<s>[INST] The variation of a Ukrainian Olympiad problem: 10982 
Given a recursion $a_{n+ 1}= \dfrac{a_{n}}{n}+ \dfrac{n}{a_{n}}$ with $a_{1}= 1.$ Prove that
$$\lim a_{n}^{2}- n= \frac{1}{2}$$
Source: StachMath/@RiverLi _ The limit and asymptotic analysis of $a_n^2 - n$ from $a_{n+1} = \frac{a_n}{n} + \frac{n}{a_n}$

The original problem already has an answer, I'll suggest this way of thinking, which is not mine, but @twelve_sakuya
Let $b_{n}:=a_{n}^{2}- n,$ so
$$a_{n+ 1}= \frac{a_{n}}{n}+ \frac{n}{a_{n}}\Leftrightarrow b_{n+ 1}- \frac{1}{2}= -\frac{n}{b_{n}+ n}\left ( b_{n}- \frac{1}{2} \right )+ \frac{b_{n}}{2\left ( b_{n}+ n \right )}+ \frac{b_{n}}{n^{2}}+ \frac{1}{n}$$
That means $\left | b_{n+ 1}- \dfrac{1}{2} \right |\leq\dfrac{n}{\left | b_{n}+ n \right |}\left | b_{n}- \dfrac{1}{2} \right |+ \dfrac{\left | b_{n} \right |}{2\left | b_{n}+ n \right |}+ \dfrac{\left | b_{n} \right |}{n^{2}}+ \dfrac{1}{n}.$ Therefore, if we can get the evaluations of $\left | b_{n} \right |$ or $\dfrac{\left | b_{n} \right |}{n},$ maybe there exists a number $\beta\in\left ( 0, 1 \right )$ so that
$$\left | b_{n}- \frac{1}{2} \right |\leq\beta^{n}B\left ( n \right )\rightarrow 0\,{\rm as}\,n\rightarrow\infty$$
My friend has no confidence to continue, he also said that is the variation of a Ukrainian Olympiad problem #10982 (I searched and got the result, but I couldn't access it). I need to the help, thanks a real lot !
 [/INST] Note that, for $n\ge 1$, $$ n^6+10n^5+38n^4+68n^3+57n^2+18n>n^6+10n^5+37n^4+62n^3+46n^2+12n+1$$  and, factorising,
$$n(n+1)^2(n+2)(n+3)^2>(n^3+5n^2+6n+1)^2.$$
Also note that the function $F_n$ such that $$F_n(x)=\frac{n}{x}+\frac{x}{n}$$ is monotonic decreasing for $0<x\le n$.
Now suppose that $$\frac {n+2}{\sqrt {n+1}}\ge a_{n+2}\ge \frac{n+1}{\sqrt n} \tag 1$$ which is true for $n=1$. Then $$F_{n+2}\left(\frac {n+1}{\sqrt {n}}\right)\ge F_{n+2}\left(a_{n+2}\right)\ge  F_{n+2}\left(\frac{n+2}{\sqrt {n+1}}\right)$$
$$\frac {n^3+5n^2+6n+1}{\sqrt {n}(n+1)(n+2)}\ge a_{n+3}\ge \frac{n+2}{\sqrt {n+1}} $$
$$\frac {n+3}{\sqrt {n+2}}\ge a_{n+3}\ge \frac{n+2}{\sqrt {n+1}} $$ Therefore, by induction, Equation (1)  is true for all positive integers $n$.
Let $a_n^2=n+x_n$ for all $n$. Then, for $n\ge 1$, $$\frac{(n+2)^2}{n+1}\ge {a_{n+2}}^2\ge\frac{(n+1)^2}{n} $$ $$\frac{n+2}{n+1}\ge x_{n+2}\ge \frac{1}{n}.$$
We can now use this bound on $x_n$. $${a_{n+1}}^2=\frac{{a_n^2}}{n^2}+\frac{n^2}{{a_n}^2}+2$$
$$n+1+x_{n+1}=\frac{n+x_n}{n^2}+\frac{n^2}{n+x_n}+2$$
$$x_{n+1}=1+\frac{1}{n}+n\left(1+\frac{x_n}{n}\right)^{-1}-n+O\left(\frac{1}{n^2}\right)$$
$$x_{n+1}=1+\frac{1}{n}-{x_n}+\frac{{x_n}^2}{n}+O\left(\frac{1}{n^2}\right)$$
Then $x_n+x_{n+1}=1+O\left(\frac{1}{n}\right)$ and so $x_n+x_{n+1}$ tends to $1$. We also have
$$x_{n+2}=1+\frac{1}{n}-{x_{n+1}}+\frac{{x_{n+1}}^2}{n}+O\left(\frac{1}{n^2}\right)$$ and therefore $$|x_{n+2}-x_{n+1}|=\frac{n-1}{n}|x_{n+1}-x_{n}|+O\left(\frac{1}{n^2}\right)$$
$$|x_{2n}-x_{2n-1}|=\frac{(n-1)n...(2n-3)}{n(n+1)...(2n-2)}|x_{n+1}-x_{n}|+O\left(\frac{1}{n}\right)=\frac{1}{2}|x_{n+1}-x_{n}|+O\left(\frac{1}{n}\right)$$
Then $x_{n+1}-x_n$ tends to $0$ and so $x_n$ tends to $\frac{1}{2}.$
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