to $0$ because
every term is between $-1$ and $1$
and at least one in every three terms has an absolute value below $\sin(\frac \pi 2\pm\frac12) \approx 0.8776$
so the absolute value of the product of $n$ terms will be below $0.8776^{n/3}$ for $n \ge 3$,
and that converges to $0$ as $n$ increases
Henry
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Expanding on @MaximilianJanisch's comment, for infinitely many integers $q>0$ there is an integer $p>0$ with $\left|\frac{p}{q}-\pi\right|<\frac{\sqrt{5}}{q^2}$, whence$$|\sin p|=\left|\sin\left[q\left(\frac{p}{q}-\pi\right)\right]\right|<\frac{\sqrt{5}}{q}.$$Such sines ensures the desired infinite product is $0$.
J.G.
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