The following is a problem that I can't solve, and I need some tips:
Problem: For $x>-1$, define $f_1(x) = x,\ f_{n+1}=\dfrac{1}{1+f_n(x)}$. Find the limit function $f(x)$ and all the subsets of $A=(-1,\infty)$ where the convergence is uniform.
First, I waste hours trying to find a closed form for $f_{n}(x)$. What I've found is that, for $n\ge 3$,$$\ f_n(x)=\dfrac{a_{n-2} + a_{n-3}x}{a_{n-1}+a_{n-2}x}$$ where $a_n$ is the n-th Fibonacci number (starting with $a_0=a_1=1$). If this is correct, then the limit function is $f(x)=\dfrac{1}{\psi}$ ($\psi$ is the golden number)
Then, I try to find where the convergence is uniform. I use the derivative in case it helps, but $$f_n'(x)= \dfrac{(-1)^n}{(a_{n-1}+a_{n-2}x)^2}$$ Cassini's Identity appears and $f_{2n}$ is increasing but $f_{2n+1}$ is decreasing. So it feels the wrong direction.
As always, any help will be appreciated.
1) You can do in this way:
If $\lim_{n\to \infty}f_n(x)$ exists, say $S$, then $S=\frac{1}{1+S}$.
Can you proceed futher?