$a_n$ is bounded and decreasing

Question

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

Answer 1

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, $a_n\gt\frac2\pi$.

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Another Approach

Since $\sin(x)\le x$, $\cos(x)\ge\left(1-x^2\right)^{1/2}\ge\left(1-x^2\right)$. Therefore,

$$ \begin{align} \prod_{k=2}^\infty\cos\left(\frac\pi{2^k}\right) &\ge\prod_{k=2}^\infty\left(1-\frac{\pi^2}{4^k}\right)\\ &=\prod_{k=2}^\infty\left(1+\frac{\frac{\pi^2}{4^k}}{1-\frac{\pi^2}{4^k}}\right)^{-1}\\ &\ge\exp\left(\sum_{k=2}^\infty\frac{\frac{\pi^2}{4^k}}{1-\frac{\pi^2}{4^k}}\right)^{-1}\\ &\ge\exp\left(\sum_{k=2}^\infty\frac{\frac{\pi^2}{4^k}}{1-\frac{\pi^2}{16}}\right)^{-1}\\ &=\exp\left(\frac{-4\pi^2}{48-3\pi^2}\right)\\[9pt] &\doteq0.116881538733185 \end{align} $$

Answer 2

That is correct. Also, you can deduce that from the fact that, since $(a_n)$ is decreasing, $a_n\le a_2$ for every $n\ge2$.

Answer 3

$a_n \leq 1,\forall n \geq 2$.