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



About $\lim_{n\to +\infty} n\prod_{k=1}^{n-1}\cos^2(k)$
It follows from the irrationality of $\pi$ and [Weyl's criterion](https://en.wikipedia.org/wiki/Equidistributed_sequence#Weyl%27s_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 $\prod\_{k=1}^{n-1}\cos^2(k)\leq 4^{-(1+o(1))n}$.


