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 }) $.
Best Answer
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: $$ L=\lim_{n\to\infty}\frac{F_{n+1}}{F_n}= \lim_{n\to\infty}\left(1+\frac{F_{n-1}}{F_n}\right)=1+\frac{1}{L} $$