Choose a valid substitution for a square-root estimate

Use $(1-x)^{1/2}=1-x/2-x^2/8+\cdots$ and $(1+2x)^{-1}=1-2x+4x^2+\cdots$. Thus

$\begin{aligned}&\frac{\sqrt{1-x}}{1+2x}\\ & =1-\frac52x+\frac{39}{8}x^2+O(x^3).\end{aligned}$

The two source binomial/geometric conditions are $|x|<1$ and $|2x|<1$, so $-1/2<x<1/2$. The proposed $x=-1$ is outside this interval and cannot justify the estimate.

At $x=1/9$, the function is $\sqrt{8/9}/(11/9)=6\sqrt2/11$, while the quadratic is $1-5/18+39/648=169/216$. Hence $6\sqrt2/11\approx169/216$, giving $\sqrt2\approx1859/1296$ and $p=1859$. This is the printed low-order estimate, not an exact identity.