Prove a sequence converges Question: Let $f$ be differentiable on R with $a = \sup{|f′(x)| : x ∈ R} < 1$.
(a) Select $s_0 \in \mathbb{R}$ and define $s_n =f(s_{n−1})$ for $n \geq 1$. Thus $s_1 = f (s_0 ), s_2 = f (s_1 )$, etc. Prove $(s_n )$ is a convergent sequence.
Hint:
(a) To show that $(s_n)$ is Cauchy, first show $|s_{n+1} − s_n| \leq a|s_n − s_{n−1}|$ for $n ≥ 1$.
(b) Show f has a fixed point, i.e., f(s) = s for some s in R.
My attempt:
I know f is uniformly continuous on R since its derivative is bounded. 
I also proved the hint for part a using the mean value theorem. I am having troubles showing that $a|s_{n+1}-s_n| < \epsilon$ for $n$ large enough. 
I know that $|s_{n+1}-s_n|<\epsilon$  if $|f(s_n)-f(s_{n-1}| < \epsilon$, which is true if $f$ converges. 
Edit
Here is my attempt to solve part b. 
Let $(x_n)$ be a sequence of real numbers. There exists a fixed point $\iff$ $|f(x_n)-(x_n)|=0$ for some n $\iff$ $|f(x_{n+1}-f(x_{n-1})=0 \iff |f(x_{n+1}-f(x_{n-1})= f'(x)*|x_{n+1}-x_{n-1}|=0$ for some n, some $x \in (x_{n-1}, x_{n+1})$.
This happens if $f'(x)=0$ for some $x \in (x_{n-1}, x_{n+1})$. 
But we know this is true because $[x_{n-1}, x_{n+1}]$ is continuous and f is differentiable in that closed interval. My worry that is that the max or min might occur at the endpoint, which means that the derivative is not 0 somewhere in that closed interval.  
 A: Using the mean value theorem we can see $f$ is a contraction, because
$$|f(x) - f(y)| = |f'(c)||x-y| \leq a|x-y|$$
With $a < 1$.
I'll leave it as a hint how to prove its a fixed point, if you want afull proof I'll edit.
Now to see it is a cauchy sequence, select $x_0$ and note $x_n = f(x_{n-1})$, what can you say of $$|x_n - x_m|$$ for arbitrary $m, n$?
HINT: Use triangle inequality and $|x_n - x_{n+1}| \leq a^n|x_0 - x_1| $ (last part is valid because its a contraction)
EDIT: Suppose $m > n$ then:
$$|x_n - x_m| \leq |x_n - x_{n+1}| + |x_{n+1} - x_{n+2}| + \dots + |x_{m-1} + x_{m}|$$
$$\leq \sum_{k = n}^m a^k |x_0 - x_1| \leq \sum_{k = n}^{\infty} a^k |x_0 - x_1| < +\infty $$
Now, because this last series converges ($a < 1$), by the Cauchy criterion, given $\varepsilon >0$, there is some $N$ such that $$\sum_{k = N}^{\infty} a^k |x_0 - x_1| < \varepsilon $$
If we pick $n,m \geq N$ we're done, now we know $x_n$ is a Cauchy sequence in $\mathbb{R}$, and thus has a limit $x$, now $$x = \lim x_n = \lim f(x_{n-1}) = f(\lim x_{n-1}) = f(x)$$ hence $x$ is a fixed point. To see its uniquenes suppose $x_1, x_2$ are two fixed points, then:
$$|x_1 - x_2| = |f(x_1) - f(x_2)| \leq a|x_1 - x_2|   $$
With $a < 1$ this is only true if $x_1 = x_2$.
