Sequence of Best Rational Approximations to Square Root of 2

Theorem
A sequence of best rational approximations to the square root of $2$ starts:
 * $\dfrac 1 1, \dfrac 3 2, \dfrac 7 5, \dfrac {17} {12}, \dfrac {41} {29}, \dfrac {99} {70}, \dfrac {239} {169}, \dfrac {577} {408}, \ldots$

where:
 * the numerators are half of the Pell-Lucas numbers, $\dfrac 1 2 Q_n$
 * the denominators are the Pell numbers $P_n$

starting from $\dfrac {\tfrac12 Q_1} {P_1}$.

Proof
Let $\tuple {a_0, a_1, \ldots}$ be the continued fraction expansion of $\sqrt 2$.

By Continued Fraction Expansion of Root 2:
 * $\sqrt 2 = \sqbrk {1, \sequence 2} = \sqbrk {1, 2, 2, 2, \ldots}$

From Convergents are Best Approximations, the convergents of $\sqbrk {1, \sequence 2}$ are the best rational approximations of $\sqrt 2$.

Let $\sequence {p_n}_{n \mathop \ge 0}$ and $\sequence {q_n}_{n \mathop \ge 0}$ be the numerators and denominators of the continued fraction expansion of $\sqrt 2$.

Then $\dfrac {p_n} {q_n}$ is the $n$th convergent of $\sqbrk {1, \sequence 2}$.

By Convergents of Simple Continued Fraction are Rationals in Canonical Form, $p_n$ and $q_n$ are coprime for all $n \ge 0$.

It remains to show that for all $n \ge 1$:
 * $Q_n = 2 p_{n - 1}$
 * $P_n = q_{n - 1}$

It is sufficient to prove that they satisfy the same recurrence relation.

By definition:

so that:
 * $\tuple {Q_1, P_1} = \tuple {2, 1} = \tuple {2 p_0, q_0}$

so that:
 * $\tuple {Q_2, P_2} = \tuple {6, 2} = \tuple {2 p_1, q_1}$

The result follows by definition of Pell numbers and Pell-Lucas numbers.