Is there a direct connection between Viète's formula and the sequence $a_{n+1}= a_n +\sqrt{a_n^2 +1}$ ($\lim_{n \to \infty } \frac{a_n}{c_n}=1$)?

Question

I saw the following problem

if $a_0=0$ and $a_{n+1}= a_n +\sqrt{a_n^2 +1}$ prove that $\lim_{n \to \infty}\frac{a_n}{2^n} =\frac{2}{\pi}$

since $\pi$ is transcendental number all of my "tricks" won't work and I have to search for a connection between the sequence $a_n$ and some transcendental functions like trigonometric functions via induction and I noticed this sequence is very similar Viète's formula as they have the the same limit and the both have the same denominator

so I want to know if there is a direct connect between the two sequence, in other words I want to show that if $b_1=\sqrt{2}$ and $b_k=\sqrt{2+b_{k-1}}$ and $c_n =\prod_{k=1} ^n b_k$ then $\lim_{n \to \infty } \frac{a_n}{c_n}=1 $

I am sure there are many simpler ways to prove $\lim_{n \to \infty}\frac{a_n}{2^n} =\frac{2}{\pi}$ than proving $\lim_{n \to \infty } \frac{a_n}{c_n}=1 $ but I want to see a direct connect between the two sequence.

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EDIT: Thank you @ayan for suggesting that my question is a dup to this question I want to clarify that I want a proof that $\lim_{n \to \infty } \frac{a_n}{c_n}=1 $ and not $\lim_{n \to \infty } \frac{a_n}{2^n}=\frac{2}{ \pi }$ of course one can make an argument $\lim_{n \to \infty }\frac{a_n}{c_n}=\lim_{n \to \infty } \frac{a^n}{2^n} \frac{2^n}{c^n}$ but this is not "direct" proof as it relay on the fact that you know $\lim_{n \to \infty}\frac{a_n}{2^n}$ and $\lim_{n \to \infty}\frac{c_n}{2^n}$.

Answer 1

First, we study the sequence $(b_k)_k$ by using the formula $$1+\cos(\theta)=2\cos^2\left(\frac{\theta}{2} \right)$$

Let's denote $b_k=2\cos(\theta_k)$, then: $$\begin{align} b_1&=\sqrt{2} = 2\cos\left(\frac{\pi}{2^2} \right)=2\cos(\theta_1)\\ b_k&=\sqrt{2+b_{k-1}}=\sqrt{2(1+\cos(\theta_k))}=2\cos\left(\frac{\theta_{k-1}}{2} \right) \end{align}$$ $$\implies \color{red}{\theta_k} =\frac{\theta_{k-1}}{2} =...=\frac{\theta_{1}}{2^{k-1}}= \color{red}{\frac{\pi}{2^{k+1}}} $$

Next, we study the sequence $(c_n)_n$, we have:

$$c_{n} = b_n\cdot c_{n-1}=2\cos\left(\frac{\pi}{2^{n+1}} \right)\cdot c_{n-1} = \frac{\sin\left(\frac{\pi}{2^{n}} \right)}{\sin\left(\frac{\pi}{2^{n+1}} \right)}\cdot c_{n-1}$$ $$\implies \sin\left(\frac{\pi}{2^{n+1}}\right)\cdot c_{n} =\sin\left(\frac{\pi}{2^{n}}\right)\cdot c_{n-1} =..=\sin\left(\frac{\pi}{2^{1+1}}\right)\cdot c_{1} = \sin\left(\frac{\pi}{4}\right) \cdot b_1 = 1$$ $$\implies \color{red}{c_{n} = \frac{1}{\sin\left(\frac{\pi}{2^{n+1}}\right)}}$$

As $a_n = \cot\frac{\pi}{2^{n+1}}$, we can deduce the connect between $a_n$ and $c_n$: $$\color{red}{\frac{a_n}{c_n}= \frac{\cot\left(\frac{\pi}{2^{n+1}}\right)}{1/ \sin\left(\frac{\pi}{2^{n+1}}\right)} = \cos\left(\frac{\pi}{2^{n+1}}\right)} \xrightarrow{n\to+\infty} 1$$