Given a recursion $a_{n+ 1}= \sqrt{1+ \dfrac{1}{1+ a_{n}}}$ with $a_{1}> 1.$ We have that $$a_{n}\sim\sqrt{1+ \frac{1}{1+ a_{n}}}+ \mathcal{O}\left ( \frac{1}{n} \right )\,{\rm as}\,n\rightarrow\infty$$ Source: It's an my own problem that I posted before and that received many solutions.
Call $\lim a_{n}= a,$ if I want to give an asymptotic analysis for $a_{n+ 1}- a_{n},$ that's no good since $a\not\in\mathbb{Q}.$ I think about Groebner Basis, is there way to find $\mathcal{O}_{\mathscr{L}\left ( a \right )}\left ( \dfrac{1}{n} \right )$ with $a$ is a root of the irreducible polynomial $\mathscr{L}\left ( a \right ),$ how can we do that ?
Note that $$ \left| {a_{n + 1} - a_n } \right| = \frac{{\left| {a_n - a_{n - 1} } \right|}}{{(1 + a_n )(1 + a_{n - 1} )\left( {\sqrt {1 + \frac{1}{{1 + a_n }}} + \sqrt {1 + \frac{1}{{1 + a_{n - 1} }}} } \right)}} \le \frac{{\left| {a_n - a_{n - 1} } \right|}}{8} $$ since $a_n>1$ for all $n\geq 1$. Thus, $$ \left| {a_{n + 1} - a_n } \right| \le \frac{{\left| {a_2 - a_1 } \right|}}{{8^{n - 1} }} \le \frac{{4\sqrt 6 + 8a_1 }}{{8^n }}, $$ i.e., $$ a_{n + 1} = a_n +\mathcal{O}\!\left( {\frac{1}{{8^n }}} \right). $$