Show there is an infinite number of parameters for which the sequence converges We have $f(x)=1-2|x|$. Let $a_1=a, a_{n+1}=f(a_n)$ for $n = 1, 2, 3, \dots$. We want to show there are infinitely many numbers $a \in [-1; 1]$ for which the series $(a_n)$ converges. I have checked by inspection that $a_{n+1}$ is stable for big $n$s and $a_{n+1} \to -1$ How could I show it formally?
 A: Note that $-1=f(1)$, hence $a=1$ gives you convergence.
Note that $1=f(0)$, hence $a=0$ gives you convergence.
Note that $0=f(1/2)$, hence $a=1/2$ gives you convergence.
Note that $1/2=f(1/4)$, hence $a=1/4$ gives you convergence.
And so on. Formally, you define a sequence $(x_k)_{k \ge 1}$ by
$$x_1=1 \\ x_{k+1}= \frac{1}{2}(1-x_k)\ \ \ \mbox{ for } k \ge1$$
You can prove by induction that such a sequence is strictly decreasing, positive, contined in $(0,1]$, and such that $$x_k=f(x_{k+1})$$. Hence, all these numbers give you convergence of the sequence $a_n$.
A: This is a strange one! The function $f(x)=1-2|x|$ has two fixed points, $x=1/3$ and $x=-1$. We have:
$$
a=\frac{q}{2^m\cdot 3}\implies a_n\to 1/3
$$
whenever $q$ is an integer relatively prime to $2$ and $3$. Moreover
$$
a=\frac{q}{2^m}\implies a_n\to-1
$$
whenever $q$ is relatively prime to $2$. This can be shown inductively working backwards from $1/3$ and $-1$ respectively and applying the principle:
$$
f(x)=r\iff x=\pm\tfrac12(1-r)\quad\text{(note that $f$ is an even function)}
$$
Strange! Note that the fact that computers work in binary may be the reason you see the limit $-1$ overrepresented in your numerical tests, since any binary rational $a=\sum_{i=0}^t 2^{-i}\cdot b_i$ results in the limit $a_n\to-1$.
