Numerical Analysis: $x_{n}\rightarrow x_{*}$ with order $\alpha$ implies limit converges. It is given that $x_{n}\rightarrow x_{*}$ with order $\alpha >1$, and we need to prove: $$\lim_{n\rightarrow \infty }\frac{|x_{n+1}-x_{n}|}{|x_{n}-x_{n-1}|^{\alpha}}<\infty $$
So by the given condition, we have $\lim_{n \rightarrow \infty}\frac{|x_{*}-x_{n+1}|}{|x_{*}-x_{n}|^{\alpha}}<\infty$, but I am not sure how to proceed from here, after trying to use triangle inequality and didn't work out.
 A: Let $p$ denote the order and let $\alpha$ denote the limit. Let $e_n$ denote the error after the $n$th iteration, i.e.,
$$ e_n = \alpha - x_n.$$
and let $f_n$ denote the difference between successive approximations
$$ f_n = x_{n-1} - x_n,$$
By assumption, there exists $c > 0$ such that
$$ \frac{|e_{n+1}|}{|e_n|^p}\rightarrow c, \quad n \rightarrow \infty.$$
Our task is to investigate the behavior of
$$ \frac{|f_{n+1}|}{|f_n|^p},$$
as $n$ tends to infinity.
We have
$$ f_n = \alpha - x_n - (\alpha - x_{n-1}) = e_n - e_{n-1} = e_{n-1} \left( \frac{e_n}{e_{n-1}} - 1 \right).$$
We observe that
$$ \frac{e_n}{e_{n-1}} = \left(\frac{e_n}{(e_{n-1})^p}\right) (e_{n-1})^{p-1} \rightarrow 0, \quad n \rightarrow \infty,$$
because $p > 1$.
It follows that
$$ \frac{x_{n+1}-x_n}{(x_n-x_{n-1})^p} = \frac{f_{n+1}}{f_n^p} = \frac{e_n(e_{n+1}/e_n - 1)}{e_{n-1}^p(e_n/e_{n-1} - 1)^p}$$
In particular, we have
$$ \frac{|x_{n+1}-x_n|}{|x_n - x_{n-1}|^p} = \frac{|e_n|}{|e_{n-1}|^p}\left(\frac{|1-e_{n+1}/e_n|}{|1 - e_n/e_{n-1}|^p} \right) \rightarrow c, \quad n \rightarrow \infty. $$

New expressions can be investigated by rewriting them to expose expressions whose behavior is known. A typical example is the proof of the product rule of differentiation. Finding the right connection can be difficult and the experience gained by solving a large number of small problems and studing diffent proofs is an asset. Here the error $e_n$ was the only expression we had any real knowledge about. It did not appear explicitly in the expression for $f_n = x_{n-1} - x_n$, but we exposed it by writing $$f_n = e_n - e_{n-1}.$$
