A conjecture concerning Fibonacci

Question

First of all, I'm aware that this conjecture has probably already been made and proved. I was just playing around, and I'd like to know whether it is in fact true and perhaps a hint as to how to prove it.

The conjecture, simply put, is that $$\lim_{n\rightarrow\infty}\frac{F_n}{F_{n-k}}=\Phi^k$$ where $k$ is a nonnegative integer and $\Phi$ is the golden ratio.

I believe this to be true based on the definition of the Fibonacci sequence itself - $F_n = F_{n-1} + F_{n-2}$ - and the well-known fact that $\lim F_n/F_{n-1}=\Phi$. Dividing our first expression through by $F_{n-1}$ lands us with $F_n/F_{n-1}=\Phi+o(n) = 1 + F_{n-2}/F_{n-1}$.

Subtracting $1$ from both sides: $\Phi+o(n) - 1 = F_{n-2}/F_{n-1}$. As I'm sure we all know, $\Phi - 1 = 1/\Phi$, so it makes sense that $F_{n-1}/F_{n-2} = \Phi$ (again, as $n$ increases without bound).

Now for the final step. We have $F_n/F_{n-1} = F_{n-1}/F_{n-2} = \Phi+o(n)$, so $F_n/F_{n-1}\times F_{n-1}/F_{n-2} = F_n/F_{n-2} = \Phi^2+o(n)$. Replacing the integer index in the denominator and the power on $\Phi$ by an arbitrary $k$, and including the limit notation, we arrive at what I conjectured.

Again, I realize that this has probably already been done. It's just a bit of fun. That's also why I don't want to go looking for some dry, technical proof online (before one of you tells me to Google it).

Cheers!

Answer 1

Your proof is not very rigorous. A rigorous proof proceeds through Binet's formula for $F_n$. Letting $\psi=-1/\varphi$ be the golden ratio conjugate: $$\lim_{n\to\infty}\frac{F_n}{F_{n-k}}=\lim_{n\to\infty}\frac{\sqrt5}{\sqrt5}\cdot\frac{\varphi^n-\psi^n}{\varphi^{n-k}-\psi^{n-k}}=\lim_{n\to\infty}\frac{\varphi^n}{\varphi^{n-k}}=\varphi^k$$ where the second-last equality is because $|\psi|<1$, so the contribution of the $\psi^n$ term vanishes in the limit.

Answer 2

You could try looking at it like this

$$ \begin{align} \frac{F_{n}}{F_{n-k}} & = \frac{F_{n}}{F_{n-1}} \frac{F_{n-1}}{F_{n-2}} \frac{F_{n-2}}{F_{n-3}} \dots \frac{F_{n-(k-1)}}{F_{n-k}} \\ & = \frac{F_{n}}{F_{n-1}} \frac{F_{n-1}}{F_{(n-1)-1}} \frac{F_{n-2}}{F_{(n-2)-1}} \dots \frac{F_{n-(k-1)}}{F_{n-(k-1)-1}} \end{align} $$

where there are $k$ terms in the product on the right hand side and each term in the numerator has an index that is one more than the index in the denominator.

If you wanted to prove this formally (i.e. without the somewhat hand-wavey "$k$ terms") you could do it by induction on $k$, and it would be basically the same argument.