# The limit and asymptotic analysis of $a_n^2 - n$ from $a_{n+1} = \frac{a_n}{n} + \frac{n}{a_n}$

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 $$a_1 = 1,\quad a_{n+1} = \frac{a_n}{n} + \frac{n}{a_n},\quad n\ge 1$$.

1. Prove that $$\lim_{n\to \infty} (a_n^2 - n) = \frac{1}{2}$$;
2. Give the asymptotic analysis of $$a_n^2 - n - \frac{1}{2}$$.

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?

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 $$b_n \sim \sqrt{2n} + \frac{\sqrt{2}}{4\sqrt{n}}\ln n + o(\frac{\ln n}{\sqrt{n}})$$.

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 $$c_n = \frac{13}{4n} + \frac{13}{8n^2} + \frac{157}{16n^3}$$.

• If $b_n = \sqrt n + 1/(4 \sqrt n)$, then it appears that $a_{2 n} - b_{2 n} \sim C_1 n^{-3/2}$ and $a_{2 n + 1} - b_{2 n + 1} \sim C_2 n^{-3/2}$, where $C_{1, 2}$ depend on $a_1$. – Maxim Aug 24 at 13:59
• @Maxim I did some numerical experiments. It looks nice. Thank you! – River Li Aug 24 at 14:13
• @Maxim Suppose $a_{2n} \sim f(n) = \sqrt{2n} + \frac{1}{4\sqrt{2n}} + \frac{c_1}{(2n)^{3/2}}$ and $a_{2n+1} \sim g(n) = \sqrt{2n+1} + \frac{1}{4\sqrt{2n+1}} + \frac{c_2}{(2n+1)^{3/2}}$. I considered $\frac{f(n)}{2n} + \frac{2n}{f(n)} - g(n) = -(c_1 + c_2 - \frac{9}{16})(2n)^{-3/2} + o(n^{-3/2})$ (Maple asympt command) and got $c_1 + c_2 = \frac{9}{16}$. I also did some numerical experiments which resembles this result. – River Li Aug 26 at 8:44