Binet's formula to golden ratio 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?
 A: 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 
$$ \lim_{n\to\infty}{1\over 2} \left(1+\sqrt 5
{(1+\sqrt5)^n+(1-\sqrt5)^n\over
(1+\sqrt5)^n - (1-\sqrt5)^n}
\right)
= \lim_{n\to\infty}{1\over 2} \left(1+\sqrt 5
{1+{(1-\sqrt5)^n\over (1+\sqrt5)^n}\over
1 - {(1-\sqrt5)^n\over (1+\sqrt5)^n}}\right) = {1+\sqrt5\over 2}
$$
A: 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 $\frac{b^{n+1}}{a^n} \to 0$
A: 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!
