I came up with the following question which is the follow up of How to prove that for $a_{n+1}=\frac{a_n}{n} + \frac{n}{a_n}$ , we have $\lfloor a_n^2 \rfloor = n$?
Problem: Let $a_1 = 1,\quad a_{n+1} = \frac{a_n}{n} + \frac{n}{a_n},\quad n\ge 1$.
- Prove that $\lim_{n\to \infty} (a_n^2 - n) = \frac{1}{2}$;
- Give the asymptotic analysis of $a_n^2 - n - \frac{1}{2}$.
Edit (2021/02/16) I also posted in https://mathoverflow.net/questions/384047/asymptotic-analysis-of-x-n1-fracx-nn2-fracn2x-n-2
For 1), I use the mathematical induction to prove the claim $$n + \frac{1}{2} - \frac{2}{n} < a_n^2 < n + \frac{1}{2} + \frac{13}{4n} + \frac{13}{8n^2} + \frac{157}{16n^3}, \quad n \ge 22. \tag{1}$$ However, we need to verify it for $n = 22$ (a computer is required). Are there simpler solutions?
$\color{blue}{\textbf{Edit}}$ 2021/02/15:
For 1), there is a solution in [1] (I know it from @haidangel's post The variation of a Ukrainian Olympiad problem: 10982). The authors proved that
$\frac{n^2}{n-1/2} \le a_n^2 \le \frac{(n-1/2)^2}{n-3/2}$ for all $n\ge 3$.
[1] Yuming Chen, Olaf Krafft and Martin Schaefer, “Variation of a Ukrainian Olympiad Problem: 10982”, The American Mathematical Monthly, Vol. 111, No. 7 (Aug. - Sep., 2004), pp. 631-632
For 2), I have no idea currently. I want to find something like: for example, for the recurrence relation $b_0 = 1, b_{n+1} = b_n + \frac{1}{b_n}, n\ge 0$, we have $b_n \sim \sqrt{2n} + \frac{\sqrt{2}}{8\sqrt{n}}\ln n + o(\frac{\ln n}{\sqrt{n}})$. (Thank @Diger for pointing out the mistake. See the comment.)
About how to construct the claim (1): I want to find $d_n, c_n$ such that, for sufficiently large $n$, $$n + \frac{1}{2} - d_n < a_n^2 < n + \frac{1}{2} + c_n.$$ To use the the mathematical induction, we need $$a_{n+1}^2 = \frac{a_n^2}{n^2} + \frac{n^2}{a_n^2} + 2 < \frac{n + \frac{1}{2} + c_n}{n^2} + \frac{n^2}{n + \frac{1}{2} - d_n} + 2 < n + 1 + \frac{1}{2} + c_{n+1},$$ $$a_{n+1}^2 = \frac{a_n^2}{n^2} + \frac{n^2}{a_n^2} + 2 > \frac{n + \frac{1}{2} - d_n}{n^2} + \frac{n^2}{n + \frac{1}{2} + c_n} + 2 > n + 1 + \frac{1}{2} - d_{n+1}$$ which results in $$c_{n+1} - \frac{c_n}{n^2} > \frac{n + \frac{1}{2}}{n^2} + \frac{n^2}{n + \frac{1}{2} - d_n} + \frac{1}{2} - n,$$ $$c_n < \frac{n^2}{n - \frac{1}{2} - d_{n+1} - \frac{n + \frac{1}{2} - d_n}{n^2}} - n - \frac{1}{2}.$$ We first choose $d_n$, then determine $c_n$. For example, $d_n = \frac{2}{n}$ and $c_n = \frac{13}{4n} + \frac{13}{8n^2} + \frac{157}{16n^3}$.