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Suppose that $0 \leq x_1 < 1$ and $x_{n+1} = 1 - \sqrt{1 - x_n}$ for all natural $n$. Prove that $x_n$ is decreasing and bounded below as $n$ converges.

Attempt: Suppose that $0 \leq x_1 < 1$ and $x_{n+1} = 1 - \sqrt{1 - x_n}$ for all natural $n$. Then we wish to show $x_{n+1} < x_n$ for all $n$ is decreasing. Then $x_{n+1} < x_n$ implies $1 -\sqrt{1 - x_n} < x_n$ implies $1 - x_n < \sqrt{1 - x_n}$. Now by our assumption we know $0 \leq x_n < 1$ then $0 < 1- x_n \leq 1$. Then this shows the sequence is decreasing. Can someone please help me? I would really appreciate it. I don't know how to continue. Thank you.

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  • $\begingroup$ a decreasing bounded-from-below sequence must converge. $\endgroup$ – Alex R. Sep 18 '14 at 2:51
  • $\begingroup$ That is what I am trying to prove. $\endgroup$ – user94463 Sep 18 '14 at 2:52
  • $\begingroup$ It's bounded from below, so it's liminf is finite. Show it converges to the liminf. Alternatively suppose by contradiction that it has no limit, the contradiction will be that you'll break boundedness. $\endgroup$ – Alex R. Sep 18 '14 at 2:54
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If $x_n$ is non increasing and $x_n \ge B$, then $x_n$ converges.

Let $x = \inf_n x_n$, we have $x_n \ge x \ge B$, of course.

Let $\epsilon>0$, by definition of $\inf$, there is some $n$ such that $x_n < x+\epsilon$.

Can you finish it from here?

We have $x_n \ge x_k$ for all $k \ge n$. Hence $|x-x_k| < \epsilon$ for all $k \ge n$. It follows that $x_n \to x$.

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