Convergence acceleration of a recursively defined sequence $a_{n+1} = (1-a_n)^{\frac1p},\ p>1$ On this question of recursive sequence, I have proved its convergence. As the sequence oscillates around its limit, the convergence rate can be accelerated. Here is the definition of sequence (series) convergence acceleration. How would one accelerate the convergence rate of this sequence? Please provide proof and references.
 A: From the previous question, we know that
$$a_{2n}<a<a_{2n+1}$$
And so, for any $m>0$, the weighted average of consecutive terms is closer to $a$.
$$a_{2n}<\frac{a_{2n}+ma_{2n+1}}{1+m}<a_{2n+1}$$
So $\frac{a_{2n}+ma_{2n+1}}{1+m}$ is closer to $a$ than $a_{2n}$ and $a_{2n+1}$, yielding faster convergence. Specifically,
\begin{align}\small\lim_{n\to\infty}\frac{a-\frac{a_n+ma_{n+1}}{1+m}}{a-a_n}&=\frac1{1+m}+\frac m{1+m}\lim_{n\to\infty}\frac{a-(1-a_n)^{1/p}}{a-a_n}\\&=\frac1{1+m}-\frac{ma}{p(1+m)(1-a)}\end{align}
For convergence to be fastest, we're interested in
$$0=\frac1{1+m}-\frac{ma}{p(1+m)(1-a)}$$
Which has the solution
$$m=\frac{p(1-a)}a$$
However, in order for this to be useful, you'll need to know what $a$ is. Replacing it with $a_n$ is good enough,
$$a=\lim_{n\to\infty}\frac{a_n^2+p(1-a_n)^{(p+1)/p}}{a+p(1-a)}$$
which converges quadratically. (as fast as Newton's method)
The technique of weighting terms such that the resulting limit is zero can easily be extended to more generally things such as
$$f(x)=\frac{x+m_1(1-x)^{1/p}+m_2(1-(1-x)^{1/p})^{1/p}+\dots}{1+m_1+m_2+\dots}$$
where you'll want
$$f^{(k)}(a)=0$$
for as many $k$ as you are interested in. (It results in quadratic, cubic, quartic, etc. convergence depending on how many terms you use.)
