Fibonacci Sequence Proof Using limits

Question

I was trying to prove the fibonacci sequence for the golden section. But I am stuck at the following step.

$$\lim _{ n\rightarrow \infty }{ \frac { F_{ n+1 } }{ F_{ n } } } =\lim _{ n\rightarrow \infty }{ \frac { F_{ n } }{ F_{ n-1 } } } =L$$

Also the recursive property says that $$F_{ n+1 }=F_{ n }+F_{ n-1 }$$

Therefore (this step I do not understand why is happening) $$\lim _{ n\rightarrow \infty }{ \frac { F_{ n }+F_{ n-1 } }{ F_{ n } } } =1+\frac { 1 }{ L } $$

This part I can not explain or understand. I understand what is happening but I can not really make sense of this step in my head. Is there any explanation for how I can get ($1+\frac { 1 }{ L }) $.

Answer 1

Assuming $$ \lim_{n\to\infty}\frac{F_{n+1}}{F_n} $$ exists and equals $L>0$, you have $$ L=\lim_{n\to\infty}\frac{F_{n+1}}{F_n}= \lim_{n\to\infty}\frac{F_{n}}{F_{n-1}} $$ for obvious reasons. Now $$ \frac{F_{n+1}}{F_n}=\frac{F_n+F_{n-1}}{F_n}= 1+\frac{F_{n-1}}{F_n} $$ Since $$ \lim_{n\to\infty}\frac{F_{n-1}}{F_n}=\frac{1}{L} $$ by the theorems on limits, you're done, again by the theorems on limits: $$ L=\lim_{n\to\infty}\frac{F_{n+1}}{F_n}= \lim_{n\to\infty}\left(1+\frac{F_{n-1}}{F_n}\right)=1+\frac{1}{L} $$

Answer 2

In addition to egreg's answer, we have

$$a_n=\frac{F_{n+1}}{F_n}$$

So,

$$a_n=1+\frac1{a_{n-1}}$$

with $a_0=1$.

So, we get the neat expansion:

$$a_n=1+\dfrac1{1+\dfrac1{1+\dfrac1{1+\dfrac1\ddots}}}$$

where there are $n$ fraction bars. It then gives the continued fraction expansion of the golden ratio as $n\to\infty$.