Given a recursion $\frac{a_{n+1}}{4a_{n+1}+n+1}=\frac{2(a_n+n)^2}{9n(4a_n+n)}$ with $a_1=1.$ Show that $a_n\sim\frac n2\,{\rm as}\,n\rightarrow\infty$ 
Given a recursion $\dfrac{a_{n+ 1}}{4a_{n+ 1}+ n+ 1}= \dfrac{2\left ( a_{n}+ n \right )^{2}}{9n\left ( 4a_{n}+ n \right )}$ with $a_{1}= 1.$ Prove that
$$a_{n}\sim\frac{n}{2}\,{\rm as}\,n\rightarrow\infty$$

I tried to solve the nonlinear asymptotic equation I posted before again and I made a few mistakes.. so I fixed them. I think the difficulty level is vicious. On the other hand_ Estimations of some new recurrence sequences, why the first expansions between both of them are similar ? I can't explain it, I need to the help, thank you so much.
 A: By regrouping $a_{n+1}$ terms we get:
$$a_{n+1}=-\frac{2(n+1)(a_n+n)^2}{-20\,n\,a_n-n^2+8{a_n}^2}$$
Since we want $a_n\sim \dfrac n2$ let set $a_n=\dfrac n2(1+u_n)$ and hope to transform the problem into proving $u_n\to 0$
After substitution we get $$u_{n+1}=\frac{3{u_n}^2}{9+6u_n-2{u_n}^2}$$
Let set $f(x)=\frac{3x^2}{9+6x-2x^2}$, then, $f(x)=x$ has three fixed points $x=-\frac 32,0,3$.
https://www.desmos.com/calculator/cskpxqtrmp
And effectively from the seed $u_1=\frac {2a_1}1-1=2-1=1$, it appears to actually converge to $0$ while decreasing.
You can show easily that the quadratic $D(x)=(9+6x-2x^2)>0$ on the interval $[0,1]$
$f'(x)=\dfrac{18x(x+3)}{D(x)^2}\ge 0$ when $x\ge 0$ so $f\nearrow$ on $\mathbb R^+$
In particular $f([0,1])\subset[f(0),f(1)]=[0,\frac 3{13}]\subset[0,1]$
It means that since $u_1=1\in[0,1]$ then by induction $\forall n,\ u_n\in[0,1]$
Let now study $f(x)-x=\dfrac{x(2x+3)(x-3)}{D(x)}<0$ for $x\in[0,1]$ therefore $u_{n+1}\le  u_n$
We can conclude that $u_n\searrow$ in compact $[0,1]$ so it has a limit, and since the only fixed point of $f$ in this interval is $0$ then $u_n\to 0$.
