Regarding obtaining a series expansion at infinity

I am trying to obtain an approximate expression for the behaviour of the following function for large $x$

$$F(x)=xfrac{ text{erf}(x)^2}{text{erf}(2 x)}$$

I know that the $lim_{xrightarrowinfty}frac{1}{x}F(x)=1$ and $lim_{xrightarrowinfty}F(x)-x=0$ so that basically one can say that for large $x$, $F(x)approx x$.

I would like to obtain this result from Series, but invoking Series[F, {x,∞,1}] I obtain

$$frac{e^{2 x^2} left(e^{2 x^2} left(x+Oleft(left(frac{1}{x}right)^2right)right)+e^{x^2} left(-frac{2}{sqrt{pi }}+Oleft(left(frac{1}{x}right)^1right)right)+left(frac{1}{pi x}+Oleft(left(frac{1}{x}right)^2right)right)right)}{left(-frac{1}{2 sqrt{pi } x}+Oleft(left(frac{1}{x}right)^2right)right)+e^{4 x^2}}$$

Why do I get a fraction?
How can I get an expansion at infinity which reflects the linear behaviour of the function plus higher order corrections?

AsymptoticSolve[y == x Erf[x]^2/Erf[2x], y, {x, Infinity, 3}]

ser10 = Series[F[1/y], {y, 0, 0}, Assumptions -> y > 0] // Normal // 
          FullSimplify

(*     (2 E^(2/y^2) (-E^((1/y^2)) Sqrt[Pi] + y)^2)/(
        2 E^(4/y^2) Pi y - Sqrt[Pi] y^2)     *)

ser11 = ExpandAll[ser10]

(Series[ser11, {y, 0, 0}] // Normal) /. y -> 1/x // FullSimplify

(*     -((2 E^-x^2)/Sqrt[Pi]) + x     *)