TITLE: The limit of a recurrence relation (with resistors)
QUESTION [6 upvotes]: Background to problem (not too important):

My proposed solution: 
The infinitely long element,

, however complex, can be represented as a single resistor of resistance $R$. 
Remembering the initial resistor near $A$, we know $R_{AB}= r+R$. However, as this is an infinitely long element, it is equivalent to a resistor of resistance $R$ attached  to the right of two resistors of resistance $r$ (the resistance $R$ is an intrinsic property of the element, so is unaffected by the fact that the further to the right it is, the lower the current passing through it).
Thus, taking $R$ in series with $r$, then the result in parallel with $r$, then in series with $r$: $$R_{AB}=  r + \left (\frac{1}{\frac{1}{R+r}+\frac{1}{r}}  \right )$$
On the second iteration (moving $R$ further to the right):
$$R_{AB}= r+ \frac{1}{\frac{1}{r}+\frac{1}{r + \left (\frac{1}{\frac{1}{R+r}+\frac{1}{r}}  \right )}}$$
Ad infinitum.
I understand this may not be the fastest solution, but I'd like to know a little more about it nonetheless.
The mathematics
$$u_{1}=r+R$$
$$\large u_{n+1}=r+\frac{1}{\frac{1}{r}+\frac{1}{u_n}}$$
Does $\lim_{n \rightarrow \infty} (u_n)$ exist (important: is the limit a function of $R$?), and, if so, what is it?
First cases
$$u_{2}=\frac{3r^2+2rR}{R+2r}$$
$$u_{3}=\frac{8r^2+5rR}{3R+5r}$$
$$u_{4}=\frac{21r^2+13rR}{8R+13r}$$
$$u_{4}=\frac{55r^2+34rR}{21R+34r}$$
$$\lim _{n \rightarrow \infty} (u_n)\stackrel{?}{=}\varphi r$$
Seems the doing of Fibonnaci. How does one take the limit of this (I assume it requires knowledge of knowledge of the form of $f(n)=F_n$).
Intuitively, why does Fibonnaci appear here? What are the rabbits in this case?
Link to Wolfram's computation.

REPLY [2 votes]: I didn't dare look at this earlier, but here's another (transcripted) solution (I prefer Aryabhata's):
Algebraically this equivalence can be written as
$$R_{AB}=r+\frac{1}{\frac{1}{r}+\frac{1}{R_{AB}}}$$
Thus
$$R_{AB}^2-rR_{AB}-r^2=0$$
This equation has two solutions:
$$R_{AB}=\frac{1}{2}(1 \pm \sqrt{5})r$$
 The solution corresponding to “-“ in the above formula is negative, while resistance must be positive. So, we reject it. Finally we receive
$$R_{AB}=\frac{1}{2}(1 + \sqrt{5})r= \varphi r$$