Continued Fraction Expansion of Golden Mean/Successive Convergents

Theorem
Consider the continued fraction expansion to the golden mean:
 * $\phi = \left[{1, 1, 1, 1, \ldots}\right] = 1 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 {\ddots}}}$

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 $P \left({n}\right)$ be the proposition:
 * $C_n = \dfrac {F_{n + 1} } {F_n}$

$P \left({1}\right)$ is the case:
 * $C_1 = \dfrac {F_{n + 1} } {F_n}$

Thus $P \left({1}\right)$ is seen to hold.

Basis for the Induction
$P \left({2}\right)$ is the case:

Thus $P \left({2}\right)$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k + 1}\right)$ 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 $n$.

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

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