Answer to Question #280114 in Real Analysis for abc

Question

Check the convergence of the sequence defined by 𝑢𝑛+1 = 𝑎/ 1+𝑢𝑛 where 𝑎 > 0, 𝑢1 > 0.

Answer 1

$u_{n+1}=1+\frac{1}{u}$

map $u\to 1+\frac{1}{u}$ can be extended to a Moebius transformation of the Riemann sphere

$C\cup \{\infin\}:$

$z\to \frac{z+1}{z},T(0)=\infin,T(\infin)=1$

Its fixed points are:

$a=\frac{(1+\sqrt 5)}{2},b=\frac{(1-\sqrt 5)}{2}$

obtained by solving the equation

$z^2-z-1=0$

We now introduce a new complex coordinate w on C, related to z via

$w=\phi(z)=\frac{z-a}{z-b}\implies z=\phi^{-1}(w)=\frac{a-bw}{1-w}$

The fixed points now are w = 0 and $w=\infin$

in terms of the new coordinate w the transformation T appears as

$\tilde{T}=\phi \circ T \circ \phi^{-1}$ , then:

$\tilde{T}: w\to \frac{b}{a}w,\tilde{T}(0)=0,\tilde{T}(\infin)=\infin$

since

$\frac{b}{a}=\frac{3-\sqrt 5}{2}=-0.382$

we can infer that the fixed point 0 is attracting with basin of attraction all of C, while $\infin$ is repelling. This allows to conclude that in the original setting all initial points $u_0\neq b$

lead to $\displaystyle \lim_{n\to \infin} u_n=a$

So, the sequence converges.