Binet's formula to golden ratio

Question

Given that the Binet's formula is $$F_n = \frac{(1+\sqrt 5)^n - (1-\sqrt 5)^n}{2^n\sqrt 5}$$ I want to verify that $$ \lim_{n\to \infty}\frac{F_{n+1}}{F_n} = \frac{1+\sqrt 5}{2} $$ I reached this step for the LHS $$ \lim_{n\to\infty}\frac 1 2 *\biggl(1+\sqrt5\frac{(1+\sqrt5)^n+(1-\sqrt5)^n}{(1+\sqrt5)^n - (1-\sqrt5)^n}\biggl)$$ but how do I prove the following$$ \lim_{n\to\infty}\frac{\sum_{i=0}^{n/2}{n \choose 2i}\sqrt5^{2i}}{\sum_{i=0}^{n/2}{n \choose 2i+1}\sqrt5^{2i+1}} = 1$$ I feel like there should be a formula...or maybe I am wrong with my calculation?

Answer 1

You don't have to do binomial expansion. Instead you use the fact that $|1+\sqrt5|>|1-\sqrt5|$. This means that

$$\lim {(1-\sqrt 5)^n\over (1+\sqrt 5)^n} = 0$$

In your penultimate expression we get

$$ \lim_{n\to\infty}{1\over 2} \left(1+\sqrt 5 {(1+\sqrt5)^n+(1-\sqrt5)^n\over (1+\sqrt5)^n - (1-\sqrt5)^n} \right) = \lim_{n\to\infty}{1\over 2} \left(1+\sqrt 5 {1+{(1-\sqrt5)^n\over (1+\sqrt5)^n}\over 1 - {(1-\sqrt5)^n\over (1+\sqrt5)^n}}\right) = {1+\sqrt5\over 2} $$

Answer 2

Let $a=1+\sqrt{5}$ and $b=1-\sqrt{5}$. Then

$\frac{F_{n+1}}{F_n}= \frac{1}{2} \frac{a^{n+1}-b^{n+1}}{a^n-b^n}=\frac{1}{2} \frac{a-\frac{b^{n+1}}{a^n}}{1-\frac{b^n}{a^n}}$.

Since $|\frac{b}{a}|<1$, we have $\frac{b^n}{a^n} \to 0$ and $\frac{b^{n+1}}{a^n} \to 0$

Answer 3

Use $\frac{F_{n+2}}{F_{n+1}}=1+\frac{F_n}{F_{n+1}}.$

Let $a_n=\frac{F_{n+1}}{F_n}$, where $a_1=1$.

Thus, $a_{n+1}=1+\frac{1}{a_n}>1$ and $$\left|a_{n+1}-\frac{1+\sqrt5}{2}\right|=\left|\frac{1}{a_n}-\frac{1}{\frac{1+\sqrt5}{2}}\right|=\frac{\left|a_{n}-\frac{1+\sqrt5}{2}\right|}{\frac{1+\sqrt5}{2}a_n}<$$ $$<\left(\frac{1+\sqrt5}{2}\right)^{-1}\left|a_{n}-\frac{1+\sqrt5}{2}\right|,$$ which gives $$\left|a_{n}-\frac{1+\sqrt5}{2}\right|<\left(\frac{1+\sqrt5}{2}\right)^{-n+1}\left|a_{1}-\frac{1+\sqrt5}{2}\right|$$ and since $\frac{1+\sqrt5}{2}>1$, we are done!