1. ## Newton method help

I have an exam on Tuesday 27th. I need to know well the bisection method, secant method, newton method and a lot more.
I have a sheet of around 25 exercises. I could do some pretty easy, but others are too hard for me.
I have a lot of questions in relation with these methods. (That I will post in other threads if I feel the need to).
Here comes an exercise I couldn't do.
1)Demonstrate that if f is a function that has a double root in a, then the Newton method applied to f has an order of convergence of 1.
Demonstrate also that the modified Newton has a convergence order of 2.
2)Study what happens if $p>2$.
As p is not defined, I guess it's the number of multiple roots.
End of the problem.
My approach was to write $f$ as $f(x)=(x-a)^2$. $f'(x)=2x-2a
$
and $f^{(2)}(x)=2$. Therefore the Newton method will converges regardless of the starting point.
From it, I supposed $x_0=0$. I calculated $x_1$, which is worth $\frac{a}{2}$. Also $x_2=\frac{3a}{4}$ and lastly $x_3=\frac{7a}{8}$.
So our intuition tells us that the method converges to a. But how can I determine the order? And how can I prove that it effectively converges to a?
I'd be glad if you could start me up for the common Newton method, this way I hope to finish the rest alone.

2. I calculated $x_4$ and guessed well $x_5$. In fact I noticed the coefficient in front of $a$ can be written as a sequence that tends to 1 when n tends to positive infinite.
I found that $x_{n+1}=2x_n+\frac{1}{2^n}$. I set $x_0=0$. I'm completely despaired. I need help on this problem. How can I prove that the sequence $x_n$ tends to $1$? The first terms are $0$, $\frac{1}{2}$, $\frac{3}{4}$, $\frac{7}{8}$, etc. I know it's a famous sequence, where each step is half closer to 1 than the precedent.
But I need to be rigorous, I can't say " I'm seeing that the coefficients of the method are like this sequence". I must prove it and I'm not able.
Is there any other way I'm not seeing to do the first question? I'm lost.