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?

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