Vrftsjtryk

Yesterday a friend challenged me to proof that
$$lim_{nrightarrowinfty}frac{a_n}{a_{n-1}}=varphi; ,$$
where $varphi$ is the golden ratio, for the Fibonacci series.

I started rewriting the limit as

$$lim_{nrightarrowinfty}frac{a_n}{a_{n-1}}=lim_{nrightarrowinfty}frac{a_{n-1}+a_{n-2}}{a_{n-1}}=lim_{nrightarrowinfty}1+frac{a_{n-2}}{a_{n-1}}; .$$

If the sequence $b_n=frac{a_n}{a_{n-1}}$ is convergent,

$$lim_{nrightarrowinfty}frac{a_{n-2}}{a_{n-1}}=left(lim_{nrightarrowinfty}frac{a_n}{a_{n-1}}right)^{-1}; .$$

Renaming the desired limit $x$, we obtain the quadratic equation

$$x=1+frac{1}{x}$$
$$x^2-x-1=0$$

if $xneq 0$. 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?

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 $(a_n)_{ngeq 0}$ is any sequence which satisfies the recurrence $a_n=a_{n-1} + a_{n-2}$ then there exist $A$ and $B$ such that
$$a_n=Acdot left(frac{1+sqrt{5}}{2}right)^n+Bcdot left(frac{1-sqrt{5}}{2}right)^n$$
where $A$ and $B$ depend on the initial terms $a_0$ and $a_1$.

So what is $lim_{nrightarrowinfty}frac{a_n}{a_{n-1}}$ in the general case?

Consider for example the case when $A=0$ and $Bnot=0$. What is the limit?

Yes, $a_nover a_{n-1}$ is convergent for any Fibonacci-esque sequence, and this happen to be the golden ratio, varphi .
The limit $1-sqrt{5}over 2$ will never occur for a Fibonacci-esque sequence.
The negative solution crops up because you multiply both sides by $x$ when you solve $x=1+frac{1}{x}$ leading to an extra solution. But this value is useful for calculating the value of the $n^{th}$ term of the Fibonacci sequence.
$$ F_n=varphi^n - (frac{-1}{varphi^n})over sqrt{5}$$
Hope this helps.