# 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 $n\ge1$?

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.

• $+1$ for well-written question with sufficient context. Commented Jan 28, 2023 at 17:10
• As $x\to 0,$ $$\sqrt{1+x}=1+\frac{x}2-\frac{x^2}8+O(x^3)$$ So $$\frac{\sqrt{1+x}-\sqrt{1-x}}2=\frac{x}2+O(x^3).$$ Not sure if that helps. Commented Jan 28, 2023 at 18:16
• @ThomasAndrews I think that gives a proof that $x_{n+1}/x_n \to 1/2$, no? (if one shows first that $x_{n+1}\to 0$).
– Pedro
Commented Jan 28, 2023 at 18:40

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 $$x_1=\sin t_1$$.
• Not the product, but $\sqrt{1-x_n^2}$ suggests polar transformation which is compatible with $x_n\in(0,1)$ Commented Jan 29, 2023 at 19:15