Using the Taylor expansion it is easy to show that the convergence of Newton's method for a root $\alpha$ is quadratic when $f'(\alpha)\neq0$ and $f''(\alpha)\neq0$. If instead $f'(\alpha)=0$ and $f''(\alpha)\neq0$ we have linear convergence.
What is the case however if $f'(\alpha)\neq0$ and $f''(\alpha)=0$?
I have found using Taylor's Theorem that the limit of $\Big\lvert\frac{x_{k+1}-\alpha}{x_k-\alpha}\Big\rvert $ is $0$ using $f(x_k)=f'(\alpha)(x_k-\alpha)+\frac{1}{2}f''(\xi_k)(x_k-\alpha)^2$ which fails to conclude linear convergence. Surely we need more information about the value of $f'''(\alpha)$ to deduce more about the convergence of the method? Is it possible that the method does not converge in this case?
Any hints would be great!
Regardless of any assumptions, the first step of the local analysis of Newton's method for $f$ at least $C^2$ goes like
$$x_{n+1}=x_n-\frac{f'(\alpha) r_n + f''(\alpha) r_n^2/2 + o(r_n^2)}{f'(\alpha)+f''(\alpha) r_n+o(r_n)}$$
since $f(\alpha)=0$, so this gives the two Taylor expansions. When $f'(\alpha) \neq 0$, regardless of anything else, you can then write
$$x_{n+1}=x_n - r_n \frac{1+\frac{f''(\alpha)}{2f'(\alpha)} r_n+o(r_n)}{1+\frac{f''(\alpha)}{f'(\alpha)} r_n + o(r_n)}.$$
Then you can further write
$$x_{n+1}=x_n-r_n \left ( 1+\frac{f''(\alpha)}{2f'(\alpha)} r_n + o(r_n) \right ) \left ( 1-\frac{f''(\alpha)}{f'(\alpha)} r_n + o(r_n) \right ) \\ = x_n - r_n \left ( 1-\frac{f''(\alpha)}{2f'(\alpha)} r_n + o(r_n) \right ) \\ = \alpha + \frac{f''(\alpha)}{2f'(\alpha)} r_n^2 + o(r_n^2).$$
You can still do this "further" computation if $f''(\alpha)=0$. It is just that then the result reads
$$x_{n+1}=\alpha + o(r_n^2)$$
which is not completely telling you the behavior of the error. Instead it is only telling you that the error behaves as $o(r_n^2)$ (i.e. faster than quadratic convergence).