# Proof the golden ratio with the limit of Fibonacci sequence [duplicate]

Let $F_n=F_{n-1}+F_{n-2}$ the Fibonacci numbers, and $\phi=\frac{1+\sqrt5}{2}$

The exercise asks me to prove that: $\lim\limits_{n \to \infty}\frac{F_{n+1}}{F_n}=\phi$.

Sorry as can be proceed??

• And how to get the sequence that converges to 1,618 ... May 17 '15 at 17:36

$$F_{n+1}=F_{n}+F_{n-1} \Rightarrow \frac{F_{n+1}}{F_n}=1+\frac{F_{n-1}}{F_n}$$

Let $$x_n:= \frac{F_{n+1}}{F_n}$$

Then $$x_n=1+\frac{1}{x_{n-1}}$$

You can now prove that $$1 \leq x_n \leq 2$$ and by induction that

$$x_1 \leq x_3 \leq x_5 \leq .. \leq x_{2n+1} \leq x_{2n} \leq x_{2n-2} \leq .. \leq x_2$$

From here you get that $$x_{2n-1}$$ and $$x_{2n}$$ converge, and you can use their definitions to get their limits. You prove that both limits are the same, which yields the desired result.

• Great, it was the answer that i was looking for.. May 17 '15 at 17:51
• Aren't your inequalities the other way around? For instance, $x_1 = 1$ and $x_2 = 2$. Also, it is the odd subsequence that increases, not the even one. Mar 7 '17 at 9:49

Let $R_n=\frac{F_{n+1}}{F_n}$. Since: $$F_n^2-F_{n-1}F_{n+1} = (-1)^n\tag{1}$$ (it is easy to prove by induction) we have that: $$\left|R_{n+1}-R_n\right|=\frac{1}{F_n F_{n+1}}, \tag{2}$$ hence $\{R_n\}_{n\in\mathbb{N}}$ is a Cauchy sequence and $R_n$ converges to some $L>1$ that satisfies $L=1+\frac{1}{L}$, since $R_n>1$ and $R_{n+1}=1+\frac{1}{R_n}$.

• Is that enough to show that $R_n$ is a Cauchy sequence? if yes, can you elaborate please. Wikipedia article about Cauchy sequences states that " it is not sufficient for each term to become arbitrarily close to the preceding term". Aug 22 '16 at 5:08
• @ Robert William HanksIt is, if you know additional fact that if $\sum^\infty_{n=1} |R_{n+1} - R_n| < \infty$ then $(R_n)$ is Cauchy. Here it is true, since $F_n >n$, and if $\sum^\infty_{n=1} |R_{n+1} - R_n| \leq \sum^\infty_{n=1} \frac{1}{n(n+1)} < \infty$ . Dec 10 '17 at 9:30

Find $\alpha,\beta$ such that $F_0=\alpha+\beta$ and $F_1=\alpha\phi+\beta(1-\phi)$ and show by induction that $$F_n=\alpha\phi^n+\beta(1-\phi)^n.$$ Then compute the desired limit (and use that $|1-\phi|<1$).