Let $E(p):=\overbrace{(n-2)(\sqrt{2}-2)}^Ap^2+\overbrace{(3n-2)}^Bp\overbrace{-n}^C$, $n\geq 3$ Since $A:=(n-2)(\sqrt{2}-2)$ is negative, the quadratic function $E$ opens downwards, thus when $p<p^*$ where $p^*=\frac{-(3n-2)+\sqrt{(3n-2)^2+4(\sqrt{2}-2)n(n-2)}}{2(n-2)(\sqrt{2}-2)}$ is the left-hand zero of $E(p)=0$
First let's check the asymptotic of $p^*$, it actually equals to $Z:=\frac{-3+\sqrt{1+4\sqrt{2}}}{2(\sqrt{2}-2)}$ asymptotically, $n\rightarrow\infty$.
And the convergence speed can be obtained by looking at $p^*-Z$,
$$p^*-Z\sim \frac{10}{-1+8\sqrt{2}}\cdot\frac{1}{n}$$
Thus $$\begin{aligned} p^*&\sim Z+\frac{10}{-1+8\sqrt{2}}\cdot\frac{1}{n}\\ &=\frac{-3+\sqrt{1+4\sqrt{2}}}{2(\sqrt{2}-2)}+\frac{10}{-1+8\sqrt{2}}\cdot\frac{1}{n} \end{aligned} $$
So, I expect that if $p$ is something decaying faster than $p^*$ then
(1) $$E(p)<0$$ and
(2) for $$p=\{\text{a constant}\} + g(n)$$ where $g(n)$ is a decreasing function that $$\lim _{n\rightarrow\infty}g(n)=0.$$ according to the analysis here. (If I didn't do it wrong)
So let $$p=\frac{-3+\sqrt{1+4\sqrt{2}}}{2(\sqrt{2}-2)}+\frac{10}{-1+8\sqrt{2}}\cdot\frac{1}{n^\alpha}$$ where $1\ll\alpha$.
But I put this $p$ into mathematica, it doesn't look like so:
However I add a little bit value on p, it indeed asymptotically equals to $n^2$.
Weird, why is this? Should I instead write
$$p=(1+\epsilon)\frac{-3+\sqrt{1+4\sqrt{2}}}{2(\sqrt{2}-2)}+\frac{10}{-1+8\sqrt{2}}\cdot\frac{1}{n^\alpha}, 1\ll\alpha?$$
I feel this is a basic thing in asymptotical analysis but I don't have experience on it.. Thus I really hope someone could give me a hint.
I also put my question in pdf form here.