Accuracy of Convergents of Convergent Simple Infinite Continued Fraction
(Redirected from Bound for Difference of Limit of Simple Infinite Continued Fraction with Convergent)
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Theorem
Let $C = \tuple {a_0, a_1, \ldots}$ be an simple infinite continued fraction in $\R$.
Let $C$ converge to $x \in \R$.
For $n \ge 0$, let $C_n = \dfrac {p_n} {q_n}$ be the $n$th convergent of $C$, where $p_n$ and $q_n$ are the $n$th numerator and denominator.
Then for all $n \ge 0$:
- $\size {x - \dfrac {p_n} {q_n} } < \dfrac 1 {q_n q_{n + 1} }$
Proof
We show that either:
- $x \in \closedint {C_n} {C_{n + 1} }$
or:
- $x \in \closedint {C_{n + 1} } {C_n}$
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 $n \ge 1$ be odd.
By Limit of Subsequence equals Limit of Sequence:
- $x = \ds \lim_{k \mathop \to \infty} C_{2 k}$
For all $2 k \ge n + 1$, by:
- Even Convergents of Simple Continued Fraction are Strictly Increasing
- Even Convergent of Simple Continued Fraction is Strictly Smaller than Odd Convergent
we have:
- $C_{n + 1} \le C_{2 k} < C_n$
By Lower and Upper Bounds for Sequences:
- $x \in \closedint {C_{n + 1} } {C_n}$
$\Box$
Even case
Let $n \ge 0$ be even.
By Limit of Subsequence equals Limit of Sequence:
- $x = \ds \lim_{k \mathop \to \infty} C_{2 k + 1}$
For all $2 k + 1 \ge n + 1$, by:
- Odd Convergents of Simple Continued Fraction are Strictly Decreasing
- Even Convergent of Simple Continued Fraction is Strictly Smaller than Odd Convergent
we have:
- $C_n < C_{2 k + 1} \le C_{n + 1}$
By Lower and Upper Bounds for Sequences:
- $x \in \closedint {C_n} {C_{n + 1} }$
$\blacksquare$