Sequence of real numbers $(b_n)$ is given recursively:
$$b_1=0, b_{n+1}=\frac{3-(b_n)^2}{2} \text{ for } n \gt1$$
Prove that it isn't monotonic, but that it still converges and has limit.
I have proven that it's not monotonic, bu I'm not sure how to prove that it converges. Thanks in advance.
We can see $b_1 = 0$ implies $ 0 <b_2= \frac{3}{2} < \sqrt{3}$. Now assume $0 <b_k < \sqrt{3}$ for some $k \geq 2$, then $$\sqrt{3}>\frac{3}{2} > b_{k+1}=\frac{3-b_k ^2}{2} > 0 .$$ Therefore by induction $$0 <b_n < \sqrt{3} \quad \forall n \geq 2 .$$Now take $\epsilon = -1+\sqrt{3} > 0$ then $b_n < 1- \epsilon=2-\sqrt{3}.$ Otherwise $b_{n+1}=\frac{3-b_n^2 }{2}\leq 0$ as $b_n \geq 1- \epsilon \Rightarrow 3-b_n^2 \leq 0.$ Now $$\frac{|1+b_n|}{2} < r=\frac{3-\sqrt{3}}{2} < 1 $$ for sufficiently large $n$ . Now $$|b_{n+1}-1|=\frac{|1+b_n|}{2} |1-b_n|< r|1-b_n| \quad \forall n \geq 2.$$ Therefore $$|b_{n+1}-1|< r|1-b_n|<r^2|1-b_{n-1}|<r^3|1-b_{n-2}|\ldots.$$ Now as $r<1,$ the limit exists and is equal to $1.$ Now the non monotonicity follows from the relation $b_{n+1}-b_n = \frac{b_{n-1}^2 -b_n^2}{2}$ and the positivity of $b_n.$