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