Limit of the ratio of consecutive Fibonacci numbers [duplicate]

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

Solution 2:

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).