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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??

Martin Sleziak
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Jianluca
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3 Answers3

16

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

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.

N. S.
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7

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 $R_{n+1}=1+\frac{1}{R_n}$.

Jack D'Aurizio
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    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". – Robert William Hanks Aug 22 '16 at 05:08
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    @ 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 $\sum^\infty_{n=1} |R_{n+1} - R_n| \leq \sum^\infty_{n=1} \frac{1}{n(n+1)} < \infty$ . – wroobell Dec 10 '17 at 09:30
5

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 $|1-\phi|<1$).

Hagen von Eitzen
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