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Let $$u_n=\prod_{k=1}^{n-1}\cos^2(k)$$ then $$\frac1n \ln(u_n) = \frac1n\sum_{k=0}^{n-1} \ln(\cos^2(k)) \underset{n\to\infty}\longrightarrow \frac1{2\pi} \int_0^{2\pi} \ln(\cos^2(x))\,{\rm d}x = -\ln(4)$$ We deduce that $$\ln(u_n)\sim -n\ln(4)$$ . I think that the sequence $v_n=\ln(4^n u_n))$ have not limit , So there is no constant c such that $u_n \sim c\,4^{-n}$. Can we find better then $$ \prod_{k=1}^{n-1}\cos^2(k)\sim 4^{-n}e^{o(n)}$$ . What is $\limsup 4^n u_n$ and $\liminf 4^n u_n$ ?

additional comments

wolfram does not confirm that $\liminf 4^n u_n =0$enter image description here

GH from MO
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Pascal
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    You should link to this: https://mathoverflow.net/questions/383866 – François Brunault Feb 14 '21 at 14:30
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    You need to be careful with the limit in your second display. If $k\bmod\pi$ gets very close to $\pi/2$ for some $k\in{1,\dotsc,n-1}$, then $\ln(u_n)/n$ can be much smaller than $-\ln(4)$. This is why, in my response to your previous question, I only stated $\prod_{k=1}^{n-1}\cos^2(k)\leq 4^{-(1+o(1))n}$ instead of $\prod_{k=1}^{n-1}\cos^2(k)=4^{-(1+o(1))n}$. Note that $f(x):=\ln(\cos^2(x))$ is not Riemann integrable on $[0,\pi]$, because it is not bounded there. Instead, $\int_0^\pi f$ exists as an improper Riemann integral, i.e. as $\lim_{h\to 0+}(\int_0^{\pi/2-h} f+\int_{\pi/2+h}^\pi f)$. – GH from MO Feb 15 '21 at 05:36
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    Thanks GH I didn't realize that f is not Riemann integrable on $I = [0,2 \ pi]$ – Pascal Feb 15 '21 at 13:43

1 Answers1

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I presume that exchanging the cosine by the sine will not matter for the large-$n$ behavior of the product, so let me consider $$a_n^2=4^n\prod_{k=1}^{n}\sin^2 k=\left(\prod_{k=1}^n\left|1-e^{k\alpha \pi i}\right|\right)^2\;\;\text{with}\;\;\alpha=2/\pi.$$ The convergence of $a_n$ was determined in this MO posting from five years ago, with lim inf $a_n$ equal to zero and lim sup $a_n$ equal to infinity for irrational $\alpha$.

Carlo Beenakker
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  • Thank you Carlo Beenakker $u_n=4^n \prod_{k=1}^{n-1}\cos^2(k)=4\left(\prod_{k=1}^{n-1}\left|1+e^{2k i}\right|\right)^2$I'm trying to see how to apply the result of your link if we replace the - with the + – Pascal Feb 14 '21 at 21:12