# Accuracy of Convergents of Convergent Simple Infinite Continued Fraction

## Contents

## Theorem

Let $C = (a_0, a_1, \ldots)$ be an infinite simple continued fraction in $\R$.

Let $C$ converge to $x \in \R$.

For $n\geq0$, let $C_n = p_n/q_n$ be the $n$th convergent of $C$, where $p_n$ and $q_n$ are the $n$th numerator and denominator.

Then for all $n\geq 0$:

- $\left\vert x - \dfrac {p_n}{q_n}\right\vert < \dfrac 1{q_nq_{n+1}}$.

## Proof

We show that either:

- $x \in \left[ C_n..C_{n+1}\right]$ or
- $x \in \left[ C_{n+1}..C_n\right]$

so that the result follows from:

- Difference between Adjacent Convergents of Simple Continued Fraction
- Distance between Point of Real Interval and Endpoint is at most Length

#### Odd case

Let $n \geq 1$ be odd.

By Limit of Subsequence equals Limit of Sequence:

- $x = \displaystyle\lim_{k \to\infty} C_{2k}$

For all $2k \geq n+1$, by:

- Even Convergents of Simple Continued Fraction are Strictly Increasing
- Even Convergent of Simple Continued Fraction is Strictly Smaller than Odd Convergent

we have $C_{n+1} \leq C_{2k} < C_n$.

By Lower and Upper Bounds for Sequences, $x \in \left[ C_{n+1}..C_n\right]$.

#### Even case

Let $n \geq 0$ be even.

By Limit of Subsequence equals Limit of Sequence:

- $x = \displaystyle\lim_{k \to\infty} C_{2k+1}$

For all $2k+1 \geq n+1$, by:

- Odd Convergents of Simple Continued Fraction are Strictly Decreasing
- Even Convergent of Simple Continued Fraction is Strictly Smaller than Odd Convergent

we have $C_{n} < C_{2k+1} \leq C_{n+1}$.

By Lower and Upper Bounds for Sequences, $x \in \left[ C_n..C_{n+1}\right]$.

$\blacksquare$