Let $x_0>0$, $a>0$, $b>0$ be given and define $$x_{n+1}:= a+b x_{n}^{1/4}$$.
Question: What is this speed of convergence of $x_n$ to the unique solution $x>0$ of $x=a+bx^{1/4}$?
Lacking contractivity of $x\mapsto a+b x^{1/4}$ I don't see how to apply the usual fixed point iteration theorems.
It's going to be asymptotically linear with coefficient $f'(r)=\frac{b}{4} r^{-3/4}$ where $r$ is the solution. One can note that the positive solution is bigger than $b^{4/3}$ (which is the positive solution to the equation when $a=0$), so the coefficient is less than $\frac{1}{4}$ irrespective of $a$ or $b$. The coefficient is also less than $\frac{b a^{-3/4}}{4}$ (what you get by ignoring the $b$ term and then plugging the resulting solution into the derivative). This is a better estimate if $b<a^{3/4}$.