Asymptotic analysis of $x_{n+1} = \frac{x_n}{n^2} + \frac{n^2}{x_n} + 2$

Consider the substitutions \begin{equation*} x_n=n+1/2+y_n/n,\quad y_n=u_n+5/8. \end{equation*} Then $u_1=-9/8$ and \begin{equation*} u_{n+1}=f_n(u_n) \end{equation*} for $n\ge1$, where \begin{equation*} f_n(u):=\frac{-64 n^4 u-8 n^3 (4 u-13)+n^2 (56 u+115)+n (96 u+76)+4 (8 u+5)}{8 n^2 \left(8 n^2+4 n+8 u+5\right)}. \end{equation*} Define $c_n(u)$ by the identity \begin{equation*} f_n(u)=-u+\frac{13}{8n}+\frac{c_n(u)}{n^2}, \end{equation*} so that \begin{equation*} c_n(u)=\frac{n^2 \left(64 u^2+96 u+63\right)+n (11-8 u)+4 (8 u+5)}{8 \left(8 n^2+4 n+8 u+5\right)}. \end{equation*} Then for $n\ge1$ \begin{equation*} u_{n+1}+u_n=\frac{13}{8n}+\frac{c_n(u)}{n^2} \tag{1} \end{equation*} and for $n\ge2$ \begin{equation*} u_{n+1}=f_n(f_{n-1}(u_{n-1}))=u_{n-1}-\frac{13}{8n(n-1)}+\frac{c_n(u_n)}{n^2} -\frac{c_{n-1}(u_{n-1})}{(n-1)^2}. \tag{2} \end{equation*}

Note that \begin{equation*} u_{101}=-0.54\ldots,\quad u_{102}=0.56\ldots, \tag{3} \end{equation*} and \begin{equation*} 0\le c_n(u)\le3 \end{equation*} if $n\ge10$ and $u\in[-6/10,8/10]$. Therefore and because for natural $m\ge102$ we have \begin{equation*} \sum_{n=m}^\infty\Big(\frac{13}{8n(n-1)}+\frac3{(n-1)^2}\Big)<\frac5{m-2}\le0.05, \end{equation*} it follows from (2) and (3) by induction that for all $n\ge101$ we have $u_n\in[-6/10,8/10]$ and hence $0\le c_n(u_n)\le3$. So, again by (2), the sequences $(u_{2m})$ and $(u_{2m+1})$ are Cauchy-convergent and hence convergent. Moreover, by (1), $u_{n+1}+u_n\to0$.

Thus, indeed \begin{equation*} y_{n+1}+y_n\to5/4, \end{equation*} and the sequences $(y_{2m})$ and $(y_{2m+1})$ are convergent. (The limits of these two sequences can in principle be found numerically with any degree of accuracy -- controlled by (2), say.)

Something about my ${\color{blue}{\textbf{GUESS}}}$ for $q_1 = \frac{61}{112}, q_2 = \frac{79}{112}$:

Using the approach of @Iosif Pinelis, we consider the substitutions: for $n = 1, 2, \cdots$ \begin{align} x_{2n} &= 2n + \frac{1}{2} + \frac{\frac{61}{112} + u_n}{2n},\\ x_{2n+1} &= (2n+1) + \frac{1}{2} + \frac{\frac{79}{112} + v_n}{2n+1}. \end{align} We have $u_n + v_n \to 0$ as $n \to \infty$.

According to $x_{2n+1} = \frac{x_{2n}}{(2n)^2} + \frac{(2n)^2}{x_{2n}} + 2$ and $x_{2n+2} = \frac{x_{2n+1}}{(2n+1)^2} + \frac{(2n+1)^2}{x_{2n+1}} + 2$, we have \begin{align} v_n &= f(n, u_n), \\ u_{n+1} &= g(n, v_n) \end{align} where \begin{align} f(n, u) &= \frac{f_1(n,u)}{401408n^5 + 100352n^4 + (100352u+54656)n^3}, \\ f_1(n,u) &= -401408 n^5 u+(-100352 u+326144) n^4+(180096 u+223528) n^3 \\ &\quad +(150528 u+94528) n^2+(25088 u^2+52416 u+21106) n \\ &\quad +12544 u^2+13664 u+3721,\\ g(n, v) &= \frac{g_1(n,v)}{g_2(n,v)},\\ g_1(n, v) &= -401408 n^5 v+(-1103872 v+326144) n^4+(-1008000 v+916136) n^3 \\ &\quad +(-207424 v+1003708) n^2+(25088 v^2+182560 v+509856) n \\ &\quad +25088 v^2+91280 v+104375, \\ g_2(n, v) &= 401408 n^5+1103872 n^4+(100352 v+1275008) n^3\\ &\quad +(150528 v+758464) n^2 +(75264 v+228704) n+12544 v+27664. \end{align} We have $u_1 = 275/112$ and $$u_{n+1} = g(n, f(n, u_n)), \ n\ge 1.$$ I use Maple to get $u_{101} \approx 0.00342$, $u_{1001} \approx 0.000676$, $u_{10001} \approx 0.0004$.

Is it $\lim_{n\to \infty} u_n = 0$?