Let $f(x)=x^p$ with $p>2$. The steps for the newton method are known as $x_{k+1}=x_k+d_k$ and $d_k = -\frac{\nabla f(x_k)}{\nabla^2 f(x_k)}$
I want to check if the Newton method converges superlinearly for this case.
It'd converges superlinearly (according to lecture) if $$\|x_{k+1}-\bar{x}\| \leq \epsilon_k \|x_k-\bar{x}\|$$ where $\{\epsilon_k\}_k \rightarrow 0$ and $\bar{x}$ is a stationary point.
Obviously, the stationary point of $f(x)=x^p$ is at $x=0$. So I thought I should prove... $$\frac{\|x_{k+1}\|}{\|x_{k}\|}\leq \epsilon_k \rightarrow 0 \Rightarrow \frac{\|x_{k+1}\|}{\|x_{k}\|}\rightarrow 0$$
This is not the case, because... $$\frac{\|x_{k+1}\|}{\|x_{k}\|} = \frac{\|x_{k}+d_k\|}{\|x_{k}\|} = \frac{\|x_{k}-\frac{\nabla f(x_k)}{\nabla^2 f(x_k)}\|}{\|x_{k}\|}=\frac{\|x_{k}-\frac{px_k^{p-1}}{p(p-1)x_k^{p-2}}\|}{\|x_{k}\|}=\frac{\|(1-\frac{1}{p-1})x_{k}\|}{\|x_{k}\|}=|(1-\frac{1}{p-1})| \nrightarrow 0$$
Is this correct?