I understand what is done below, however I don't understand the induction used, especially not when applying $0\leq n$.
If $s_n+1=f(s_n)$ with $|f′(x)|\leq 1/2$ prove that the sequence $s_n$ converges for $0\leq n$.
Proof: For $x, y \in\mathbb{R}$ we may apply the Mean Value Theorem to find $c \in (x, y)\subseteq\mathbb{R}$ such that $$\frac{|f(y) − f(x)|}{|y − x|}= |f(c)| \leq a<1. $$ Rearranging, we have $|f(x)−f(y)|\leq a|x−y|$ for all $x, y \in\mathbb{R}$. That is, $f$ is a "strict Banach contraction."
For any $n\in\mathbb{N}$ we have $|s_n −s_{n+1}| = |f(s_{n−1})−f(s_n)|\leq a|s_{n−1} −s_n|$, according to the hypothesis on $f$. Iterating this process and applying induction yields the estimate $|s_n − s_{n+1}|\leq a^n|s_0 − s_1|$.
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So, we form sequential, iterative steps as follows
$$\begin{align} |s_{n+1}-s_n|&\le a |s_{n}-s_{n-1}|\\\\ &\le a\cdot \left(a\,\,|s_{n-1}-s_{n-2}|\right)\\\\ &= a^2|s_{n-1}-s_{n-2}| \\\\ &\le a^3|s_{n-2}-s_{n-3}| \\\\ &\le \cdots \\\\ &\le a^n |s_{n-(n-1)}-s_{n-(n)}|\\\\ &=a^n |s_{1}-s_{0}| \end{align}$$
If $|a|<1$, then the sequence converges. For the problem here, $|f'|\le 1/2<1$.