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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.

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    $\begingroup$ $+1$ for well-written question with sufficient context. $\endgroup$ Commented Jan 28, 2023 at 17:10
  • $\begingroup$ 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. $\endgroup$ Commented Jan 28, 2023 at 18:16
  • $\begingroup$ @ThomasAndrews I think that gives a proof that $x_{n+1}/x_n \to 1/2$, no? (if one shows first that $x_{n+1}\to 0$). $\endgroup$
    – Pedro
    Commented Jan 28, 2023 at 18:40

1 Answer 1

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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 $x_1=\sin t_1$.

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  • $\begingroup$ Wow! Nice solution! How did you mind using trigonometry? Because of the product of sqaure roots? $\endgroup$ Commented Jan 29, 2023 at 16:55
  • $\begingroup$ Not the product, but $\sqrt{1-x_n^2}$ suggests polar transformation which is compatible with $x_n\in(0,1)$ $\endgroup$
    – TheSimpliFire
    Commented Jan 29, 2023 at 19:15
  • $\begingroup$ Oh I see, thanks! $\endgroup$ Commented Jan 29, 2023 at 21:02

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