Let $ (x_n)_{n=1}^{\infty} $ be a sequence given by $ 0 \lt x_1 < 1 $, $ x_{n+1} = 1- \sqrt{1-x_n} $. I need to prove that $ (x_n) $ is well-defined, convergent and find its limit, I already proved that $ \forall x \in (0,1 ), x\gt 1 - \sqrt{1-x} $.
Can you verify my answer?
$ \underline{\text{Well-Defined:}} $
In order to prove that $ (x_n) $ is well-defined, I will use induction in to show that $ 0 \lt x_n \lt 1, \forall n $. For $ n = 1,\ 0 < x_1 < 1 $. Assume the statement is true for some natrual number $ n $. Therefore, $ 0 < x_n < 1 \Rightarrow 0 \gt -x_n \gt -1 \Rightarrow 1 \gt 1 - x_n \gt 0 \Rightarrow 1 \gt \sqrt{1 - x_n} \gt 0 \Rightarrow -1 \lt -\sqrt{1 - x_n} \lt 0 \Rightarrow 0 \lt 1 - \sqrt{1-x_n} \lt 1 \Rightarrow 0 \lt x_{n+1} \lt 1. $ Therefore, $ (x_n) $ is well-defined.
$ \underline{\text{Convergent:}} $
We've already proven $ (x_n) $ is bounded below by 0. We will now show its decreasing. Let $ n \in \mathbb{N} $. Therefore, $ 0 \lt x_n \lt 1 \Rightarrow x_n \gt 1 - \sqrt{1- x_n} = x_{n+1} $.
$ (x_n) $ is monotone and bounded below and therefore convergent.
$ \underline{\text{Limit:}} $
As a convergent sequence, $ \displaystyle \lim_{n \to \infty} x_n = \lim_{n \to \infty} x_{n+1} $, and therefore, if $ \lim_{n \to \infty} x_n = L $, then $ L = 1- \sqrt{1-L} $.
$$ L-1=-\sqrt{1-L} $$
$$ L^2 -2L + 1 = 1-L $$
$$ L^2 - L = 0 $$
$$ L = 0 \lor L = 1 $$
$ (x_n) $ is monotonically decreasing, and therefore, for all $ n > 1 $, $ x_n < x_1 < 1 $, which means, for $ \varepsilon = \frac{x_1}{2}, \forall N, x_{N+1} \notin (1 - \varepsilon, 1 + \varepsilon) $ and therefore $ L \neq 1 $. Hence $ L = 0 $.
