About $\lim_{n\to +\infty} n\prod_{k=1}^{n-1}\cos^2(k)$

Question

In my research work, I need to show that the sequence $(nu_n)$ tends to 0 where $ (u_n)$ is defined by $$u_{n+1}=u_{n} \cos^{2}(n),\quad u_{0}=1$$ $(u_n)$ is a positive and decreasing sequence. My adempt \begin{align*} u_{n+1}&= \prod_{k=1}^n\cos^2(k) =(\prod_{k=1}^n \frac{e^{k i } + e^{- k i }}{2} )^2 \\ &= 4^{-n}e^{-2(1+2+\cdots+n) \cdot i } \prod_{k=1}^n \left( 1 + e^{2 k i } \right)^2 \\ &=4^{-n}e^{-n(n+1)\cdot i } \prod_{k=1}^n \left( 1 + e^{2 k i } \right)^2 \end{align*} Afterwards, I don't see how to continue

Answer 1

It follows from the irrationality of $\pi$ and Weyl's criterion that the positive integers are equidistributed modulo $\pi$. In particular, asymptotically one-third of the integers $k\in\{1,\dotsc,n-1\}$ satisfy that $k\bmod\pi$ lies in $[\pi/3,2\pi/3]$. It follows, for $n$ large, that $\prod_{k=1}^{n-1}\cos^2(k)\leq 2^{-n/2}$. Hence the limit in the original question equals zero.

Remark. From $\int_0^\pi\ln(\cos^2(x))\,dx=-\pi\ln 4$ it even follows that $\prod_{k=1}^{n-1}\cos^2(k)\leq 4^{-(1+o(1))n}$.

Answer 2

It is graphically evident and easy to prove that if for $\theta\in\mathbb R$ one has $\cos^2\theta \ge c:=\cos^2\frac12=0,7701\dotso<1$, then $|\cos^2(\theta+1)|\le c$. As a consequence the number of integers $k\in\{0,\dotsc,n-1\}$ such that $\cos^2 k\ge c $ is at most $n/2$ (in fact it is even larger, due to the uniform distribution of $e^{2ki}$ on the circle). This ensures the product for $c_n$ an exponential decay, so that $nc_n=o(1)$.