Ratio of Consecutive terms in a generalized Fibonacci Sequence tending to phi

Question

I'm trying to prove that the ratio of consecutive terms in a generalized fibonacci sequence tends to phi as n approaches infinity. So far I have got to the point where I obtain an equation for the limit as n tends to infinity of two successive terms x(n-1) and x(n) where I let L be the limit which looks like L^2 -L -1 =0. However on working out the solutions I get phi and -1/(phi) from the use of the quadratic equation. My question is why is the answer phi and not -1/(phi) for any generalized fibonacci sequence. I have tried an exhaustive list to show that different starting values in a sequence will end up having their limit as phi since after a point in the sequence, every term will have the same sign. This way of doing it doesn't feel very efficient and I would like to know if there is an easier way of always showing phi is the limit.

Answer 1

Hint:

$$F_{n+2}=F_{n+1}+F_n$$

Divide both sides by $F_{n+1}$ to get

$$\frac{F_{n+2}}{F_{n+1}}=1+\frac{F_n}{F_{n+1}}$$

Let $L=\lim_{n\to\infty}\frac{F_{n+1}}{F_n}$. Thus,

$$L=1+\frac1L,\ L>0$$

$$\implies L=\phi$$

The last step is to show convergence, which may be done by showing it is bounded above and monotonically increasing.

Answer 2

the set of solutions to $$x_{n+2} = x_{n+1} + x_n $$ is a vector space of dimension two over the reals. The basis is evident, indeed, there are fixed real numbers $A,B$ such that any solution sequence is $$ x_n = A \phi^n + B (1-\phi)^n. $$

The set of solutions where the limiting ratio of consecutive terms is $(1-\phi)$ is one dimensional, that is $A = 0,$ with sequence taking $B=1$ being $$ 1, \; \frac{-1}{\phi}, \; \frac{1}{\phi^2}, \cdots $$

As soon as we have $A \neq 0,$ the ratio $x_{n+1}/x_n$ becomes $$ \frac{ \phi + \frac{B}{A} \frac{(-1)^{n+1}}{\phi^{2n+1}}}{1 + \frac{B}{A} \frac{(-1)^{n}}{\phi^{2n}}} $$