Proving a sequence converges based on a differentiable function If we suppose $f : \mathbb{R} \to \mathbb{R}$ is differentiable and that $0 < f'(x) < 1$ $\forall x$ and also that it has a fixed point $f(a) = a$. Then if I define a sequence $x_n$ by selecting an initial value $x_1$ and then, for $n > 1$, setting $x_n = f(x_{n-1})$ how can I show the sequence converges? And also show that the limit is the fixed point mentioned before? 
I have proved that the function has only one fixed point, and that if $x < a$ then $f(x) > x$ and vice versa. But I am not sure where to go next?
 A: Suppose $y < a$. Then $y < f(y) < a$ because $0 < f’(x) < 1$.
That means, that if $x_1 < a$, then one can prove by induction  that the sequence of $x_n$ is monotonously increasing and $x_n < a$ for each natural n. That means, the sequence converges according to the Weierstrass theorem.
Suppose $y > a$. Then as $y > f(y) > a$ because $0 < f’(x) < 1$.
That means, that if $x_1 > a$, then one can prove by induction  that the sequence of $x_n$ is monotonously increasing and $x_n > a$ for each natural n. That means, the sequence converges according to the Weierstrass theorem.
If $x_1 = a$ the sequence is constant, and thus converges.
Suppose that b is the limit of the sequence. Then $b = \lim_{n \to \infty} {a_n} = f(b)$, because f is continuous. That means that if a is not equal to b, then, according to the mean value theorem, there exists such c between a and b, that $f’(c) = \frac {f(b) - f(a)} {b - a} = 1$. But that contradicts to $0 < f’(x) < 1$, which makes b = a the only possible solution.
