What is the value of $\lim\limits_{n\to\infty}2^{n}x_{n}$ if $x_1\in(0,1)$ and $x_{n+1}=\frac{\sqrt{1+x_n}-\sqrt{1-x_n}}2$ for every $n\ge1$?

Question

We have the sequence $(x_{n})_{n\geq1}$ with $x_{1}\in(0,1)$ and

$x_{n+1}=\dfrac{\sqrt{1+x_{n}}-\sqrt{1-x_{n}}}{2}\;$ for every $n\geq1$.

What is the value of $$\lim_{n \rightarrow \infty}2^{n}x_{n}=\;?$$ We can easily find that $\lim\limits_{n \rightarrow\infty}x_{n}=0$ and $\lim\limits_{n \rightarrow\infty}\dfrac{x_{n+1}}{x_{n}}=\dfrac{1}{2}$.

I tried using Stolz-Cèsaro once, twice but it does not work. I tried using the ratio test but again nothing. I tried taking the natural logarithm of the $z_{n}=2^{n}x_{n}$ and try calculating $\lim\limits_{n \rightarrow\infty}\big(n\ln 2+\ln x_{n}\big)$ but nothing.

Answer 1

Writing as $2x_{n+1}^2=1-\sqrt{1-x_n^2}$, let $x_n=\sin t_n$ so $\sin^2t_{n+1}=(1-\cos t_n)/2=\sin^2(t_n/2)$.

Thus $t_n=t_1/2^{n-1}$ so $$\lim_{n\to\infty}2^nx_n=\lim_{n\to\infty}2^n\sin\frac{t_1}{2^{n-1}}=2\arcsin x_1$$ since $\sin x=x+\mathcal O(x^3)$ and $x_1=\sin t_1$.