Original problem. Find the $\mathcal{o}\left(n\right)$ so that $a_n\sim\frac{\mathcal{o}\left(n\right)}{n}+\mathcal{O}\left(\frac{1}{n}\right){\rm as}\lim a_n=\infty$
Given a recursion $a_{n+ 1}= \dfrac{1}{2}\left ( a_{n}+ \dfrac{1}{a_{n}} \right )$ with $a_{1}> 0.$ First, we have $\lim a_{n}= 1,$ let $a_{n}= b_{n}+ 1,$ therefore $$\ln\frac{1}{b_{n+ 1}}- \ln\frac{1}{b_{n}}= \ln b_{n}- \ln\left ( \frac{1}{2}\left ( b_{n}+ 1+ \frac{1}{b_{n}+ 1} \right )- 1 \right )\sim\ln\frac{2}{b_{n}}+ b_{n}+ \mathcal{O}\left ( b_{n}^{2} \right )\,{\rm as}\,n\rightarrow\infty$$ New problem. Find the $o\left ( 1 \right )$ so that $2b_{n+ 1}\sim b_{n}^{2}+ o\left ( 1 \right ).$ I still can't find such a way to approach this. To have the above formula I used the Moubinool Omarjee's technique for differential equation ${y}'= -y,$ if I chose ${y}'= -y+ \frac{y^{2}}{2}\Rightarrow x= \ln\left ( \frac{2}{y}- 1 \right ),$ that's not simple, so I need to a new way of thinking which is much more effective. Thank you so much.
This is based on the answer to the previous question. Let $$ c: = \frac{{a_1 - 1}}{{a_1 + 1}},\quad \left| c \right| < 1. $$ Then $$ b_n = a_n - 1 = \frac{{1 - c^{2^{n - 1} } }}{{1 + c^{2^{n - 1} } }} - 1 = - 2\frac{{c^{2^{n - 1} } }}{{1 + c^{2^{n - 1} } }}. $$ Thus, $$ 2b_{n + 1} - b_n^2 = - 4\frac{{c^{2^n } }}{{1 + c^{2^n } }} - 4\frac{{c^{2^n } }}{{1 + 2c^{2^{n - 1} } + c^{2^n } }} = - 4c^{2^n } \left( {\frac{1}{{1 + c^{2^n } }} + \frac{1}{{1 + 2c^{2^{n - 1} } + c^{2^n } }}} \right). $$ By the geometric series, $$ \frac{1}{{1 + c^{2^n } }} + \frac{1}{{1 + 2c^{2^{n - 1} } + c^{2^n } }} = 2 + \mathcal{O}(c^{2^{n - 1} } ) $$ Therefore $$ 2b_{n + 1} - b_n^2 \sim - 8c^{2^n } $$ as $n\to +\infty$. Note that by the Taylor series $$ \frac{{1 - x}}{{1 + x}} = 1 + 2\sum\limits_{k = 1}^\infty {( - 1)^k x^k } ,\quad \left| x \right| < 1, $$ it holds that $$ a_n = 1 + 2\sum\limits_{k = 1}^\infty {( - 1)^k c^{k2^{n - 1} } } . $$