Limit of the ratio of consecutive Fibonacci numbers

Question

I have read in a book that the limit of the ratio of consequent Fibonacci numbers is the golden ratio. However, it was just mentioned thus not justified. So, my question is how would you derive the following limit: $$\lim_{x\to\infty}{\frac{F_n}{F_{n+1}}}=?$$ Where $F_n$ is the nth Fibonacci number?

Answer 1

HINT:

Let $\displaystyle\lim_{n\to\infty}\frac{F_{n+1}}{F_n}=u$

Clearly, $u\not<0$

By definition, we have $\displaystyle F_{n+1}=F_n+F_{n-1}$

$\displaystyle\implies \frac{F_{n+1}}{F_n}=1+\frac1{\frac{F_n}{F_{n-1}}}$

Setting $\displaystyle n\to\infty, u=1+\frac1u$

Answer 2

One way to show that the limit exists is to prove (by induction) that

$${F_n\over F_{n+1}}=1-{1\over1\cdot2}+{1\over2\cdot3}-{1\over3\cdot5}+\cdots+{(-1)^n\over F_nF_{n+1}}$$

The expression on the right is an alternating sum of terms tending to $0$, hence has a limit. With existence established, the value of the limit can be found as in lab bhattacharjee's answer. (Note, I'm using the convention $F_0=F_1=1$.)

Answer 3

It is well known that if $\phi$ is the golden ratio and $\overline\phi$ the other root of $x^2-x-1$, then $$F_n=\frac{\phi^n-\overline\phi^n}{\phi-\overline\phi}.$$ There is a whole theory behind this kind of thing, but the formula is easily verified using induction, and if you do so you also see why the roots of $x^2-x-1$ pop up. Now to derive the limit from this you only need to know that $\left|\overline\phi\right|<\left|\phi\right|$.

A friend of mine who is a runner and a mathematician has pointed out that the golden ratio is conveniently close to $1.609$ so that you can use the Fibonacci sequence $1,1,3,5,8,13,21,34,\ldots$ to approximately convert between miles and kilometres. Three miles are about five kilometres (error less than half a lap), thirteen miles are about 21km (slightly less than a half marathon).