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emphasized text$$\left[\, \prod_{n=1}^\infty \;\frac{3}{1+2 \cos\left(\frac{\pi}{3^n}\right)}\, \right]\, =\;? $$ Where $\;[\, .]\;$ denotes the integral part function.

$\mathbf {My Attempt}$
I tried to confine the n-th term to get $$\frac12 \le \cos\left(\frac{\pi}{3^n}\right) \lt 1$$ $$ 1 \lt \;\frac{3}{1+2 \cos\left(\frac{\pi}{3^n}\right)}\; \le \frac32 $$ But this didn't help as the product now is bounded below but unbounded above $\rightarrow \infty$
Using Wolfram Alpha, the product approaches 1.5708. So, the floor is 1.

I tried also to bound product from above using $$\left(\frac{\pi}{3^n}\right) \lt \left(\frac{\pi}{2^n}\right)$$ $$\cos\left(\frac{\pi}{3^n}\right) \gt \cos\left(\frac{\pi}{2^n}\right) \quad(\cos x \text { is decreasing on } ]0, \frac{\pi}{3}] )$$ And use the Telescoping product $$\cos\left(\frac{\pi}{2^n}\right)=\frac{\sin\left(\frac{\pi}{2^{n-1}}\right)}{2\sin\left(\frac{\pi}{2^n}\right)}$$ But this doesn't help much.
Any hint?

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$$1+2\cos\alpha=e^{i\alpha}+1+e^{-i\alpha}=\frac{e^{3i\alpha/2} -e^{-3i\alpha/2}}{e^{i\alpha/2}-e^{i\alpha/2}}=\frac{\sin3\alpha/2} {\sin\alpha/2}.$$ Therefore $$\prod_{n=1}^N\frac3{1+2\cos(\pi/3^n)} =\frac{3^N\sin(3^{-N}\pi/2)}{\sin(\pi/2)}\to\frac\pi2$$ as $N\to\infty$.

The integer part of $\pi/2$ is $1$.

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$$1+2\cos2x=1+2(1-2\sin^2x)=\dfrac{\sin3x}{\sin x}=\dfrac{f(n+1)}{f(n)}$$

where $f(m)=\sin(3^mx)$

Here $3^mx=?$

Related:$\cos x(2\cos2x-1)=\cos3x$

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