In a Fibonacci sequence 1, 1, 2, 3, 5, …, F_n, F_n+1
define R_n = F_n/ F_n-1

Prove that lim (R_n) as n approaches infinity
is .5*(1+sqrt(5))=1.618...
a.k.a. the golden ratio, φ (phi).

Point taken, Steve. I didn't think Ady wanted all the gory details. If you can stand the sight of blood, here are the details  :-)

Let a = ( 1+ sqr(5))/2 and b = (1 - sqr(5) )/2

a and b are the roots of x = 1 + 1/x

From the Euler-Binet formula, F(n) = (a^n - b^2 ) / sqr(5)

So, F(n+1)/F(n) = (a^(n+1) - b^(n+1)) / (a^n - b^n)

            = [(a^(n+1) - ab^n)) + (ab^n - b^(n+1) ] / (a^n - b^n) 

            = a + b^n (a - b) / (a^n - b^n)

            = a + sqr(5) / (k^n - 1)    where k = a/b

Now |k| = |1+sqr(5) / (1-sqr(5))| > 1

Hence, lim, n-> inf of F(n+1)/F(n) = a + 0 = a

But a is just phi (for easier notation above.)

So, the required limit exists and = (1+sqr(5))/2 = phi

QED

       

Edited on September 18, 2015, 3:59 pm

Posted by JayDeeKay on 2015-09-18 15:58:36