How to prove that $\infty$ is an attractive fixed point for the function in Newton's method for the given function? I'd like to ask to check for a solution to a homework problem, below.
Define $\forall n \in \mathbb{N}, g_n(x):=\frac{x}{(1-x)^n}$. Let $F_n$ be the associated Newton function, i.e. $F_n(x):= x - \frac{g_n(x)}{g_n'(x)}$. 
How do we show that $\infty$ is an attracting fixed point of $F_n$. I can see that $F_n(\infty)=\infty$, since $F_n(x)= x [1 - \frac{x-1}{x(1-n)-1)}] = x [1 - \frac{1-1/x}{(1-n)-1/x)}]$. But, how would define $F_n'(\infty)$, as a limit of $F_n'(x)$ as $x \to \infty$? If yes, how do we show that $|F_n'(\infty)|< 1$, in order to prove that $\infty$ is attracting? Thanks in advance!
EDIT: if I'm not mistake, my camculations show that:
$F_n'(x)= x(x-1)^n[  \frac{ (1-n)(x-1)  - (n+1)(x-1-nx) }{(x-1-nx)^2}    ]$.
This seems to go to infinity as $x \to \infty$, because it's of the form $(x-1)^n \frac{O(x^2)}{O(x^2)}$ as $x\to \infty$, if my calculations are right; so $\infty$ seems to be a repelling fixed point. Could someone please check it? 
 A: $F'(\infty)$ does not make any sense, because $\infty$ is not a point in the domain of $F'$. 
For similar reasons, the behavior of $F'(x)$ as $x \to \infty$ cannot help you to determine whether $\infty$ is an attracting fixed point. In particular, even if you did successfully prove that $F'(x) \to \infty$ as $x \to \infty$, it would not follow that $\infty$ is a repelling fixed point.
So you've got to do something else.
This does raise the important question of what it even means for $\infty$ to be an attracting fixed point, as brought up by @Ian. The guess in your comment works: you could work in the 1-point compactification $\mathbb R^* = \mathbb R \cup \{\infty\}$ (which is indeed homeomorphic to $\mathbb S^1$). But one still has to refine that guess into a working mathematical strategy.
Here's a strategy that works: instead of thinking about $\mathbb S^1$ directly, let's use the "reciprocal" coordinate $y = \frac{1}{x}$ defined on the set $\mathbb R^* - \{0\}$, which actually does define a valid coordinate system around $\infty$ in the 1-point compactification $\mathbb R^*$. 
Working in the reciprocal coordinate, let's rewrite the formula for $F_n(x)$:
$$G_n(y) = \frac{1}{F_n(1/y)} = \frac{1}{\frac{1}{y}[1 - \frac{1-y}{1-n-y}]} = \frac{y(1-n-y)}{(1-n-y)-(1-y)} = \frac{y(1-n-y)}{-n} = \frac{1}{n}(y^2+(n-1)y)
$$
Now compute $G'_n(y) = \frac{1}{n}(2y + n - 1)$, and so $G'_n(0) = \frac{n-1}{n}$ and hence $|G'_n(0)| < 1$. This proves that $0$ is a attracting fixed point of $G_n(y)$, hence $\infty$ is an attracting fixed point of $F_n(x)$.
A: Since
$\begin{array}\\
F_n(x)
&=x-\dfrac{g_n'(x)}{g_n(x)}\\
&=x-\dfrac{\frac{ (n - 1) x + 1}{(1 - x)^{n + 1}}}{\frac{x}{(1-x)^n}}\\
&=x-\dfrac{(n - 1) x + 1}{x(1-x)}\\
&=x+\dfrac{(n - 1) x + 1}{x(x-1)}\\
&=x+\dfrac{n - 1}{x-1}+\dfrac{1}{x(x-1)}\\
&>x+\dfrac{n - 1}{x-1}\\
\end{array}
$
so
$F_n(x) > x$.
