Given the equation $x^2-p=0, p>0$, one has to show that Newton's method will always converge for every initial value $x_0>0$.
I have found the sequence
$x_{k+1}=x_k-\frac{f(x_k)}{f'(x_k)}=x_k-\frac{x_k^2-p}{2x_k}=\frac{x_k^2+p}{2x_k}$.
The bit I'm struggling with now is to show that this sequence does indeed converge. How can I show this or is there another way to prove this task?
Here's a start. The algebra to make things like this work can sometimes be a little tricky.
$\begin{array}\\ x^2_{k+1}-p &=\dfrac{(x_k^2+p)^2}{4x^2_k}-p\\ &=\dfrac{x_k^4+2px_k^2+p^2-4x_k^2p}{4x^2_k}\\ &=\dfrac{x_k^4-2px_k^2+p^2}{4x^2_k}\\ &=\dfrac{(x_k^2-p)^2}{4x^2_k}\\ \text{so}\\ \dfrac{x^2_{k+1}-p}{x_k^2-p} &=\dfrac{x_k^2-p}{4x^2_k}\\ \end{array} $
so once $x_k$ is close enough to $\sqrt{p}$ (and I'll leave you to come up with a more precise estimate), the iteration will converge.
Note that this shows, no matter what $x_k$ is, that $x_{k+1} > \sqrt{p}$.