Let $0<x_{1}<1$ and $x_{n+1}=1-\sqrt{1-x_{n}} $.Show that $(x_{n})$ is decreasing.Find the limit of $x_{n}$ and the limit of $\frac{x_{n+1}}{x_{n}}$. I showed that $\frac{x_{n+1}}{x_{n}}$=$\frac{1}{1+\sqrt{1-x_{n}}}$ ,that is smaller or equal to $1$,so the sequence is decreasing. Then I supposed that $x_{n} \to l \in R$.From $x_{n+1}=1-\sqrt{1-x_{n}} $ we find that $l=0$ or $l=1$.But,how can I accept one solution and reject the other one?Or can I find the limit in another way?
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Since $\{x_n\}$ is decreasing and $0<x_1<1$, you have $x_n<x_1<1$ for all $n$. In particular the limit must be $\le x_1$.
Julián Aguirre
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1Could I also say that the sequence is decreasing and bounded from below,so it converges to its inf?? – evinda Feb 07 '14 at 16:43
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2Yes, you can say that. – Julián Aguirre Feb 07 '14 at 16:46
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Simply use the fact that $0\lt x_1\lt 1$, and the fact that the sequence is decreasing $(l\lt \cdots < x_3 \lt x_2 \lt x_1 \lt 1)$ to show that the limit therefore cannot be $1$, and hence, the limit must then be $0$, given the possibilities that $l = 1 \lor l = 0$.
amWhy
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1Could I also say that the sequence is decreasing and bounded from below,so it converges to its inf?? – evinda Feb 07 '14 at 16:42
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