# Limit of the ratio of consecutive Fibonacci numbers [duplicate]

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?

• Reciprocal of golden ratio? Commented Dec 15, 2013 at 11:42
• It turns out that the three answers here mirror the answers given to this question. Commented Dec 15, 2013 at 13:21
• @CarstenSchultz, the one-to-one correspondence is remarkable! Commented Dec 15, 2013 at 13:54
• proofwiki.org/wiki/Ratio_of_Consecutive_Fibonacci_Numbers Commented Dec 15, 2013 at 14:31

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

• assuming that the limit $u$ exists. But I think the existence itself would require some more effort like actual formula of $F_{n}$ or some technique to show that $F_{n + 1}/F_{n}$ is monotone and bounded Commented Dec 15, 2013 at 11:50
• @ParamanandSingh, mathhelpforum.com/number-theory/… Commented Dec 15, 2013 at 14:55

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).

• Note, this uses the convention $F_1=F_2=1$. Commented Dec 15, 2013 at 13:14
• @BarryCipra, yes, thanks. One sees immediately that the formula yields $F_0=0$ and $F_1=1$. Commented Dec 15, 2013 at 13:17
• There existed six proofs for the fact that the limit of the ratio of two consecutive Fibonacci numbers is the golden ratio in the following paper: Chao-Ping Chen, Ai-Qi Liu, and Feng Qi, Proofs for the limit of ratios of two consecutive terms in Fibonacci sequence, Cubo Matematica Educacional Vol. 5 (2003), no. 3, 23--30. This paper can be downloaded freely from the site revistas.ufro.cl/ojs/index.php/cubo/article/view/1663. Commented Sep 12, 2021 at 10:40

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 $F_0=F_1=1$.)