Continued Fraction Expansion of Irrational Number Converges to Number Itself

Theorem
Let $x$ be an irrational number.

Then the continued fraction expansion of $x$ converges to $x$.

Proof
The Continued Fraction Algorithm produces a sequence of (non-simple) finite continued fractions as follows:
 * $x = x_1 = \left[{a_1, x_2}\right] = \left[{a_1, a_2, x_3}\right] = \cdots = \left[{a_1, a_2, \ldots, a_n, x_{n+1}}\right]$

where:
 * $\forall n \ge 1: a_n = \left \lfloor {x_n} \right \rfloor$

and the step from one step to the next is done by:
 * $x_n = a_n + \dfrac 1 {x_{n + 1}}$

First we need to show that the ICF built up thus has a limit $x$.

To do that we show that the sequence of simple finite continued fractions:
 * $\left[{a_1}\right], \left[{a_1, a_2}\right], \ldots, \left[{a_1, a_2, \ldots, a_n}\right], \ldots$

converges to $x$.

Using standard notation for numerators and denominators:
 * $\left[{a_1, a_2, \ldots, a_k}\right] = \dfrac {p_k} {q_k}$

we have, for $n \ge 2$:
 * $\left|{x - \dfrac {p_k} {q_k}}\right| < \dfrac 1 {q_n q_{n + 1} }$

by Bound for Difference of Limit of Simple Infinite Continued Fraction with Convergent.

We have that the sequence of the denominators $\left \lfloor {q_n} \right \rfloor$ is strictly increasing.

So from Basic Null Sequences and the Squeeze Theorem:
 * $\dfrac 1 {q_n q_{n+1} } \to 0$

as $n \to \infty$.

So the SICF determined from $x$ by the Continued Fraction Algorithm does converge to $x$.

Also see

 * Irrational Number is Limit of Unique Simple Infinite Continued Fraction