Yesterday a friend challenged me to prove that
$$\lim_{n\rightarrow\infty}\frac{a_n}{a_{n-1}}=\varphi\; ,$$
where $\varphi$ is the golden ratio, for the Fibonacci series.
I started rewriting the limit as
$$\lim_{n\rightarrow\infty}\frac{a_n}{a_{n-1}}=\lim_{n\rightarrow\infty}\frac{a_{n-1}+a_{n-2}}{a_{n-1}}=\lim_{n\rightarrow\infty}1+\frac{a_{n-2}}{a_{n-1}}\; .$$
If the sequence $b_n=\frac{a_n}{a_{n-1}}$ is convergent,
$$\lim_{n\rightarrow\infty}\frac{a_{n-2}}{a_{n-1}}=\left(\lim_{n\rightarrow\infty}\frac{a_n}{a_{n-1}}\right)^{-1}\; .$$
Renaming the desired limit $x$, we obtain the quadratic equation
$$x=1+\frac{1}{x}$$
$$x^2-x-1=0$$
if $x\neq 0$. Therefore, if $b_n$ is convergent, it must be equal to $\frac{1+\sqrt{5}}{2}$ or $\frac{1-\sqrt{5}}{2}$.
Since $a_n>0$, $b_n>0, \forall n$, so the limit must be equal to $\varphi=\frac{1+\sqrt{5}}{2}$.
This proof made me think that I didn't make use of the initial values of the sequence, so it must hold true for any sequence where $a_{n}=a_{n-1}+a_{n-2}$. The first question is, is $a_{n}/a_{n-1}$ convergent for all Fibonacci-like sequences?
The second and most intriguing for me is, is there any Fibonacci-like sequence where the limit is $\frac{1-\sqrt{5}}{2}$? Since this solution is negative, $a_n$ should change its sing with each $n$, but I couldn't find any values for $a_0$ and $a_1$, which would lead me to this case. If the answer to this question is no, what mathematical sense does this negative solution have?
Best Answer
You showed that if a limit exists for $a_{n}/a_{n-1}$ and $a_n>0$, then it is $\frac{1+\sqrt{5}}{2}$. Actually if $(a_n)_{n\geq 0}$ is any sequence which satisfies the recurrence $a_n=a_{n-1} + a_{n-2}$ then there exist $A$ and $B$ such that $$a_n=A\cdot \left(\frac{1+\sqrt{5}}{2}\right)^n+B\cdot \left(\frac{1-\sqrt{5}}{2}\right)^n$$ where $A$ and $B$ depend on the initial terms $a_0$ and $a_1$.
So what is $\lim_{n\rightarrow\infty}\frac{a_n}{a_{n-1}}$ in the general case?
Consider for example the case when $A=0$ and $B\not=0$. What is the limit?