Problem Let $f(x)$ be a differentiable function on $\Bbb R$ with $\left|\,f ' (x)\right| \leq r < 1$, where $r$ is constant. Then consider the sequence $\{x_n\}$ such that $x_1 = 0$, $x_{n+1} = f(x_n)$, $n\geq1$. Show that $x_n \to x^*$ as $n$ approaches infinity. Moreover, $x^* = f(x^*)$.
Attempt I tried showing that when $f ' (x) = r$, then eventually you end up with $x_{n+1} = \sum_{n=1}^{\infty}r^n$. That sum converges by the $p$ test, so then $f(x)$ should converge to some number $x^*$. Then for $\epsilon>0$ there exists $N$ such that $n\geq N$ such that $$ \left|\,f(x^*)-x^*\right| \leq \left|x_{n+1} - x_n\right| \leq \epsilon. $$ Then to show for $f ' (x) < r$ use direct comparison test to say the bigger series converges so the smaller one must converge as well.
I'm not sure if I can just do this or not, can someone let me know if I'm using the correct method or not?
The trick is to show the sequence is Cauchy, i.e. given any $\varepsilon >0$ there is some $N$ s.t.
$$|x_n-x_m|<\varepsilon$$
whenever $n,m>N$. The trick here is that you know that $\sum_i r^i$ converges, so you end up with an upper bound which is the tail $\sum_{i=N}^\infty r^i$. This converges to zero as $N\to \infty$. This should be enough hints to let you fill in the details.