Given a recursion $a_{n+ 1}= \dfrac{a_{n}^{2}+ 1}{2}$ with $a_{1}= \dfrac{1}{2}.$ Find the $\mathcal{O}\left ( \dfrac{1}{n} \right )$ so that $$a_{n}- 1\sim\mathcal{O}\left ( \dfrac{1}{n} \right )\,{\rm as}\,n\rightarrow\infty$$
Remark. By this recurrence sequence, Ji Chen proposed 2 inequalities on AoPS, which I can't prove, of course $$1- \frac{2}{n}+ \frac{2}{n^{2}}\ln\frac{n}{3}+ \frac{417}{128n^{2}}\leq a_{n}\leq 1- \frac{2}{n}+ \frac{5\ln n+ 3}{n^{2}}$$ and for non-negative $x_{1\div n}$ quite not related $$\sum_{i= 1}^{n}\frac{x_{i}}{\left ( 1+ \sum_{j= 1}^{i}x_{j} \right )^{2}}\leq a_{n}^{2}$$ Return to the OP, I realise that $$a_{n+ 1}- 1= \frac{1}{2}\left ( a_{n}- 1 \right )^{2}+ a_{n}- 1= \frac{1}{2}\left ( a_{n}- 1 \right )^{2}+ \omega\left ( \left ( a_{n}- 1 \right )^{2} \right )$$ As same as what I told on this topic * Find the $\mathcal{o}\left(n\right)$ so that $a_n\sim\frac{\mathcal{o}\left(n\right)}{n}+\mathcal{O}\left(\frac{1}{n}\right){\rm as}\lim a_n=\infty$ *, I still can't find such an $\mathcal{o}$ satisfying my wish. I need to your help, thanks.
First, we have the sequence $a_n$ satisfying $a_{n+1}= \frac{ a_{n}^2+1}{2}$ (1) converges to $1$. Indeed, we have $a_n$ is a increasing sequence ($a_{n+1}-a_n = \frac{1}{2}(a_{n}- 1)^2)$ and the sequence is bounded above by $1$ (we deduce that by proof of induction, as $a_{n+ 1}- 1= \frac{a_{n}^2- 1}{2}<1$ and $a_1 <1$).
Now, with the technique "The second term in asymptotic expansion" of Moubinool OMARJEE provided by RiverLi, we can use the function $f(y)=\frac{1}{1-y}$ for the second order expansion. Indeed, we find this function by noticing that $a_{n+1}-a_n = \frac{1}{2}(a_{n}- 1)^2$ and from the differential equation $\frac{dy}{dx} = \frac{1}{2}(y- 1)^2$ we deduce that $x=\frac{2}{1-y}$).
We have
\begin{align} (1) &\iff 1-a_{n+1} = \frac{ 1-a_{n}^2}{2}\\ & \iff \frac{1}{1-a_{n+1}}=\frac{1}{1-a_{n}}+\frac{1}{1+a_{n}} \\ & \iff \frac{1}{1-a_{n+1}}-\frac{1}{1-a_{n}}=\frac{1}{1+a_{n}} \end{align} then $$ \frac{1}{1-a_{n+1}}-\frac{1}{1-a_{n}}=\frac{1}{1+a_{n}}\xrightarrow{n \rightarrow +\infty} \frac{1}{2} $$ or $$\frac{1}{1-a_{n}} \sim \frac{n}{2} \tag{2}$$
And from (2) we can deduce that $$a_n \sim 1-\frac{2}{n}$$
Let's denote $a_n=1-\frac{2}{n}+\frac{1}{n}b_n$, then $b_n = \mathcal{o}(1)$. We have \begin{align} (1) &\iff 1-\frac{2}{n+1}+\frac{b_{n+1}}{n+1} = \frac{1}{2}((1-\frac{2}{n}+\frac{b_n}{n})^2+1) \\ &\iff b_{n+1} =2+\frac{n+1}{2}\left( 2 \frac{-2+b_n}{n} +\left(\frac{-2+b_n}{n} \right)^2 \right) \\ &\iff b_{n+1} = 2+\frac{n+1}{2}\left( -\frac{4}{n} +\frac{2b_n}{n} +\frac{4}{n^2} -\frac{4b_n}{n^2}+\frac{b_n^2}{n^2}\right)\\ &\iff b_{n+1} = 2+\frac{n+1}{2}\left( -\frac{4}{n} +\frac{4}{n^2} \right) +\frac{n+1}{2}\left( \frac{2b_n}{n} -\frac{4b_n}{n^2}\right) +\frac{n+1}{2}\left( \frac{b_n^2}{n^2}\right)\\ &\iff b_{n+1} = \frac{2}{n^2}+(1+\frac{1}{n})(1-\frac{2}{n})b_n + \frac{1}{2n}(1+\frac{1}{n})b_n^2 \\ &\iff b_{n+1} = (1-\frac{1}{n})b_n +\frac{b_n^2}{2n}+ \frac{2}{n^2}-\frac{2}{n^2}b_n+\frac{1}{2n^2}b_n^2 \tag{2}\\ \end{align} We guess that $b_n\sim \alpha \frac{\ln(n)}{n}$, we denote $b_n=\frac{c_n}{n}$. Because $b_n = \mathcal{o}(1)$ then $c_n = \mathcal{o}(\ln(n))$. \begin{align} (2) &\iff \frac{c_{n+1}}{n+1} = (1-\frac{1}{n})\frac{c_n}{n} +\frac{b_n^2}{2n}+ \frac{2}{n^2}-\frac{2}{n^2}b_n+\frac{1}{2n^2}b_n^2 \\ &\iff c_{n+1}= (1-\frac{1}{n^2})c_n +\frac{n+1}{n}\frac{b_n^2}{2}+ \frac{2(n+1)}{n^2} + (n+1)\left(-\frac{2}{n^2}b_n+\frac{1}{2n^2}b_n^2 \right) \\ &\iff c_{n+1} -c_n= -\frac{c_n}{n^2} +\frac{n+1}{n}\frac{b_n^2}{2}+ \frac{2(n+1)}{n^2} + (n+1)\left(-\frac{2}{n^2}b_n+\frac{1}{2n^2}b_n^2 \right) \tag{3} \\ \end{align}
We notice that $$-\frac{c_n}{n^2} = \mathcal{o}(\frac{1}{n})$$ $$\frac{n+1}{n}\frac{b_n^2}{2} = \mathcal{O^*}(\frac{c_n^2}{n^2})$$ $$\frac{2(n+1)}{n^2} \sim \frac{2}{n}$$ $$(n+1)\left(-\frac{2}{n^2}b_n+\frac{1}{2n^2}b_n^2 \right) = \mathcal{o}(\frac{1}{n})$$
Hence, $$(3) \iff c_{n+1} -c_n \sim \frac{2}{n} + \mathcal{O^*}(\frac{c_n^2}{n^2})+ \mathcal{o}(\frac{1}{n}) \tag{4}$$
Here, if we can prove that $$b_n^2 = \frac{c_n^2}{n^2} = \mathcal{o}(\frac{1}{n}) \tag{5}$$ , then $$(4) \iff c_{n+1} -c_n \sim \frac{2}{n} + \mathcal{o}(\frac{1}{n})$$ or $$c_n = 2\ln(n)$$
And $$b_n = 2\frac{\ln(n)}{n}$$
Note: I haven't proven yet (5). According to the topic owner, Ji Chen proposed 2 inequalities on AoPS $$1- \frac{2}{n}+ \frac{2}{n^{2}}\ln\frac{n}{3}+ \frac{417}{128n^{2}}\leq a_{n}\leq 1- \frac{2}{n}+ \frac{5\ln n+ 3}{n^{2}} \tag{6}$$ I guess that directly proving (5) is more difficult than proving the 2 inequalities (6) of Ji Chen.
But if we succeed to prove these 2 inequalities (6), then we can conclude (5) holds true also (because $2\frac{\ln(n)}{n} +\frac{p}{n^2} \le b_n \le 5\frac{\ln(n)}{n} +\frac{q}{n^2} \implies b_n^2 = \mathcal{o}(\frac{1}{n}) $) and consequently we have the 3rd order expansion of $a_n$ as follows $$a_n = 1-\frac{2}{n}+2\frac{\ln(n)}{n^2} $$ And with this expansion, we can deduce a stronger inequality than the second one of (6).