I have this hint from old question of mine if someone could help me to understand it
Sequence $0\leq a_{n}-l\leq \dfrac{\pi^{2} }{2^{2n+1}}$
for the first question it's easy to see that a_n is decreasing :
note that $ \forall x\in [0,\pi/2],\quad \cos(x) \geq 0$ since $\pi/2^n \in [0,\pi/2]$ then $\cos(\pi/2)\geq 0$ then $a_n\geq 0 $
let $a_n=a_{n−1}\cos(\pi2n)\quad \forall n\geq 3$
then $a_{n+1}−a_n=u_n(cos(\pi/2^n)-1)$ which is negative since $(cos(\pi/2^n)-1)$ 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 $0\leq a_n \leq 1$
for $b_{n}=a_{n}\cos\left(\dfrac{\pi }{2^{n}}\right)$ is increasing
$b_{n+1}-b_{n}=a_{n+1}\cos\left(\dfrac{\pi }{2^{n+1}}\right)-a_{n}\cos\left(\dfrac{\pi }{2^{n}}\right)=a_{n}\cos\left(\dfrac{\pi }{2^{n}}\right)\cos\left(\dfrac{\pi }{2^{n+1}}\right)-a_{n}\cos\left(\dfrac{\pi }{2^{n}}\right)$
or $\dfrac{b_{n+1}}{b_n}=\cos(\pi/2^{n+1})$
im stuck here and after we show $b_n$ is increasing is that can help us to study $c_n$
Thanks for any help
Hint: $\cos \left(\dfrac{\pi}{2^k}\right)=\dfrac{\sin \left(\dfrac{\pi}{2^{k-1}}\right)}{2\sin \left(\dfrac{\pi}{2^k}\right)}$. Now let $k$ runs from $2$ to $n$ and simplify the product.