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Carlo Beenakker
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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 considereddetermined in this MO postingMO posting from five years ago, with lim inf $a_n$ equal to zero and lim sup $a_n$ equal to infinity for irrational $\alpha$.

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 considered 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 $\alpha$.

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 $\alpha$.

Source Link
Carlo Beenakker
  • 188.1k
  • 18
  • 448
  • 651

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 considered 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 $\alpha$.