Continued Fraction Expansion of Golden Mean

Theorem

The golden mean has the simplest possible continued fraction expansion, namely $\sqbrk {1, 1, 1, 1, \ldots}$:

$\phi = 1 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 {\ddots} } }$

Successive Convergents

The $n$th convergent is given by:

$C_n = \dfrac {F_{n + 1} } {F_n}$

where $F_n$ denotes the $n$th Fibonacci number.

Rate of Convergence

This continued fraction expansion has the slowest rate of convergence of all simple infinite continued fractions.

Proof

Let:

$x = 1 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 {\ddots} } }$

Then:

 $\ds x$ $=$ $\ds 1 + \frac 1 x$ substituting for $x$ $\ds \leadsto \ \$ $\ds x^2$ $=$ $\ds x + 1$ $\ds \leadsto \ \$ $\ds x^2 - x - 1$ $=$ $\ds 0$

The result follows from Golden Mean as Root of Quadratic.

$\blacksquare$