let sequence $\{a_{n}\}$,such $a_{1}=1$,and $$a_{n+1}=\sqrt{a^2_{n}-2a_{n}+2}-1$$
prove or disprove :there exists a constant $c$ such $$a_{2n}<c<a_{2n+1}?$$
my idea: since $$a_{n+1}+1=\sqrt{(a_{n}-1)^2+1}$$ so $$(a_{n+1}+1)^2=(a_{n}-1)^2+1$$ then I can't it. Thank you
By the way this problem seems the form is simple, but when I try solve this problem, I fell is very hard.
Consider the function $f\colon [0,1] \to [0,1]$ given by $f(x)= \sqrt{(1-x)^2+1}-1$. $f$ is differentiable on $[0,1]$, with
$$f'(x) = \frac{x-1}{\sqrt{(1-x)^2+1}} < 0,$$
except for $x = 1$, where the derivative is zero. So $f$ is strictly decreasing on $[0,1]$, and has a unique fixed point $c = \frac{1}{4}$. Since $f$ is strictly decreasing, we have
$$f(x) < c \iff x > c,$$
and vice versa. Since $a_{n+1} = f(a_n)$ and $a_1 = 1 > c$, we have indeed
$$a_{2n} < c < a_{2n+1}$$
for all $n$.
Since $\lvert f'\left(x\right)\rvert \leqslant \frac{1}{\sqrt{2}} < 1$ for all $x\in [0,1]$, $f$ is a contraction, so the sequence converges to the fixed point, and the subsequence $(a_{2n})$ is strictly increasing, while $(a_{2n+1})$ is strictly decreasing.
