Accuracy of Convergents of Convergent Simple Infinite Continued Fraction

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

$\blacksquare$


Also see