# Continued Fraction Expansion of Golden Mean/Successive Convergents

## Theorem

Consider the continued fraction expansion to the golden mean:

$\phi = \sqbrk {1, 1, 1, 1, \ldots} = 1 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 {\ddots} } }$

The $n$th convergent is given by:

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

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

## Proof

The proof proceeds by induction.

Listing the first few convergents, which can be calculated:

$C_1 = \dfrac 1 1$
$C_2 = \dfrac 2 1$
$C_3 = \dfrac 3 2$
$C_4 = \dfrac 5 3$

and so on.

For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:

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

$\map P 1$ is the case:

$C_1 = \dfrac {F_{n + 1} } {F_n}$
 $\ds C_1$ $=$ $\ds \dfrac 1 1$ $\ds$ $=$ $\ds \dfrac {F_2} {F_1}$ Definition of Fibonacci Numbers

Thus $\map P 1$ is seen to hold.

### Basis for the Induction

$\map P 2$ is the case:

 $\ds C_2$ $=$ $\ds 1 + \cfrac 1 1$ $\ds$ $=$ $\ds \dfrac 2 1$ $\ds$ $=$ $\ds \dfrac {F_3} {F_2}$ Definition of Fibonacci Numbers

Thus $\map P 2$ is seen to hold.

This is the basis for the induction.

### Induction Hypothesis

Now it needs to be shown that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.

So this is the induction hypothesis:

$C_k = \dfrac {F_{k + 1} } {F_k}$

from which it is to be shown that:

$C_{k + 1} = \dfrac {F_{k + 2} } {F_{k + 1} }$

### Induction Step

This is the induction step:

Let $C_n$ be expressed as $\dfrac {p_n} {q_n}$ for any given $n$.

 $\ds C_{k + 1}$ $=$ $\ds \dfrac {p_{k + 1} } {q_{k + 1} }$ $\ds$ $=$ $\ds \dfrac {p_k + p_{k - 1} } {q_k + q_{k - 1} }$ Definition of Numerators and Denominators of Continued Fraction $\ds$ $=$ $\ds \dfrac {F_{k + 1} + F_k} {F_k + F_{k - 1} }$ Induction Hypothesis $\ds$ $=$ $\ds \dfrac {F_{k + 2} } {F_{k + 1} }$ Definition of Fibonacci Numbers

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:

$\forall n \in \Z_{>0}: C_n = \dfrac {F_{n + 1} } {F_n}$

$\blacksquare$