[Math] Decreasing sequence, bounded below.

Question

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.

Answer 1

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 $x_n \to x$.