# Thread: Proof: limit of ratios of fibonacci numbers is convergent

1. ## Proof: limit of ratios of fibonacci numbers is convergent

So I was given a proof: Let $F_n$ be the n-th term of the Fibonacci sequence. Show that $\lim_{n\to\infty}\frac{F_n}{f_{n-1}}=\frac{1+\sqrt{5}}{2}$

I figured out how to do the proof starting with the assumption that the limit exists for $\lim_{n\to\infty}\frac{F_n}{f_{n-1}}$. My problem is that I can't figure out how to show that the limit is convergent. My instructor suggested using something like the squeeze theorem, but I can't figure out how to use the squeeze theorem on a sequence defined recursively.

I'm convinced that the ration decreases to the limit for odd values of n, and increases to the limit for even values of n, but don't know how to prove this, especially without assuming that the limit exists beforehand.

I would appreciate avenues to pursue, but I am trying not to find an already complete proof, so don't give me the answer.

Thanks!

2. ## Re: Proof: limit of ratios of fibonacci numbers is convergent

I'm not sure if this will help or not, but a proof for the "ratio" test of series seems to be similar to what you're doing. Maybe worth a shot. Basically using the logic that having the n+1 term divided by the nth term is the same as having the nth term divided by the n-1 term. Here is a link to the proof for the ratio test, although I haven't yet checked it out myself. Again, I'm still in highschool, so I'm not quite sure if this will help or not.

3. ## Re: Proof: limit of ratios of fibonacci numbers is convergent

Start by showing that $\displaystyle x := 1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \dots}}}} } = \frac{1 + \sqrt{5}}{2}$. To do this, notice that since the fraction is continuing ad infinitum, everything after the first vinculum is exactly what we are trying to find. Therefore, we have

\displaystyle \begin{align*} x &= 1 + \frac{1}{x} \\ x^2 &= x + 1 \\ x^2 - x - 1 &= 0 \\ x^2 - x + \left(-\frac{1}{2}\right)^2 - \left(-\frac{1}{2}\right)^2 - 1 &= 0 \\ \left(x - \frac{1}{2}\right)^2 - \frac{1}{4} - 1 &= 0 \\ \left(x - \frac{1}{2}\right)^2 - \frac{5}{4} &= 0 \\ \left(x - \frac{1}{2}\right)^2 &= \frac{5}{4} \\ x - \frac{1}{2} &= \pm \frac{\sqrt{5}}{2} \\ x &= \frac{1}{2} \pm \frac{\sqrt{5}}{2} \end{align*}

And since it's clear that we are looking for a positive solution, that means $\displaystyle x = 1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \dots}}}} } = \frac{1 + \sqrt{5}}{2}$.

Now, notice what happens when we examine this continued fraction in more detail...

\displaystyle \begin{align*} 1 = \frac{1}{1} &= \frac{F_0}{F_1} \\ 1 + \frac{1}{1} = \frac{2}{1} &= \frac{F_1}{F_2} \\ 1 + \frac{1}{1 + \frac{1}{1}} = \frac{3}{2} &= \frac{F_3}{F_2} \\ 1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1}}} = \frac{5}{3} &= \frac{F_4}{F_3} \\ \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1}}}} = \frac{8}{5} &= \frac{F_5}{F_4} \\ &\vdots \\ 1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \dots}}}} } &= \frac{F_n}{F_{n-1}} \end{align*}

I think that you can see that the ratio of consecutive Fibonnaci numbers is upon \displaystyle \begin{align*} \frac{1 + \sqrt{5}}{2} \end{align*}

4. ## Re: Proof: limit of ratios of fibonacci numbers is convergent

Thank you! It hadn't occured to me to use continuing fractions.

5. ## Re: Proof: limit of ratios of fibonacci numbers is convergent

Hello, beebe!

Are we allowed the closed form? . $F_n \:=\:\frac{\left(1 + \sqrt{5}\:\!\right)^n - \left(1-\sqrt{5}\:\!\right)^n}{2^n\sqrt{5}}$

$\text{Let }F_n\text{ be the }n^{th}\text{ term of the Fibonacci sequence.}$
$\text{Show that: }\lim_{n\to\infty}\frac{F_n}{F_{n-1}}\;=\;\frac{1+\sqrt{5}}{2}$

$\frac{F_n}{F_{n-1}} \;=\;\frac{1+\sqrt{5})^n - (1-\sqrt{5})^n}{2^n\sqrt{5}} \cdot\frac{2^{n-1}\sqrt{5}}{(1+\sqrt{5})^{n-1} - (1-\sqrt{5})^{n-1}}$

. . . . $=\;\frac{1}{2}\cdot\frac{(1+\sqrt{5})^n - (1-\sqrt{5})^n}{(1+\sqrt{5})^{n-1} - (1-\sqrt{5})^{n-1}}$

Divide numerator and denominator by $(1+\sqrt{5})^{n-1}$

. . $\frac{1}{2}\cdot \frac{\;\dfrac{(1+\sqrt{5})^n}{(1+\sqrt{5})^{n-1}} -\dfrac{1-\sqrt{5})^n}{(1+\sqrt{5})^{n-1}}\;} { \dfrac{(1+\sqrt{5})^{n-1}}{(1+\sqrt{5})^{n-1}} - \dfrac{(1-\sqrt{5})^{n-1}}{(1+\sqrt{5})^{n-1}}} \;=\; \frac{1}{2}\cdot\frac{(1 + \sqrt{5}) - (1-\sqrt{5})\left(\dfrac{1-\sqrt{5}}{1+\sqrt{5}}\right)^{n-1}} {1 - \left(\dfrac{1-\sqrt{5}}{1+\sqrt{5}}\right)^{n-1}}}$

$\lim_{n\to\infty}\frac{F_n}{F_{n-1}} \;=\;\lim_{n\to\infty} \frac{1}{2}\cdot\frac{(1 + \sqrt{5}) - (1-\sqrt{5})\left(\dfrac{1-\sqrt{5}}{1+\sqrt{5}}\right)^{n-1}} {1 - \left(\dfrac{1-\sqrt{5}}{1+\sqrt{5}}\right)^{n-1}}}$

Note that: . $\lim_{n\to\infty}\left(\frac{1-\sqrt{5}}{1+\sqrt{5}}\right)^n \:=\:0$

Therefore: . $\lim_{n\to\infty}\frac{F_n}{F_{n-1}} \;=\;\frac{1}{2}\cdot \frac{(1 + \sqrt{5}) - 0}{1 - 0} \;=\;\frac{1 + \sqrt{5}}{2}$

6. ## Re: Proof: limit of ratios of fibonacci numbers is convergent

Here's another idea.

First, prove inductively that $F_{n+1}F_{n-1}=F_n^2+(-1)^n$. It's a pretty straightforward proof: remember that $F_n=F_{n+1}-F_{n-1}$.

Then divide by $F_nF_{n-1}$ to get $\frac{F_{n+1}}{F_n}=\frac{F_n}{F_{n-1}}+\frac{(-1)^n}{F_nF_{n-1}}$.

Now use the alternating series test.

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