I'm given a sequence $x_n \rightarrow \alpha$ of a nonlinear solver such that $$\lim_{n\rightarrow\infty}\frac{x_{n+1}-\alpha}{x_n-\alpha}=c$$ converges linearly (i.e. $c\in(0,1)$).
Now, I need to determine the asymptotic rate of convergence $p^*$ and constant $c^*$ of new iterations, determined by the sequence
$$x_n^*=\frac{x_n+x_{n-1}}{2}$$
I've attempted to solve this by using the definition of linear convergence to show that as
$$\lim_{n\rightarrow\infty}\frac{\varepsilon_{n+1}}{\varepsilon_{n}} =c$$
for $\varepsilon_i \geq |x_i - \alpha|$, the asymptotic error constant of $x_n^*$
$$\lim_{n\rightarrow\infty}\frac{x_{n+1}^*-\alpha}{(x_n^* -\alpha)^{p^*}}=c^*$$
is bounded by
$$\lim_{n\rightarrow\infty}\frac{\frac{\varepsilon_{n+1}+ \varepsilon_n}{2}}{(\frac{\varepsilon_{n}+ \varepsilon_{n-1}}{2})^{p^*}}$$
which I found by substituting the values of $x_n^*$ using the variable definition.
The first problem is that I'm not sure if it's correct to simplify further by substituting the initial constant of convergence $c$ to get a result in terms of $\varepsilon_n$. If I do, I would get something like this:
$$\lim_{n\rightarrow\infty}\frac{(c+1)/2}{(\frac{c+1}{2c})^{p^*}\varepsilon_n^{p^* -1}}$$
The second problem is that I'm not sure this answers the question, which I presume is to determine $p^*$ and $c^*$ in terms of the original solver (i.e. they should not include one another).