Given a recursion $a_{n+ 1}= \dfrac{a_{n}^{2}+ 1}{2}$ with $a_{1}= \dfrac{1}{2}.$ Prove that $$1- a_{n}\sim\frac{2}{n}- \frac{2\ln n}{n^{2}}\,{\rm as}\,n\rightarrow\infty$$ Source: StachMath/@NN2 _ https://math.stackexchange.com/a/4025195/822157
I am continuing my research to the asymptotic behaviors, I summarized some idea from my friend @twelve_sakuya to find a proof :
Firstly, we let $b_{n}:= 1- a_{n},$ then we can show inductively that $$a_{1}= \frac{1}{2}, a_{n+ 1}= \frac{a_{n}^{2}+ 1}{2}, \frac{1}{2}\leq a_{n}< 1\Rightarrow b_{1}= \frac{1}{2}, \frac{1}{b_{n+ 1}}= \frac{1}{b_{n}}- \frac{1}{b_{n}- 2}, 0< b_{n}\leq\frac{1}{2}$$ Step 1. Show that $\dfrac{1}{b_{n}}\geq\dfrac{n+ 3}{2}$
Proof: $$b_{n}> 0\Rightarrow -\frac{1}{b_{n}- 2}> \frac{1}{2}\Rightarrow\frac{1}{b_{n}}= \frac{1}{b_{n- 1}}- \frac{1}{b_{n- 1}- 2}> \frac{1}{b_{n- 1}}+ \frac{1}{2}> \frac{1}{b_{1}}+ \frac{n- 1}{2}= \frac{n+ 3}{2}$$ Step 2. Show that $\dfrac{1}{b_{n}}\leq\dfrac{n+ H_{n+ 1}+ \frac{3}{2}}{2}$
Proof: $$\frac{1}{b_{n}}\geq\frac{n+ 3}{2}\Rightarrow -\frac{1}{b_{n}- 2}\leq\frac{n+3}{2\left ( n+ 2 \right )}\Rightarrow\frac{1}{b_{n}}= \frac{1}{b_{n- 1}}- \frac{1}{b_{n- 1}- 2}\leq\frac{1}{b_{n- 1}}+ \frac{n+ 2}{2\left ( n+ 1 \right )}\leq$$ $$\leq\frac{1}{b_{1}}+ \frac{1}{2}\sum_{k= 2}^{n}\frac{k+ 2}{k+ 1}= \frac{n+ H_{n+ 1}+ \frac{3}{2}}{2}$$ Because $H_{n}\sim\ln n,$ we speculate that $$\frac{1}{b_{n}}\leq\frac{n+ H_{n+ 1}+ \frac{3}{2}}{2}\Rightarrow\frac{1}{b_{n}}\geq\frac{n^{2}}{2\left ( n- \ln n \right )}+ \mathcal{O}_{1}\left ( \frac{1}{n} \right )\Rightarrow\frac{1}{b_{n}}\leq\frac{n^{2}}{2\left ( n- \ln n \right )}+ \mathcal{O}_{2}\left ( \frac{1}{n} \right )$$ He can't continue and we haven't done yet ! I need to the help to complete it ! Thanks for notice me.