Newton's method states that, if there is a root $r$ of $f(x)$ that we want to calculate and $r_0$ is an approximation, then a better approximation is $$r_1=r_0-\frac{f(r_0)}{f'(r_0)}$$We could substitute this into the formula again to get $$r_2=r_1-\frac{f(r_1)}{f'(r_1)}=r_0-\frac{f(r_0)}{f'(r_0)}-\frac{f\left(r_0-\frac{f(r_0)}{f'(r_0)}\right)}{f'\left(r_0-\frac{f(r_0)}{f'(r_0)}\right)}$$We could keep substituting like so. How can we turn this into an infinite series? I can't find an explicit formula for $a_k$ such that $$r=\sum_{k=0}^\infty a_k$$Because the $a_n$ depends on all other terms before it.
Edit: The recursive relation for $a_n$ is $a_n=\frac{f\left(\sum_{k=1}^na_{k-1}\right)}{f'\left(\sum_{k=1}^na_{k-1}\right)}$ where $a_0=r_0$, but can we find a closed form for $a_n$?