Proving convergence of a sequence for all integers Having difficulty understanding how to approach the following sequence.
Let $f$ be differentiable on $\mathbb {R}$ with $a := \sup\{ |f'(x)| : x\in \mathbb {R}\} <1$. Select $s_0 \in \mathbb {R}$ and let 
$s_1 = f(s_0)$, $s_2 = f(s_1)$ , ..., and, in general, 
$s_n = f(s_{n-1})$ for $n = 1, 2, \ldots $. 
Prove that $(s_n)_{n\in \mathbb {N}}$ is a convergent sequence.
 A: Hint: use the fact that $|f(x)-f(y)|=|f'(c)(x-y)|\le a|x-y|$ to prove that your sequence is Cauchy.
A: By the mean value theorem, $\forall x, y \in \mathbb{R}$, $|f(x)-f(y)|\leq|x-y||f'(\theta)|\leq a|x-y|$ (where $\theta$ is some value in the open interval bounded by $x$ and $y$.) We call such an $f$ a contraction. The proof of this theorem is identical to the proof of the more general "contraction mapping theorem" in a complete metric space, the statement and proof of which can be readily found in many places. But the following gives a good outline:
$1)$ We aim to show that $s_n$ is a Cauchy sequence and so converges since $\mathbb{R}$ is complete. Let $\Delta = |s_1-s_0|$. For any $N$, for all $n,m\geq N$ with $n < m$ we have:
$$|s_n-s_m|\leq|s_n-s_{n+1}|+\dots+|s_{m-1}-s_m|$$
$$|s_n-s_m|\leq|s_N-s_{N+1}|+\dots+|s_{m-1}-s_m|=\Delta(a^N+\dots+a^m)$$
$$|s_n-s_m|\leq\Delta a^N\bigg(\sum_{i=0}^{\infty}a^j\bigg)=\frac{\Delta a^N}{1-a}$$
This final upper bound tends to $0$ as $N \rightarrow \infty$ so the sequence is Cauchy, and thus it converges.
Some related results (and easier to prove, too) which you might like to think about:
$1)$ If we call the limit of the sequence $s$, show that $f(s)=s$ ("$s$ is a fixed point of $f$")
$2)$ Show $s$ is the unique fixed point of $f$.
A: From the mean value theorem, $\forall x,y\in\mathbb R$ we have that there exists $\xi$ between $x$ and $y$ such that 
$$|f(x)-f(y)|=|f'(\xi)|\cdot |x-y|\leq a|x-y|\Rightarrow$$
$$|f(x)-f(y)|\leq a|x-y|,\forall x,y\in\mathbb R$$
Now take $$|s_{n+1}-s_n|=|f(s_n)-f(s_{n-1})|\leq a|s_n-s_{n-1}|=$$
$$a|f(s_{n-1})-f(s_{n-2})|\leq a^2|s_{n-1}-s_{n-2}|=\dots$$
$$\dots \leq a^n|s_1-s_0|$$
Therefore, $$|s_{n+p}-s_n|=|s_{n+p}-s_{n+p-1}+s_{n+p-1}-\dots-s_{n+1}+s_{n+1}-s_n|\leq$$
$$|s_{n+p}-s_{n+p-1}|+\dots +|s_{n+1}-s_n|\leq$$
$$|s_1-s_0|\cdot \sum \limits_{k=n}^{n+p-1} a^k\leq $$
$$|s_1-s_0|\sum\limits_{k=n}^\infty a^k<|s_1-s_0|\cdot\varepsilon$$
Where the last inequality holds, since the sum converges($a<1$), and therefore its "tail" tends to zero. 
The above relation proves that $\{s_n\}$ is a Cauchy sequence, and therefore converges. 
