Convergence of Newton Iteration For $a>0$, I want to compute $\frac{1}{a}$ using Newton's iteration by finding a zero of $f(x)=a-\frac{1}{x}$. Newton's iteration formula reads $$x_{k+1}=x_k-\frac{f(x_k)}{f'(x_k)}=2x_k-ax_k^2$$
By the Banach Fixed Point Theorem, I can conclude that this Netwon iteration converges for starting values in the interval $I=\left(\frac{1}{2a}, \frac{3}{2a}\right)$.
Now I would like to show that we also have convergence for starting values in $\left(0,\frac{1}{a}\right]$.
To this end, it would be enough to show that at some point of the iteration, we land in the interval $I$, right? Is that the right approach? How can we show that?

The Banach fixed point theorem isn't always the best for proving convergence. Your function is convex for $x>0$ and concave for $x<0$. You might be able to use this. Try doing this exercise (Spivak's calculus book) for a better idea of what I am suggesting.
