Denoting $y_n = x_n-\frac{(x_{n+1}-x_n)^2}{x_{n+2}-2x_{n+1} +x_n}$ when $x_n\rightarrow_{n\rightarrow\infty} a $ with $p$ convergence order.
I showed that $\lim_{n\rightarrow\infty} |\frac {y_n-a}{x_n-a}| =0$
I wish to show that the method does not improve the rate of convergence asymptotically, i.e $|\frac{y_{n+1}-a}{(y_n-a)^p}| = c_1$ when $|\frac{x_{n+1}-a}{(x_n-a)^p}| = c_2$ .
I first tried to make it work with $p=1$.
I denote $e_n = x_n-x \Rightarrow x_n=e_n +x$.
$x_{n+1} - a-\frac{(x_{n+2}-x_{n+1})^2}{x_{n+3}-2x_{n+2} +x_{n+1}} = $ $e_{n+1}-\frac{(e_{n+2}-e_{n+1})^2}{e_{n+3}-2e_{n+2}+e_{n+1}} = \frac{e_{n+1}e_{n+3} - e_{n+2}^2}{e_{n+3}-2e_{n+2}+e_{n+1}}$
and
$x_n- a -\frac{(x_{n+1}-x_n)^2}{x_{n+2}-2x_{n+1} +x_n} = $ $\frac{e_{n}e_{n+2} - e_{n+1}^2}{e_{n+2}-2e_{n+1}+e_{n}}$
thus
Thus $\frac{y_{n+1}-a}{y_n-a} = \frac{e_{n+1}e_{n+3} - e_{n+2}^2}{e_{n+3}-2e_{n+2}+e_{n+1}} \cdot \frac{e_{n+2}-2e_{n+1}+e_{n}}{e_{n}e_{n+2} - e_{n+1}^2} = \frac{e_{n+2}-2e_{n+1}+e_{n}}{e_{n+3}-2e_{n+2}+e_{n+1}} \cdot \frac{e_{n+1}e_{n+3} - e_{n+2}^2}{e_{n}e_{n+2} - e_{n+1}^2} $
now using $\frac{e_{n+1}}{e_n}\rightarrow c$ then $\frac{e_{n+k}}{e_{n}} \rightarrow c^k$ lets try to divide numerator and denominator of the left expression by $e^n$ and the right expression by $e_{n+1}^2$ and take the limit:
$\frac{e_{n+2}-2e_{n+1}+e_{n}}{e_{n+3}-2e_{n+2}+e_{n+1}} \cdot \frac{e_{n+1}e_{n+3} - e_{n+2}^2}{e_{n}e_{n+2} - e_{n+1}^2} =_{n\rightarrow \infty} \frac{c^2-2c+1}{c^3-2c^2+c} \cdot \frac{c^2-c^2}{\frac{1}{c}\cdot c-1} $
the problem is that no matter how I try to manipulate the right expression (for example deviding by $e_{n+2}^2$, it results in $\frac{1}{c} \cdot \frac{0}{0}$ when taking the limits.
Other approaches (especially ones that don't demand so much writing) are more the welcome!