I've been attempting to prove that the ratio of differences in the Newton-Raphson method which produces a a sequence of second order convergence, is equal to the ratio of Absolute Errors. A way of mathematically putting this could be as follows: $$\frac{|AbsoluteError (x_{n+1})|}{(AbsoluteError(x_{n}))^2}\thickapprox k$$ as does, $$\frac{|x_{n+2}-x_{n+1}|}{(x_{n+1}-x_{n})^2}\thickapprox k$$ where $k$ is a constant.
Proving this result for methods which produce a first order sequence where the denominators of the fractions are not raised to the power $2$, proves to be quite easy. The method I used to do this is as follows:
Let the Absolute Error in the approximation of the root at $x_n$ be equal to $\mathcal{E}$. This could be written as $|AbsoluteError(x_n)| = \mathcal{E}$
Since we know that $\frac{|AbsoluteError (x_{n+1})|}{(AbsoluteError(x_{n}))}\thickapprox k$, we can gather that $|AbsoluteError (x_{n+1})| = k\mathcal{E}$, and thus $|AbsoluteError (x_{n+2})| = k^2\mathcal{E}$ and so forth.
Using these results, we find that $|x_{n+1}-x_{n}| = \mathcal{E}-k\mathcal{E}$
and, $|x_{n+2}-x_{n+1}| = k\mathcal{E}-k^2\mathcal{E}$
Combining these gives us the following result: $$\frac{|x_{n+2}-x_{n+1}|}{|x_{n+1}-x_{n}|}=k$$
The same result can be obtained from the Ratios of Differences in further terms, i.e. $\frac{x_{n+3}-x_{n+2}}{x_{n+2}-x_{n+1}} = k$
I am attempting to use this same method for proving the resut for second order sequences, and in doing so have derived the current relationships between terms:
I believe these relationships are true since when substituting them into the following expression I get the following results for all consecutive Errors in $x$: $$\frac{|AbsoluteError(x_{n+1})|}{(AbsoluteError(x_n))^2} = \frac{k}{a}$$ which is a constant value.
However, when embarking on the same method I used for the first order converging sequences, I do not find consistency between $\frac{|x_{n+2}-x_{n+1}|}{(x_{n+1}-x_{n})^2}$ and $\frac{|x_{n+3}-x_{n+2}|}{(x_{n+2}-x_{n+1})^2}$
Is there error in my methodology of tackling this problem, and if so could you provide an alternate proof to the original result I mentioned (Ratio of Differences = Ratio of Errors)?