<s>[INST] $a_n$ is bounded and decreasing my second question from 
[An inequality for the product $\prod_{k=2}^{n}\cos\frac{\pi }{2^{k}}$

Let $n\geq 2\quad a_{n}=\prod\limits_{k=2}^{n}\cos\left(\dfrac{\pi }{2^{k}}\right)$ and $b_{n}=a_{n}\cos\left(\dfrac{\pi }{2^{n}}\right)$

Show that $a_{n}$ is decreasing and bounded
note that $ \forall x\in [0,\dfrac{\pi}{2}],\quad \cos(x) \geq 0$
since $\pi/2^n \in [0,\dfrac{\pi}{2}]$ then $cos(\dfrac{\pi}{2})\geq 0$ then $a_n\geq 0  $ 
let $a_n=a_{n−1}\cos(\dfrac{\pi}{2^n})\quad  \forall n\geq 3$ 
then  $a_{n+1}−a_n=a_n(\cos(\dfrac{\pi}{2^{n+1}})-1)$ which is negative since $(\cos(\dfrac{\pi}{2^{n+1}})-1)$ its.
but for bounded :
we've already that $a_n\geq 0,\quad \forall n \geq 2$
note that $|(cos(\pi/2^k)|\leq 1\quad \forall k\geq 2$ then $\prod_{k=3}^{n}|(cos(\pi/2^k)|\leq 1 $ thus $0\leq a_n \leq 1$
am i right ?
any help would be appreciated
 [/INST] The reasoning about $a_n$ decreasing is fine; that is, for $n\ge2$, $0\lt\cos\left(\frac\pi{2^n}\right)\lt1$. This also shows that $a_n\gt0$. However, we can do a bit better.
Trigonometric Approach
Since
$$
2\sin\left(\frac\pi{2^k}\right)\cos\left(\frac\pi{2^k}\right)
=\sin\left(\frac\pi{2^{k-1}}\right)
$$
We have
$$
\prod_{k=2}^n\cos\left(\frac\pi{2^k}\right)=\frac2{2^n\sin\left(\frac\pi{2^n}\right)}
$$
and
$$
\lim_{n\to\infty}\frac2{2^n\sin\left(\frac\pi{2^n}\right)}=\frac2\pi
$$
Therefore, $a_n\gt\frac2\pi$.

Another Approach
Since $\sin(x)\le x$, $\cos(x)\ge\left(1-x^2\right)^{1/2}\ge\left(1-x^2\right)$. Therefore, 
$$
\begin{align}
\prod_{k=2}^\infty\cos\left(\frac\pi{2^k}\right)
&\ge\prod_{k=2}^\infty\left(1-\frac{\pi^2}{4^k}\right)\\
&=\prod_{k=2}^\infty\left(1+\frac{\frac{\pi^2}{4^k}}{1-\frac{\pi^2}{4^k}}\right)^{-1}\\
&\ge\exp\left(\sum_{k=2}^\infty\frac{\frac{\pi^2}{4^k}}{1-\frac{\pi^2}{4^k}}\right)^{-1}\\
&\ge\exp\left(\sum_{k=2}^\infty\frac{\frac{\pi^2}{4^k}}{1-\frac{\pi^2}{16}}\right)^{-1}\\
&=\exp\left(\frac{-4\pi^2}{48-3\pi^2}\right)\\[9pt]
&\doteq0.116881538733185
\end{align}
$$
</s>