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**