Limit of Ratio of Adjacent Fibonacci numbers $\to \phi$

Question

We define the $n^{th}$ Fibonacci number as $a_1 = a_2 = 1$ and $a_n = a_{n-1} + a_{n-2}$ for $n \geq 3$. Consider

$$ \lim_{n \to \infty} \frac{a_{n+1}}{a_n}. $$

I wrote a script and found that this limit converges to the golden ratio $\phi \approx 1.61803$. However, I'm having trouble giving a rigorous proof for this. Here's what I have so far:

$$ \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{a_n + a_{n-1}}{a_n} = \lim_{n \to \infty} \left( 1 + \frac{a_{n-1}}{a_n} \right). $$

I'm unsure of how to proceed after this. I know that expressing the limit as a certain continued fraction would imply that it converges to $\phi$ but I'm unsure of how to get there. Any ideas?

Answer 1

Call the limit $\phi$. Then from what you wrote you get $\phi=1+\frac{1}{\phi}$.

Answer 2

We know that the Formula for Fibonacci Numbers is $Fib(n) = { ((\sqrt5+1)/2)^n - ((\sqrt5-1)/2)^ n }/√5$. Now dividing by $Fib(n-1)$ and taking $n \rightarrow \infty $ gives the result.

Answer 3

Dividing the recurrence relation throughout by $a_n$ :

$$\frac{a_{n+1}}{a_{n}} = 1 + \frac{a_{n-1}}{a_{n}}\\ = 1 + \frac{1}{\frac{a_n}{a_{n-1}}}$$

As $n \to \infty$, we let $\frac{a_{n+1}}{a_{n}}, \frac{a_n}{a_{n-1}}\to k$ for some real number $k$. Then,

$$k = 1 + \frac{1}{k}\\ k^2 - k - 1 = 0$$

Whose positive solution turns out to be $\frac{1 + \sqrt{5}}{2}$, which is exactly $\phi$.

Answer 4

It's easy if you use Binet's formula and let $n$ tend to $\infty$.

Answer 5

Let $$\lim_{n\to\infty}{\frac{F_{n+1}}{F_n}} = L$$ Then $$L = \lim_{n\to\infty}{\frac{F_{n+1}}{F_{n}}} = \lim_{n\to\infty}{\frac{F_{n}}{F_{n-1}}}$$ Which is $$L = \lim_{n\to\infty}{\frac{F_{n} + F_{n-1}}{F_{n}}} = \lim_{n\to\infty}{\frac{F_{n}}{F_{n-1}}}$$ $$L =\lim_{n\to\infty}{1 + \frac{ F_{n-1}}{F_{n}}} = 1 + \lim_{n\to\infty}{\frac{ F_{n-1}}{F_{n}}} = \lim_{n\to\infty}{\frac{F_{n}}{F_{n-1}}}$$ Or $$L = 1 + \frac{1}{L}$$

Now do your job.